{"id":284,"date":"2021-07-14T15:59:06","date_gmt":"2021-07-14T15:59:06","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/answers-to-selected-exercises-6\/"},"modified":"2023-12-05T09:36:36","modified_gmt":"2023-12-05T09:36:36","slug":"answers-to-selected-exercises-6","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/answers-to-selected-exercises-6\/","title":{"raw":"Answers to Selected Exercises","rendered":"Answers to Selected Exercises"},"content":{"raw":"

Null and Alternative Hypotheses - Practice<\/h2>\r\n1.\u00a0The random variable is the mean Internet speed in Megabits per second.\r\n\r\n3 . The random variable is the mean number of children an American family has.\r\n\r\n5.\u00a0The random variable is the proportion of people picked at random in Times Square visiting the city.\r\n\r\n7.\r\n
\r\n
    \r\n \t
  1. H0<\/sub><\/em>: p<\/em> = 0.42<\/li>\r\n \t
  2. Ha<\/sub><\/em>: p<\/em> < 0.42<\/li>\r\n<\/ol>\r\n
    \r\n\r\n9.\r\n\r\n
    \r\n
      \r\n \t
    1. H0<\/sub><\/em>: \u03bc<\/em> = 15<\/li>\r\n \t
    2. Ha<\/sub><\/em>: \u03bc<\/em> \u2260 15<\/li>\r\n<\/ol>\r\n \r\n

      Outcomes and the Type I and Type II Errors \u2013 Practice<\/h2>\r\n
      11.\u00a0Type I: The mean price of mid-sized cars is $32,000, but we conclude that it is not $32,000.\u00a0<\/span>Type II: The mean price of mid-sized cars is not $32,000, but we conclude that it is $32,000.<\/span><\/section><\/div>\r\n
      15. Type I: The procedure will go well, but the doctors think it will not.\u00a0Type II: The procedure will not go well, but the doctors think it will.<\/span><\/section><\/section>
      17.\u00a00.019<\/span><\/section>
      19. 0.998<\/section><\/section><\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n
      \r\n
      \r\n

      Distribution Needed for Hypothesis Testing \u2013 Practice<\/h2>\r\n
      \r\n
      \r\n

      21.\u00a0A normal distribution or a Student\u2019s t<\/em>-distribution<\/p>\r\n

      23.\u00a0Use a Student\u2019s t<\/em>-distribution<\/p>\r\n

      25. A normal distribution for a single population mean<\/p>\r\n

      27.\u00a0It must be approximately normally distributed.<\/p>\r\n

      29.\u00a0They must both be greater than five.<\/p>\r\n

      31. Binomial distribution<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n

      \r\n

      Rare Events, the Sample, Decision and Conclusion \u2013 Practice<\/h2>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n
      \r\n
      \r\n
      33.\u00a0The outcome of winning is very unlikely.35.
      \r\n
        \r\n \t
      1. H0<\/sub><\/em>: \u03bc<\/em> > = 73<\/li>\r\n \t
      2. Ha<\/sub><\/em>: \u03bc<\/em> < 73<\/li>\r\n<\/ol>\r\nThe p<\/em>-value is almost zero, which means there is sufficient data to conclude that the mean height of high school students who play basketball on the school team is less than 73 inches at the 5% level. The data do support the claim.\r\n\r\n<\/section><\/section><\/section><\/section>
        37.\u00a0The shaded region shows a low p<\/em>-value.<\/section><\/div>\r\n
        <\/section><\/div>\r\n
        39.\u00a0Do not reject H0<\/sub><\/em><\/section><\/div>\r\n
        <\/section>\r\n
        41. Means<\/section><\/div>\r\n
        <\/section>\r\n
        43. The mean time spent in jail for 26 first time convicted burglars<\/section><\/div>\r\n
        <\/section>\r\n
        45.
        \r\n
          \r\n \t
        1. 3<\/li>\r\n \t
        2. 1.5<\/li>\r\n \t
        3. 1.8<\/li>\r\n \t
        4. 26<\/li>\r\n<\/ol>\r\n47. [latex]\\overline{X}~{N}\\left({2.5},\\frac{1.5}{\\sqrt{26}}\\right)[\/latex]\r\n\r\n<\/section><\/section><\/div>\r\n

          Additional Information and Full Hypothesis Test Examples \u2013 Practice<\/h2>\r\n
          49.\u00a0This is a left-tailed test.<\/section>
          <\/section>
          51.\u00a0This is a two-tailed test.<\/section>
          <\/section>
          <\/section>
          <\/section>
          <\/section>
          <\/section>
          <\/section>
          <\/section>
          53.\u00a0the graph of a two-tailed test.
          \r\n
          \"Normal<\/span><\/figure>\r\n \r\n
          <\/figure>\r\n<\/section>55. a right-tailed test\r\n\r\n57. a left-tailed test<\/span>\r\n\r\n59. This is a left-tailed test.<\/span>\r\n\r\n61.\u00a0This is a two-tailed test.<\/span>\r\n\r\n<\/section>\r\n

          Null and Alternative Hypotheses \u2013 Homework<\/h2>\r\n
          \r\n
          62.
          \r\n
            \r\n \t
          1. H0<\/sub><\/em>: \u03bc<\/em> = 34; Ha<\/sub><\/em>: \u03bc<\/em> \u2260 34<\/li>\r\n \t
          2. H0<\/sub><\/em>: p<\/em> \u2264 0.60; Ha<\/sub><\/em>: p<\/em> > 0.60<\/li>\r\n \t
          3. H0<\/sub><\/em>: \u03bc<\/em> \u2265 100,000; Ha<\/sub><\/em>: \u03bc<\/em> < 100,000<\/li>\r\n \t
          4. H0<\/sub><\/em>: p<\/em> = 0.29; Ha<\/sub><\/em>: p<\/em> \u2260 0.29<\/li>\r\n \t
          5. H0<\/sub><\/em>: p<\/em> = 0.05; Ha<\/sub><\/em>: p<\/em> < 0.05<\/li>\r\n \t
          6. H0<\/sub><\/em>: \u03bc<\/em> \u2264 10; Ha<\/sub><\/em>: \u03bc<\/em> > 10<\/li>\r\n \t
          7. H0<\/sub><\/em>: p<\/em> = 0.50; Ha<\/sub><\/em>: p<\/em> \u2260 0.50<\/li>\r\n \t
          8. H0<\/sub><\/em>: \u03bc<\/em> = 6; Ha<\/sub><\/em>: \u03bc<\/em> \u2260 6<\/li>\r\n \t
          9. H0<\/sub><\/em>: p<\/em> \u2265 0.11; Ha<\/sub><\/em>: p<\/em> < 0.11<\/li>\r\n \t
          10. H0<\/sub><\/em>: \u03bc<\/em> \u2264 20,000; Ha<\/sub><\/em>: \u03bc<\/em> > 20,000<\/li>\r\n<\/ol>\r\n64.\u00a0p<\/em> < 0.20\r\n

            Outcomes and the Type I and Type II Errors \u2013 Homework<\/h2>\r\n<\/section>
            66.\r\n
              \r\n \t
            1. Type I error: We conclude that the mean is not 34 years when it really is 34 years. Type II error: We conclude that the mean is 34 years, when in fact it really is not 34 years.<\/li>\r\n \t
            2. Type I error: We conclude that more than 60% of Americans vote in presidential elections when the actual percentage is at most 60%. Type II error: We conclude that at most 60% of Americans vote in presidential elections when, in fact, more than 60% do.<\/li>\r\n \t
            3. Type I error: We conclude that the mean starting salary is less than $100,000 when it really is at least $100,000. Type II error: We conclude that the mean starting salary is at least $100,000 when, in fact, it is less than $100,000.<\/li>\r\n \t
            4. Type I error: We conclude that the proportion of high school seniors who get drunk each month is not 29% when it really is 29%. Type II error: We conclude that the proportion of high school seniors who get drunk each month is 29% when, in fact, it is not 29%.<\/li>\r\n \t
            5. Type I error: We conclude that fewer than 5% of adults ride the bus to work in Los Angeles when the percentage that do is really 5% or more. Type II error: We conclude that 5% or more adults ride the bus to work in Los Angeles when, in fact, fewer than 5% do.<\/li>\r\n \t
            6. Type I error: We conclude that the mean number of cars a person owns in his or her lifetime is more than 10 when in reality it is not more than 10. Type II error: We conclude that the mean number of cars a person owns in his or her lifetime is not more than 10 when, in fact, it is more than 10.<\/li>\r\n \t
            7. Type I error: We conclude that the proportion of Americans who prefer to live away from cities is not about half, though the actual proportion is about half. Type II error: We conclude that the proportion of Americans who prefer to live away from cities is half when, in fact, it is not half.<\/li>\r\n \t
            8. Type I error: We conclude that the duration of paid vacations each year for Europeans is not six weeks, when in fact it is six weeks. Type II error: We conclude that the duration of paid vacations each year for Europeans is six weeks when, in fact, it is not.<\/li>\r\n \t
            9. Type I error: We conclude that the proportion is less than 11%, when it is really at least 11%. Type II error: We conclude that the proportion of women who develop breast cancer is at least 11%, when in fact it is less than 11%.<\/li>\r\n \t
            10. Type I error: We conclude that the average tuition cost at private universities is more than $20,000, though in reality, it is at most $20,000. Type II error: We conclude that the average tuition cost at private universities is at most $20,000 when, in fact, it is more than $20,000.<\/li>\r\n<\/ol>\r\n

              68.\u00a0Not to conclude the drug is safe when, in fact, it is safe.<\/p>\r\n

              70.\u00a0is less than seven hours<\/p>\r\n\r\n

              Distribution Needed for Hypothesis Testing \u2013 Homework<\/h2>\r\n

              72.\u00a0t<\/em>21<\/sub><\/p>\r\n\r\n

              Rare Events, the Sample, Decision and Conclusion - Homework<\/h2>\r\n<\/section><\/section><\/section><\/div>\r\n<\/div>\r\n
              N\/A<\/section>
              \r\n

              Additional Information and Full Hypothesis Test Examples - Homework<\/h2>\r\n<\/section>
              <\/section><\/section>
              <\/section>
              74.\u00a0
              \r\n
                \r\n \t
              1. H0<\/sub><\/em>: \u03bc<\/em> \u2265 50,000<\/li>\r\n \t
              2. Ha<\/sub><\/em>: \u03bc<\/em> < 50,000<\/li>\r\n \t
              3. Let [latex]\\overline{X}[\/latex]\u00a0= the average lifespan of a brand of tires.<\/li>\r\n \t
              4. normal distribution<\/li>\r\n \t
              5. z<\/em> = -2.315<\/li>\r\n \t
              6. p<\/em>-value = 0.0103<\/li>\r\n \t
              7. Check student\u2019s solution.\r\n
                  \r\n \t
                1. alpha: 0.05<\/li>\r\n \t
                2. Decision: Reject the null hypothesis.<\/li>\r\n \t
                3. Reason for decision: The p<\/em>-value is less than 0.05.<\/li>\r\n \t
                4. Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                5. (43,537, 49,463)<\/li>\r\n<\/ol>\r\n<\/section>76.\r\n\r\n
                  \r\n
                    \r\n \t
                  1. H0<\/sub><\/em>: \u03bc<\/em> = $1.00<\/li>\r\n \t
                  2. Ha<\/sub><\/em>: \u03bc<\/em> \u2260 $1.00<\/li>\r\n \t
                  3. Let [latex]\\overline{X}[\/latex]= the average cost of a daily newspaper.<\/li>\r\n \t
                  4. normal distribution<\/li>\r\n \t
                  5. z<\/em> = \u20130.866<\/li>\r\n \t
                  6. p<\/em>-value = 0.3865<\/li>\r\n \t
                  7. Check student\u2019s solution.\r\n
                      \r\n \t
                    1. Alpha: 0.01<\/li>\r\n \t
                    2. Decision: Do not reject the null hypothesis.<\/li>\r\n \t
                    3. Reason for decision: The p<\/em>-value is greater than 0.01.<\/li>\r\n \t
                    4. Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is $1. The mean cost could be $1.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                    5. ($0.84, $1.06)<\/li>\r\n<\/ol>\r\n78.\r\n\r\n
                      \r\n
                        \r\n \t
                      1. H0<\/sub><\/em>: \u03bc<\/em> = 10<\/li>\r\n \t
                      2. Ha<\/sub><\/em>: \u03bc<\/em> \u2260 10<\/li>\r\n \t
                      3. Let [latex]\\overline{X}[\/latex]\u00a0the mean number of sick days an employee takes per year.<\/li>\r\n \t
                      4. Student\u2019s t<\/em>-distribution<\/li>\r\n \t
                      5. t<\/em> = \u20131.12<\/li>\r\n \t
                      6. p<\/em>-value = 0.300<\/li>\r\n \t
                      7. Check student\u2019s solution.\r\n
                          \r\n \t
                        1. Alpha: 0.05<\/li>\r\n \t
                        2. Decision: Do not reject the null hypothesis.<\/li>\r\n \t
                        3. Reason for decision: The p<\/em>-value is greater than 0.05.<\/li>\r\n \t
                        4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean number of sick days is not ten.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                        5. (4.9443, 11.806)<\/li>\r\n<\/ol>\r\n80.\r\n\r\n
                          \r\n
                            \r\n \t
                          1. H0<\/sub><\/em>: p<\/em> \u2265 0.6<\/li>\r\n \t
                          2. Ha<\/sub><\/em>: p<\/em> < 0.6<\/li>\r\n \t
                          3. Let P\u2032<\/em> = the proportion of students who feel more enriched as a result of taking Elementary Statistics.<\/li>\r\n \t
                          4. normal for a single proportion<\/li>\r\n \t
                          5. 1.12<\/li>\r\n \t
                          6. p<\/em>-value = 0.1308<\/li>\r\n \t
                          7. Check student\u2019s solution.\r\n
                              \r\n \t
                            1. Alpha: 0.05<\/li>\r\n \t
                            2. Decision: Do not reject the null hypothesis.<\/li>\r\n \t
                            3. Reason for decision: The p<\/em>-value is greater than 0.05.<\/li>\r\n \t
                            4. Conclusion: There is insufficient evidence to conclude that less than 60 percent of her students feel more enriched.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                            5. Confidence Interval: (0.409, 0.654)\r\nThe \u201cplus-4s\u201d confidence interval is (0.411, 0.648)<\/li>\r\n<\/ol>\r\n<\/section>82.\r\n
                                \r\n \t
                              1. H0<\/sub><\/em>:\u00a0\u03bc<\/em>\u00a0= 4<\/li>\r\n \t
                              2. Ha<\/sub><\/em>:\u00a0\u03bc<\/em>\u00a0\u2260 4<\/li>\r\n \t
                              3. Let [latex]\\overline{X}[\/latex]<\/span>\u00a0the average I.Q. of a set of brown trout.<\/li>\r\n \t
                              4. two-tailed Student's t-test<\/li>\r\n \t
                              5. t<\/em>\u00a0= 1.95<\/li>\r\n \t
                              6. p<\/em>-value = 0.076<\/li>\r\n \t
                              7. Check student\u2019s solution.\r\n
                                  \r\n \t
                                1. Alpha: 0.05<\/li>\r\n \t
                                2. Decision: Reject the null hypothesis.<\/li>\r\n \t
                                3. Reason for decision: The\u00a0p<\/em>-value is greater than 0.05<\/li>\r\n \t
                                4. Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                5. (3.8865,5.9468)<\/li>\r\n<\/ol>\r\n
                                  84.
                                  \r\n
                                    \r\n \t
                                  1. H0<\/sub><\/em>: p<\/em> \u2265 0.13<\/li>\r\n \t
                                  2. Ha<\/sub><\/em>: p<\/em> < 0.13<\/li>\r\n \t
                                  3. Let P\u2032<\/em> = the proportion of Americans who have seen or sensed angels<\/li>\r\n \t
                                  4. normal for a single proportion<\/li>\r\n \t
                                  5. \u20132.688<\/li>\r\n \t
                                  6. p<\/em>-value = 0.0036<\/li>\r\n \t
                                  7. Check student\u2019s solution.\r\n
                                      \r\n \t
                                    1. alpha: 0.05<\/li>\r\n \t
                                    2. Decision: Reject the null hypothesis.<\/li>\r\n \t
                                    3. Reason for decision: The p<\/em>-value is less than 0.05.<\/li>\r\n \t
                                    4. Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have seen or sensed an angel is less than 13%.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                    5. (0, 0.0623).\r\nThe\u201cplus-4s\u201d confidence interval is (0.0022, 0.0978)<\/li>\r\n<\/ol>\r\n88.\r\n
                                        \r\n \t
                                      1. H0<\/sub><\/em>:\u00a0\u03bc<\/em>\u00a0\u2265 129<\/li>\r\n \t
                                      2. Ha<\/sub><\/em>:\u00a0\u03bc<\/em>\u00a0< 129<\/li>\r\n \t
                                      3. Let [latex]\\overline{X}[\/latex]<\/span>\u00a0= the average time in seconds that Terri finishes Lap 4.<\/li>\r\n \t
                                      4. Student's\u00a0t<\/em>-distribution<\/li>\r\n \t
                                      5. t<\/em>\u00a0= 1.209<\/li>\r\n \t
                                      6. 0.8792<\/li>\r\n \t
                                      7. Check student\u2019s solution.\r\n
                                          \r\n \t
                                        1. Alpha: 0.05<\/li>\r\n \t
                                        2. Decision: Do not reject the null hypothesis.<\/li>\r\n \t
                                        3. Reason for decision: The\u00a0p<\/em>-value is greater than 0.05.<\/li>\r\n \t
                                        4. Conclusion: There is insufficient evidence to conclude that Terri\u2019s mean lap time is less than 129 seconds.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                        5. (128.63, 130.37)<\/li>\r\n<\/ol>\r\n90.\r\n
                                            \r\n \t
                                          1. H0<\/sub><\/em>:\u00a0p<\/em>\u00a0= 0.60<\/li>\r\n \t
                                          2. Ha<\/sub><\/em>:\u00a0p<\/em>\u00a0< 0.60<\/li>\r\n \t
                                          3. Let\u00a0P\u2032<\/em>\u00a0= the proportion of family members who shed tears at a reunion.<\/li>\r\n \t
                                          4. normal for a single proportion<\/li>\r\n \t
                                          5. \u20131.71<\/li>\r\n \t
                                          6. 0.0438<\/li>\r\n \t
                                          7. Check student\u2019s solution.\r\n
                                              \r\n \t
                                            1. alpha: 0.05<\/li>\r\n \t
                                            2. Decision: Reject the null hypothesis.<\/li>\r\n \t
                                            3. Reason for decision:\u00a0p<\/em>-value < alpha<\/li>\r\n \t
                                            4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of family members who shed tears at a reunion is less than 0.60. However, the test is weak because the\u00a0p<\/em>-value and alpha are quite close, so other tests should be done.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                            5. We are 95% confident that between 38.29% and 61.71% of family members will shed tears at a family reunion. (0.3829, 0.6171). The\u201cplus-4s\u201d confidence interval (see chapter 8) is (0.3861, 0.6139)<\/li>\r\n<\/ol>\r\n

                                              Note that here the \u201clarge-sample\u201d 1 \u2013 PropZTest provides the approximate\u00a0p<\/em>-value of 0.0438. Whenever a\u00a0p<\/em>-value based on a normal approximation is close to the level of significance, the exact\u00a0p<\/em>-value based on binomial probabilities should be calculated whenever possible. This is beyond the scope of this course.<\/p>\r\n92.\r\n\r\n<\/section>\r\n

                                                \r\n \t
                                              1. H0<\/sub><\/em>:\u00a0\u03bc<\/em>\u00a0\u2265 22<\/li>\r\n \t
                                              2. Ha<\/sub><\/em>:\u00a0\u03bc<\/em>\u00a0< 22<\/li>\r\n \t
                                              3. Let [latex]\\overline{X}[\/latex]<\/span>\u00a0= the mean number of bubbles per blow.<\/li>\r\n \t
                                              4. Student's\u00a0t<\/em>-distribution<\/li>\r\n \t
                                              5. \u20132.667<\/li>\r\n \t
                                              6. p<\/em>-value = 0.00486<\/li>\r\n \t
                                              7. Check student\u2019s solution.\r\n
                                                  \r\n \t
                                                1. Alpha: 0.05<\/li>\r\n \t
                                                2. Decision: Reject the null hypothesis.<\/li>\r\n \t
                                                3. Reason for decision: The\u00a0p<\/em>-value is less than 0.05.<\/li>\r\n \t
                                                4. Conclusion: There is sufficient evidence to conclude that the mean number of bubbles per blow is less than 22.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                                5. (18.501, 21.499)<\/li>\r\n<\/ol>\r\n
                                                  94.\r\n
                                                    \r\n \t
                                                  1. H0<\/sub><\/em>:\u00a0\u03bc<\/em>\u00a0\u2264 1<\/li>\r\n \t
                                                  2. Ha<\/sub><\/em>:\u00a0\u03bc<\/em>\u00a0> 1<\/li>\r\n \t
                                                  3. Let [latex]\\overline{X}[\/latex]<\/span>\u00a0= the mean cost in dollars of macaroni and cheese in a certain town.<\/li>\r\n \t
                                                  4. Student's\u00a0t<\/em>-distribution<\/li>\r\n \t
                                                  5. t<\/em>\u00a0= 0.340<\/li>\r\n \t
                                                  6. p<\/em>-value = 0.36756<\/li>\r\n \t
                                                  7. Check student\u2019s solution.\r\n
                                                      \r\n \t
                                                    1. Alpha: 0.05<\/li>\r\n \t
                                                    2. Decision: Do not reject the null hypothesis.<\/li>\r\n \t
                                                    3. Reason for decision: The\u00a0p<\/em>-value is greater than 0.05<\/li>\r\n \t
                                                    4. Conclusion: The mean cost could be $1, or less. At the 5% significance level, there is insufficient evidence to conclude that the mean price of a box of macaroni and cheese is more than $1.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                                    5. (0.8291, 1.241)<\/li>\r\n<\/ol>\r\n96.\r\n
                                                        \r\n \t
                                                      1. H0<\/sub><\/em>:\u00a0p<\/em>\u00a0= 0.01<\/li>\r\n \t
                                                      2. Ha<\/sub><\/em>:\u00a0p<\/em>\u00a0> 0.01<\/li>\r\n \t
                                                      3. Let\u00a0P\u2032<\/em>\u00a0= the proportion of errors generated<\/li>\r\n \t
                                                      4. Normal for a single proportion<\/li>\r\n \t
                                                      5. 2.13<\/li>\r\n \t
                                                      6. 0.0165<\/li>\r\n \t
                                                      7. Check student\u2019s solution.<\/li>\r\n \t
                                                      8. \r\n
                                                          \r\n \t
                                                        1. Alpha: 0.05<\/li>\r\n \t
                                                        2. Decision: Reject the null hypothesis<\/li>\r\n \t
                                                        3. Reason for decision: The\u00a0p<\/em>-value is less than 0.05.<\/li>\r\n \t
                                                        4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of errors generated is more than 0.01.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                                        5. Confidence interval: (0, 0.094).\r\n<\/span>The\u201cplus-4s\u201d confidence interval is (0.004, 0.144).<\/li>\r\n<\/ol>\r\n<\/section>98.\r\n\r\n<\/section>\r\n
                                                            \r\n \t
                                                          1. H0<\/sub><\/em>:\u00a0p<\/em>\u00a0= 0.50<\/li>\r\n \t
                                                          2. Ha<\/sub><\/em>:\u00a0p<\/em>\u00a0< 0.50<\/li>\r\n \t
                                                          3. Let\u00a0P\u2032<\/em>\u00a0= the proportion of friends that has a pierced ear.<\/li>\r\n \t
                                                          4. normal for a single proportion<\/li>\r\n \t
                                                          5. \u20131.70<\/li>\r\n \t
                                                          6. p<\/em>-value = 0.0448<\/li>\r\n \t
                                                          7. Check student\u2019s solution.<\/li>\r\n \t
                                                          8. \r\n
                                                              \r\n \t
                                                            1. Alpha: 0.05<\/li>\r\n \t
                                                            2. Decision: Reject the null hypothesis<\/li>\r\n \t
                                                            3. Reason for decision: The\u00a0p<\/em>-value is less than 0.05. (However, they are very close.)<\/li>\r\n \t
                                                            4. Conclusion: There is sufficient evidence to support the claim that less than 50% of his friends have pierced ears.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                                            5. Confidence Interval: (0.245, 0.515): The \u201cplus-4s\u201d confidence interval is (0.259, 0.519).<\/li>\r\n<\/ol>\r\n
                                                              100.
                                                              \r\n
                                                                \r\n \t
                                                              1. H0<\/sub><\/em>: p<\/em> = 0.40<\/li>\r\n \t
                                                              2. Ha<\/sub><\/em>: p<\/em> < 0.40<\/li>\r\n \t
                                                              3. Let P\u2032<\/em> = the proportion of schoolmates who fear public speaking.<\/li>\r\n \t
                                                              4. normal for a single proportion<\/li>\r\n \t
                                                              5. \u20131.01<\/li>\r\n \t
                                                              6. p<\/em>-value = 0.1563<\/li>\r\n \t
                                                              7. Check student\u2019s solution.\r\n
                                                                  \r\n \t
                                                                1. Alpha: 0.05<\/li>\r\n \t
                                                                2. Decision: Do not reject the null hypothesis.<\/li>\r\n \t
                                                                3. Reason for decision: The p<\/em>-value is greater than 0.05.<\/li>\r\n \t
                                                                4. Conclusion: There is insufficient evidence to support the claim that less than 40% of students at the school fear public speaking.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                                                5. Confidence Interval: (0.3241, 0.4240): The \u201cplus-4s\u201d confidence interval is (0.3257, 0.4250).<\/li>\r\n<\/ol>\r\n102.\r\n\r\n
                                                                  \r\n
                                                                    \r\n \t
                                                                  1. H0<\/sub><\/em>: p<\/em> = 0.14<\/li>\r\n \t
                                                                  2. Ha<\/sub><\/em>: p<\/em> < 0.14<\/li>\r\n \t
                                                                  3. Let P\u2032<\/em> = the proportion of NYC residents that smoke.<\/li>\r\n \t
                                                                  4. normal for a single proportion<\/li>\r\n \t
                                                                  5. \u20130.2756<\/li>\r\n \t
                                                                  6. p<\/em>-value = 0.3914<\/li>\r\n \t
                                                                  7. Check student\u2019s solution.\r\n
                                                                      \r\n \t
                                                                    1. alpha: 0.05<\/li>\r\n \t
                                                                    2. Decision: Do not reject the null hypothesis.<\/li>\r\n \t
                                                                    3. Reason for decision: The p<\/em>-value is greater than 0.05.<\/li>\r\n \t
                                                                    4. At the 5% significance level, there is insufficient evidence to conclude that the proportion of NYC residents who smoke is less than 0.14.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                                                    5. Confidence Interval: (0.0502, 0.2070): The \u201cplus-4s\u201d confidence interval (see chapter 8) is (0.0676, 0.2297).<\/li>\r\n<\/ol>\r\n104.\r\n\r\n
                                                                      \r\n
                                                                        \r\n \t
                                                                      1. H0<\/sub><\/em>: \u03bc<\/em> = 69,110<\/li>\r\n \t
                                                                      2. Ha<\/sub><\/em>: \u03bc<\/em> > 69,110<\/li>\r\n \t
                                                                      3. Let [latex]\\overline{X}[\/latex]\u00a0= the mean salary in dollars for California registered nurses.<\/li>\r\n \t
                                                                      4. Student's t<\/em>-distribution<\/li>\r\n \t
                                                                      5. t<\/em> = 1.719<\/li>\r\n \t
                                                                      6. p<\/em>-value: 0.0466<\/li>\r\n \t
                                                                      7. Check student\u2019s solution.\r\n
                                                                          \r\n \t
                                                                        1. Alpha: 0.05<\/li>\r\n \t
                                                                        2. Decision: Reject the null hypothesis.<\/li>\r\n \t
                                                                        3. Reason for decision: The p<\/em>-value is less than 0.05.<\/li>\r\n \t
                                                                        4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean salary of California registered nurses exceeds $69,110.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                                                        5. ($68,757, $73,485)<\/li>\r\n<\/ol>\r\n106. Do not reject H0<\/sub><\/em>: There is not sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.\r\n\r\n<\/section>108.\u00a0There is not enough evidence to conclude that the mean number of hours is more than 4.5\r\n\r\n110.\r\n\r\n
                                                                          \r\n
                                                                            \r\n \t
                                                                          1. H0<\/sub><\/em>: p<\/em> = 0.488 Ha<\/sub><\/em>: p<\/em> \u2260 0.488<\/li>\r\n \t
                                                                          2. p<\/em>-value = 0.0114<\/li>\r\n \t
                                                                          3. alpha = 0.05<\/li>\r\n \t
                                                                          4. Reject the null hypothesis.<\/li>\r\n \t
                                                                          5. At the 5% level of significance, there is enough evidence to conclude that 48.8% of families own stocks.<\/li>\r\n \t
                                                                          6. The survey does not appear to be accurate.<\/li>\r\n<\/ol>\r\n112.\r\n\r\n
                                                                            \r\n
                                                                              \r\n \t
                                                                            1. H0<\/sub><\/em>: p<\/em> = 0.517 Ha<\/sub><\/em>: p<\/em> \u2260 0.517<\/li>\r\n \t
                                                                            2. p<\/em>-value = 0.9203.<\/li>\r\n \t
                                                                            3. alpha = 0.05.<\/li>\r\n \t
                                                                            4. Do not reject the null hypothesis.<\/li>\r\n \t
                                                                            5. At the 5% significance level, there is not enough evidence to conclude that the proportion of homes in Kentucky that are heated by natural gas is 0.517.<\/li>\r\n \t
                                                                            6. However, we cannot generalize this result to the entire nation. First, the sample\u2019s population is only the state of Kentucky. Second, it is reasonable to assume that homes in the extreme north and south will have extreme high usage and low usage, respectively. We would need to expand our sample base to include these possibilities if we wanted to generalize this claim to the entire nation.<\/li>\r\n<\/ol>\r\n114.\r\n\r\n
                                                                              \r\n
                                                                                \r\n \t
                                                                              1. H0<\/sub><\/em>: \u00b5<\/em> \u2265 11.52 Ha<\/sub><\/em>: \u00b5<\/em> < 11.52<\/li>\r\n \t
                                                                              2. p<\/em>-value = 0.000002 which is almost 0.<\/li>\r\n \t
                                                                              3. alpha = 0.05.<\/li>\r\n \t
                                                                              4. Reject the null hypothesis.<\/li>\r\n \t
                                                                              5. At the 5% significance level, there is enough evidence to conclude that the mean amount of summer rain in the northeastern US is less than 11.52 inches, on average.<\/li>\r\n \t
                                                                              6. We would make the same conclusion if alpha was 1% because the p<\/em>-value is almost 0.<\/li>\r\n<\/ol>\r\n116.\r\n\r\n
                                                                                \r\n
                                                                                  \r\n \t
                                                                                1. H0<\/sub><\/em>: \u00b5<\/em> \u2264 5.8 Ha<\/sub><\/em>: \u00b5<\/em> > 5.8<\/li>\r\n \t
                                                                                2. p<\/em>-value = 0.9987<\/li>\r\n \t
                                                                                3. alpha = 0.05<\/li>\r\n \t
                                                                                4. Do not reject the null hypothesis.<\/li>\r\n \t
                                                                                5. At the 5% level of significance, there is not enough evidence to conclude that a woman visits her doctor, on average, more than 5.8 times a year.<\/li>\r\n<\/ol>\r\n118.\r\n\r\n
                                                                                  \r\n
                                                                                    \r\n \t
                                                                                  1. H0<\/sub><\/em>: \u00b5<\/em> \u2265 150 Ha<\/sub><\/em>: \u00b5<\/em> < 150<\/li>\r\n \t
                                                                                  2. p<\/em>-value = 0.0622<\/li>\r\n \t
                                                                                  3. alpha = 0.01<\/li>\r\n \t
                                                                                  4. Do not reject the null hypothesis.<\/li>\r\n \t
                                                                                  5. At the 1% significance level, there is not enough evidence to conclude that freshmen students study less than 2.5 hours per day, on average.<\/li>\r\n \t
                                                                                  6. The student academic group\u2019s claim appears to be correct.<\/li>\r\n<\/ol>\r\n<\/section><\/section><\/section><\/section><\/section><\/section><\/section><\/section><\/section><\/section><\/section><\/section>","rendered":"

                                                                                    Null and Alternative Hypotheses – Practice<\/h2>\n

                                                                                    1.\u00a0The random variable is the mean Internet speed in Megabits per second.<\/p>\n

                                                                                    3 . The random variable is the mean number of children an American family has.<\/p>\n

                                                                                    5.\u00a0The random variable is the proportion of people picked at random in Times Square visiting the city.<\/p>\n

                                                                                    7.<\/p>\n

                                                                                    \n
                                                                                      \n
                                                                                    1. H0<\/sub><\/em>: p<\/em> = 0.42<\/li>\n
                                                                                    2. Ha<\/sub><\/em>: p<\/em> < 0.42<\/li>\n<\/ol>\n
                                                                                      \n

                                                                                      9.<\/p>\n

                                                                                      \n
                                                                                        \n
                                                                                      1. H0<\/sub><\/em>: \u03bc<\/em> = 15<\/li>\n
                                                                                      2. Ha<\/sub><\/em>: \u03bc<\/em> \u2260 15<\/li>\n<\/ol>\n

                                                                                         <\/p>\n

                                                                                        Outcomes and the Type I and Type II Errors \u2013 Practice<\/h2>\n
                                                                                        \n
                                                                                        11.\u00a0Type I: The mean price of mid-sized cars is $32,000, but we conclude that it is not $32,000.\u00a0<\/span>Type II: The mean price of mid-sized cars is not $32,000, but we conclude that it is $32,000.<\/span><\/section>\n<\/div>\n
                                                                                        \n
                                                                                        \n
                                                                                        15. Type I: The procedure will go well, but the doctors think it will not.\u00a0Type II: The procedure will not go well, but the doctors think it will.<\/span><\/section>\n<\/section>\n
                                                                                        \n
                                                                                        17.\u00a00.019<\/span><\/section>\n
                                                                                        19. 0.998<\/section>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n
                                                                                        \n
                                                                                        \n
                                                                                        \n

                                                                                        Distribution Needed for Hypothesis Testing \u2013 Practice<\/h2>\n
                                                                                        \n
                                                                                        \n
                                                                                        \n

                                                                                        21.\u00a0A normal distribution or a Student\u2019s t<\/em>-distribution<\/p>\n

                                                                                        23.\u00a0Use a Student\u2019s t<\/em>-distribution<\/p>\n

                                                                                        25. A normal distribution for a single population mean<\/p>\n

                                                                                        27.\u00a0It must be approximately normally distributed.<\/p>\n

                                                                                        29.\u00a0They must both be greater than five.<\/p>\n

                                                                                        31. Binomial distribution<\/p>\n<\/div>\n<\/section>\n<\/div>\n

                                                                                        \n
                                                                                        \n

                                                                                        Rare Events, the Sample, Decision and Conclusion \u2013 Practice<\/h2>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n
                                                                                        \n
                                                                                        \n
                                                                                        \n
                                                                                        \n
                                                                                        \n
                                                                                        33.\u00a0The outcome of winning is very unlikely.35.<\/p>\n
                                                                                        \n
                                                                                          \n
                                                                                        1. H0<\/sub><\/em>: \u03bc<\/em> > = 73<\/li>\n
                                                                                        2. Ha<\/sub><\/em>: \u03bc<\/em> < 73<\/li>\n<\/ol>\n

                                                                                          The p<\/em>-value is almost zero, which means there is sufficient data to conclude that the mean height of high school students who play basketball on the school team is less than 73 inches at the 5% level. The data do support the claim.<\/p>\n<\/section>\n<\/section>\n<\/section>\n<\/section>\n

                                                                                          37.\u00a0The shaded region shows a low p<\/em>-value.<\/section>\n<\/div>\n
                                                                                          <\/section>\n<\/div>\n
                                                                                          39.\u00a0Do not reject H0<\/sub><\/em><\/section>\n<\/div>\n
                                                                                          <\/section>\n
                                                                                          \n
                                                                                          41. Means<\/section>\n<\/div>\n
                                                                                          <\/section>\n
                                                                                          \n
                                                                                          43. The mean time spent in jail for 26 first time convicted burglars<\/section>\n<\/div>\n
                                                                                          <\/section>\n
                                                                                          \n
                                                                                          45.<\/p>\n
                                                                                          \n
                                                                                            \n
                                                                                          1. 3<\/li>\n
                                                                                          2. 1.5<\/li>\n
                                                                                          3. 1.8<\/li>\n
                                                                                          4. 26<\/li>\n<\/ol>\n

                                                                                            47. [latex]\\overline{X}~{N}\\left({2.5},\\frac{1.5}{\\sqrt{26}}\\right)[\/latex]<\/p>\n<\/section>\n<\/section>\n<\/div>\n

                                                                                            Additional Information and Full Hypothesis Test Examples \u2013 Practice<\/h2>\n
                                                                                            49.\u00a0This is a left-tailed test.<\/section>\n
                                                                                            <\/section>\n
                                                                                            51.\u00a0This is a two-tailed test.<\/section>\n
                                                                                            <\/section>\n
                                                                                            <\/section>\n
                                                                                            <\/section>\n
                                                                                            <\/section>\n
                                                                                            <\/section>\n
                                                                                            <\/section>\n
                                                                                            <\/section>\n
                                                                                            53.\u00a0the graph of a two-tailed test.<\/p>\n
                                                                                            \n
                                                                                            \"Normal<\/span><\/figure>\n

                                                                                             <\/p>\n

                                                                                            <\/figure>\n<\/section>\n

                                                                                            55. a right-tailed test<\/p>\n

                                                                                            57. a left-tailed test<\/span><\/p>\n

                                                                                            59. This is a left-tailed test.<\/span><\/p>\n

                                                                                            61.\u00a0This is a two-tailed test.<\/span><\/p>\n<\/section>\n

                                                                                            Null and Alternative Hypotheses \u2013 Homework<\/h2>\n
                                                                                            \n
                                                                                            \n
                                                                                            62.<\/p>\n
                                                                                            \n
                                                                                              \n
                                                                                            1. H0<\/sub><\/em>: \u03bc<\/em> = 34; Ha<\/sub><\/em>: \u03bc<\/em> \u2260 34<\/li>\n
                                                                                            2. H0<\/sub><\/em>: p<\/em> \u2264 0.60; Ha<\/sub><\/em>: p<\/em> > 0.60<\/li>\n
                                                                                            3. H0<\/sub><\/em>: \u03bc<\/em> \u2265 100,000; Ha<\/sub><\/em>: \u03bc<\/em> < 100,000<\/li>\n
                                                                                            4. H0<\/sub><\/em>: p<\/em> = 0.29; Ha<\/sub><\/em>: p<\/em> \u2260 0.29<\/li>\n
                                                                                            5. H0<\/sub><\/em>: p<\/em> = 0.05; Ha<\/sub><\/em>: p<\/em> < 0.05<\/li>\n
                                                                                            6. H0<\/sub><\/em>: \u03bc<\/em> \u2264 10; Ha<\/sub><\/em>: \u03bc<\/em> > 10<\/li>\n
                                                                                            7. H0<\/sub><\/em>: p<\/em> = 0.50; Ha<\/sub><\/em>: p<\/em> \u2260 0.50<\/li>\n
                                                                                            8. H0<\/sub><\/em>: \u03bc<\/em> = 6; Ha<\/sub><\/em>: \u03bc<\/em> \u2260 6<\/li>\n
                                                                                            9. H0<\/sub><\/em>: p<\/em> \u2265 0.11; Ha<\/sub><\/em>: p<\/em> < 0.11<\/li>\n
                                                                                            10. H0<\/sub><\/em>: \u03bc<\/em> \u2264 20,000; Ha<\/sub><\/em>: \u03bc<\/em> > 20,000<\/li>\n<\/ol>\n

                                                                                              64.\u00a0p<\/em> < 0.20<\/p>\n

                                                                                              Outcomes and the Type I and Type II Errors \u2013 Homework<\/h2>\n<\/section>\n
                                                                                              \n
                                                                                              66.<\/p>\n
                                                                                                \n
                                                                                              1. Type I error: We conclude that the mean is not 34 years when it really is 34 years. Type II error: We conclude that the mean is 34 years, when in fact it really is not 34 years.<\/li>\n
                                                                                              2. Type I error: We conclude that more than 60% of Americans vote in presidential elections when the actual percentage is at most 60%. Type II error: We conclude that at most 60% of Americans vote in presidential elections when, in fact, more than 60% do.<\/li>\n
                                                                                              3. Type I error: We conclude that the mean starting salary is less than $100,000 when it really is at least $100,000. Type II error: We conclude that the mean starting salary is at least $100,000 when, in fact, it is less than $100,000.<\/li>\n
                                                                                              4. Type I error: We conclude that the proportion of high school seniors who get drunk each month is not 29% when it really is 29%. Type II error: We conclude that the proportion of high school seniors who get drunk each month is 29% when, in fact, it is not 29%.<\/li>\n
                                                                                              5. Type I error: We conclude that fewer than 5% of adults ride the bus to work in Los Angeles when the percentage that do is really 5% or more. Type II error: We conclude that 5% or more adults ride the bus to work in Los Angeles when, in fact, fewer than 5% do.<\/li>\n
                                                                                              6. Type I error: We conclude that the mean number of cars a person owns in his or her lifetime is more than 10 when in reality it is not more than 10. Type II error: We conclude that the mean number of cars a person owns in his or her lifetime is not more than 10 when, in fact, it is more than 10.<\/li>\n
                                                                                              7. Type I error: We conclude that the proportion of Americans who prefer to live away from cities is not about half, though the actual proportion is about half. Type II error: We conclude that the proportion of Americans who prefer to live away from cities is half when, in fact, it is not half.<\/li>\n
                                                                                              8. Type I error: We conclude that the duration of paid vacations each year for Europeans is not six weeks, when in fact it is six weeks. Type II error: We conclude that the duration of paid vacations each year for Europeans is six weeks when, in fact, it is not.<\/li>\n
                                                                                              9. Type I error: We conclude that the proportion is less than 11%, when it is really at least 11%. Type II error: We conclude that the proportion of women who develop breast cancer is at least 11%, when in fact it is less than 11%.<\/li>\n
                                                                                              10. Type I error: We conclude that the average tuition cost at private universities is more than $20,000, though in reality, it is at most $20,000. Type II error: We conclude that the average tuition cost at private universities is at most $20,000 when, in fact, it is more than $20,000.<\/li>\n<\/ol>\n

                                                                                                68.\u00a0Not to conclude the drug is safe when, in fact, it is safe.<\/p>\n

                                                                                                70.\u00a0is less than seven hours<\/p>\n

                                                                                                Distribution Needed for Hypothesis Testing \u2013 Homework<\/h2>\n

                                                                                                72.\u00a0t<\/em>21<\/sub><\/p>\n

                                                                                                Rare Events, the Sample, Decision and Conclusion – Homework<\/h2>\n<\/section>\n<\/section>\n<\/section>\n<\/div>\n<\/div>\n
                                                                                                \n
                                                                                                N\/A<\/section>\n
                                                                                                \n

                                                                                                Additional Information and Full Hypothesis Test Examples – Homework<\/h2>\n<\/section>\n
                                                                                                <\/section>\n<\/section>\n
                                                                                                <\/section>\n
                                                                                                74.\u00a0<\/p>\n
                                                                                                \n
                                                                                                  \n
                                                                                                1. H0<\/sub><\/em>: \u03bc<\/em> \u2265 50,000<\/li>\n
                                                                                                2. Ha<\/sub><\/em>: \u03bc<\/em> < 50,000<\/li>\n
                                                                                                3. Let [latex]\\overline{X}[\/latex]\u00a0= the average lifespan of a brand of tires.<\/li>\n
                                                                                                4. normal distribution<\/li>\n
                                                                                                5. z<\/em> = -2.315<\/li>\n
                                                                                                6. p<\/em>-value = 0.0103<\/li>\n
                                                                                                7. Check student\u2019s solution.\n
                                                                                                    \n
                                                                                                  1. alpha: 0.05<\/li>\n
                                                                                                  2. Decision: Reject the null hypothesis.<\/li>\n
                                                                                                  3. Reason for decision: The p<\/em>-value is less than 0.05.<\/li>\n
                                                                                                  4. Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles.<\/li>\n<\/ol>\n<\/li>\n
                                                                                                  5. (43,537, 49,463)<\/li>\n<\/ol>\n<\/section>\n

                                                                                                    76.<\/p>\n

                                                                                                    \n
                                                                                                      \n
                                                                                                    1. H0<\/sub><\/em>: \u03bc<\/em> = $1.00<\/li>\n
                                                                                                    2. Ha<\/sub><\/em>: \u03bc<\/em> \u2260 $1.00<\/li>\n
                                                                                                    3. Let [latex]\\overline{X}[\/latex]= the average cost of a daily newspaper.<\/li>\n
                                                                                                    4. normal distribution<\/li>\n
                                                                                                    5. z<\/em> = \u20130.866<\/li>\n
                                                                                                    6. p<\/em>-value = 0.3865<\/li>\n
                                                                                                    7. Check student\u2019s solution.\n
                                                                                                        \n
                                                                                                      1. Alpha: 0.01<\/li>\n
                                                                                                      2. Decision: Do not reject the null hypothesis.<\/li>\n
                                                                                                      3. Reason for decision: The p<\/em>-value is greater than 0.01.<\/li>\n
                                                                                                      4. Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is $1. The mean cost could be $1.<\/li>\n<\/ol>\n<\/li>\n
                                                                                                      5. ($0.84, $1.06)<\/li>\n<\/ol>\n

                                                                                                        78.<\/p>\n

                                                                                                        \n
                                                                                                          \n
                                                                                                        1. H0<\/sub><\/em>: \u03bc<\/em> = 10<\/li>\n
                                                                                                        2. Ha<\/sub><\/em>: \u03bc<\/em> \u2260 10<\/li>\n
                                                                                                        3. Let [latex]\\overline{X}[\/latex]\u00a0the mean number of sick days an employee takes per year.<\/li>\n
                                                                                                        4. Student\u2019s t<\/em>-distribution<\/li>\n
                                                                                                        5. t<\/em> = \u20131.12<\/li>\n
                                                                                                        6. p<\/em>-value = 0.300<\/li>\n
                                                                                                        7. Check student\u2019s solution.\n
                                                                                                            \n
                                                                                                          1. Alpha: 0.05<\/li>\n
                                                                                                          2. Decision: Do not reject the null hypothesis.<\/li>\n
                                                                                                          3. Reason for decision: The p<\/em>-value is greater than 0.05.<\/li>\n
                                                                                                          4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean number of sick days is not ten.<\/li>\n<\/ol>\n<\/li>\n
                                                                                                          5. (4.9443, 11.806)<\/li>\n<\/ol>\n

                                                                                                            80.<\/p>\n

                                                                                                            \n
                                                                                                              \n
                                                                                                            1. H0<\/sub><\/em>: p<\/em> \u2265 0.6<\/li>\n
                                                                                                            2. Ha<\/sub><\/em>: p<\/em> < 0.6<\/li>\n
                                                                                                            3. Let P\u2032<\/em> = the proportion of students who feel more enriched as a result of taking Elementary Statistics.<\/li>\n
                                                                                                            4. normal for a single proportion<\/li>\n
                                                                                                            5. 1.12<\/li>\n
                                                                                                            6. p<\/em>-value = 0.1308<\/li>\n
                                                                                                            7. Check student\u2019s solution.\n
                                                                                                                \n
                                                                                                              1. Alpha: 0.05<\/li>\n
                                                                                                              2. Decision: Do not reject the null hypothesis.<\/li>\n
                                                                                                              3. Reason for decision: The p<\/em>-value is greater than 0.05.<\/li>\n
                                                                                                              4. Conclusion: There is insufficient evidence to conclude that less than 60 percent of her students feel more enriched.<\/li>\n<\/ol>\n<\/li>\n
                                                                                                              5. Confidence Interval: (0.409, 0.654)
                                                                                                                \nThe \u201cplus-4s\u201d confidence interval is (0.411, 0.648)<\/li>\n<\/ol>\n<\/section>\n

                                                                                                                82.<\/p>\n

                                                                                                                  \n
                                                                                                                1. H0<\/sub><\/em>:\u00a0\u03bc<\/em>\u00a0= 4<\/li>\n
                                                                                                                2. Ha<\/sub><\/em>:\u00a0\u03bc<\/em>\u00a0\u2260 4<\/li>\n
                                                                                                                3. Let [latex]\\overline{X}[\/latex]<\/span>\u00a0the average I.Q. of a set of brown trout.<\/li>\n
                                                                                                                4. two-tailed Student’s t-test<\/li>\n
                                                                                                                5. t<\/em>\u00a0= 1.95<\/li>\n
                                                                                                                6. p<\/em>-value = 0.076<\/li>\n
                                                                                                                7. Check student\u2019s solution.\n
                                                                                                                    \n
                                                                                                                  1. Alpha: 0.05<\/li>\n
                                                                                                                  2. Decision: Reject the null hypothesis.<\/li>\n
                                                                                                                  3. Reason for decision: The\u00a0p<\/em>-value is greater than 0.05<\/li>\n
                                                                                                                  4. Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four.<\/li>\n<\/ol>\n<\/li>\n
                                                                                                                  5. (3.8865,5.9468)<\/li>\n<\/ol>\n
                                                                                                                    84.<\/p>\n
                                                                                                                    \n
                                                                                                                      \n
                                                                                                                    1. H0<\/sub><\/em>: p<\/em> \u2265 0.13<\/li>\n
                                                                                                                    2. Ha<\/sub><\/em>: p<\/em> < 0.13<\/li>\n
                                                                                                                    3. Let P\u2032<\/em> = the proportion of Americans who have seen or sensed angels<\/li>\n
                                                                                                                    4. normal for a single proportion<\/li>\n
                                                                                                                    5. \u20132.688<\/li>\n
                                                                                                                    6. p<\/em>-value = 0.0036<\/li>\n
                                                                                                                    7. Check student\u2019s solution.\n
                                                                                                                        \n
                                                                                                                      1. alpha: 0.05<\/li>\n
                                                                                                                      2. Decision: Reject the null hypothesis.<\/li>\n
                                                                                                                      3. Reason for decision: The p<\/em>-value is less than 0.05.<\/li>\n
                                                                                                                      4. Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have seen or sensed an angel is less than 13%.<\/li>\n<\/ol>\n<\/li>\n
                                                                                                                      5. (0, 0.0623).
                                                                                                                        \nThe\u201cplus-4s\u201d confidence interval is (0.0022, 0.0978)<\/li>\n<\/ol>\n

                                                                                                                        88.<\/p>\n

                                                                                                                          \n
                                                                                                                        1. H0<\/sub><\/em>:\u00a0\u03bc<\/em>\u00a0\u2265 129<\/li>\n
                                                                                                                        2. Ha<\/sub><\/em>:\u00a0\u03bc<\/em>\u00a0< 129<\/li>\n
                                                                                                                        3. Let [latex]\\overline{X}[\/latex]<\/span>\u00a0= the average time in seconds that Terri finishes Lap 4.<\/li>\n
                                                                                                                        4. Student’s\u00a0t<\/em>-distribution<\/li>\n
                                                                                                                        5. t<\/em>\u00a0= 1.209<\/li>\n
                                                                                                                        6. 0.8792<\/li>\n
                                                                                                                        7. Check student\u2019s solution.\n
                                                                                                                            \n
                                                                                                                          1. Alpha: 0.05<\/li>\n
                                                                                                                          2. Decision: Do not reject the null hypothesis.<\/li>\n
                                                                                                                          3. Reason for decision: The\u00a0p<\/em>-value is greater than 0.05.<\/li>\n
                                                                                                                          4. Conclusion: There is insufficient evidence to conclude that Terri\u2019s mean lap time is less than 129 seconds.<\/li>\n<\/ol>\n<\/li>\n
                                                                                                                          5. (128.63, 130.37)<\/li>\n<\/ol>\n

                                                                                                                            90.<\/p>\n

                                                                                                                              \n
                                                                                                                            1. H0<\/sub><\/em>:\u00a0p<\/em>\u00a0= 0.60<\/li>\n
                                                                                                                            2. Ha<\/sub><\/em>:\u00a0p<\/em>\u00a0< 0.60<\/li>\n
                                                                                                                            3. Let\u00a0P\u2032<\/em>\u00a0= the proportion of family members who shed tears at a reunion.<\/li>\n
                                                                                                                            4. normal for a single proportion<\/li>\n
                                                                                                                            5. \u20131.71<\/li>\n
                                                                                                                            6. 0.0438<\/li>\n
                                                                                                                            7. Check student\u2019s solution.\n
                                                                                                                                \n
                                                                                                                              1. alpha: 0.05<\/li>\n
                                                                                                                              2. Decision: Reject the null hypothesis.<\/li>\n
                                                                                                                              3. Reason for decision:\u00a0p<\/em>-value < alpha<\/li>\n
                                                                                                                              4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of family members who shed tears at a reunion is less than 0.60. However, the test is weak because the\u00a0p<\/em>-value and alpha are quite close, so other tests should be done.<\/li>\n<\/ol>\n<\/li>\n
                                                                                                                              5. We are 95% confident that between 38.29% and 61.71% of family members will shed tears at a family reunion. (0.3829, 0.6171). The\u201cplus-4s\u201d confidence interval (see chapter 8) is (0.3861, 0.6139)<\/li>\n<\/ol>\n

                                                                                                                                Note that here the \u201clarge-sample\u201d 1 \u2013 PropZTest provides the approximate\u00a0p<\/em>-value of 0.0438. Whenever a\u00a0p<\/em>-value based on a normal approximation is close to the level of significance, the exact\u00a0p<\/em>-value based on binomial probabilities should be calculated whenever possible. This is beyond the scope of this course.<\/p>\n

                                                                                                                                92.<\/p>\n<\/section>\n

                                                                                                                                  \n
                                                                                                                                1. H0<\/sub><\/em>:\u00a0\u03bc<\/em>\u00a0\u2265 22<\/li>\n
                                                                                                                                2. Ha<\/sub><\/em>:\u00a0\u03bc<\/em>\u00a0< 22<\/li>\n
                                                                                                                                3. Let [latex]\\overline{X}[\/latex]<\/span>\u00a0= the mean number of bubbles per blow.<\/li>\n
                                                                                                                                4. Student’s\u00a0t<\/em>-distribution<\/li>\n
                                                                                                                                5. \u20132.667<\/li>\n
                                                                                                                                6. p<\/em>-value = 0.00486<\/li>\n
                                                                                                                                7. Check student\u2019s solution.\n
                                                                                                                                    \n
                                                                                                                                  1. Alpha: 0.05<\/li>\n
                                                                                                                                  2. Decision: Reject the null hypothesis.<\/li>\n
                                                                                                                                  3. Reason for decision: The\u00a0p<\/em>-value is less than 0.05.<\/li>\n
                                                                                                                                  4. Conclusion: There is sufficient evidence to conclude that the mean number of bubbles per blow is less than 22.<\/li>\n<\/ol>\n<\/li>\n
                                                                                                                                  5. (18.501, 21.499)<\/li>\n<\/ol>\n
                                                                                                                                    94.<\/p>\n
                                                                                                                                      \n
                                                                                                                                    1. H0<\/sub><\/em>:\u00a0\u03bc<\/em>\u00a0\u2264 1<\/li>\n
                                                                                                                                    2. Ha<\/sub><\/em>:\u00a0\u03bc<\/em>\u00a0> 1<\/li>\n
                                                                                                                                    3. Let [latex]\\overline{X}[\/latex]<\/span>\u00a0= the mean cost in dollars of macaroni and cheese in a certain town.<\/li>\n
                                                                                                                                    4. Student’s\u00a0t<\/em>-distribution<\/li>\n
                                                                                                                                    5. t<\/em>\u00a0= 0.340<\/li>\n
                                                                                                                                    6. p<\/em>-value = 0.36756<\/li>\n
                                                                                                                                    7. Check student\u2019s solution.\n
                                                                                                                                        \n
                                                                                                                                      1. Alpha: 0.05<\/li>\n
                                                                                                                                      2. Decision: Do not reject the null hypothesis.<\/li>\n
                                                                                                                                      3. Reason for decision: The\u00a0p<\/em>-value is greater than 0.05<\/li>\n
                                                                                                                                      4. Conclusion: The mean cost could be $1, or less. At the 5% significance level, there is insufficient evidence to conclude that the mean price of a box of macaroni and cheese is more than $1.<\/li>\n<\/ol>\n<\/li>\n
                                                                                                                                      5. (0.8291, 1.241)<\/li>\n<\/ol>\n

                                                                                                                                        96.<\/p>\n

                                                                                                                                          \n
                                                                                                                                        1. H0<\/sub><\/em>:\u00a0p<\/em>\u00a0= 0.01<\/li>\n
                                                                                                                                        2. Ha<\/sub><\/em>:\u00a0p<\/em>\u00a0> 0.01<\/li>\n
                                                                                                                                        3. Let\u00a0P\u2032<\/em>\u00a0= the proportion of errors generated<\/li>\n
                                                                                                                                        4. Normal for a single proportion<\/li>\n
                                                                                                                                        5. 2.13<\/li>\n
                                                                                                                                        6. 0.0165<\/li>\n
                                                                                                                                        7. Check student\u2019s solution.<\/li>\n
                                                                                                                                        8. \n
                                                                                                                                            \n
                                                                                                                                          1. Alpha: 0.05<\/li>\n
                                                                                                                                          2. Decision: Reject the null hypothesis<\/li>\n
                                                                                                                                          3. Reason for decision: The\u00a0p<\/em>-value is less than 0.05.<\/li>\n
                                                                                                                                          4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of errors generated is more than 0.01.<\/li>\n<\/ol>\n<\/li>\n
                                                                                                                                          5. Confidence interval: (0, 0.094).
                                                                                                                                            \n<\/span>The\u201cplus-4s\u201d confidence interval is (0.004, 0.144).<\/li>\n<\/ol>\n<\/section>\n

                                                                                                                                            98.<\/p>\n<\/section>\n

                                                                                                                                              \n
                                                                                                                                            1. H0<\/sub><\/em>:\u00a0p<\/em>\u00a0= 0.50<\/li>\n
                                                                                                                                            2. Ha<\/sub><\/em>:\u00a0p<\/em>\u00a0< 0.50<\/li>\n
                                                                                                                                            3. Let\u00a0P\u2032<\/em>\u00a0= the proportion of friends that has a pierced ear.<\/li>\n
                                                                                                                                            4. normal for a single proportion<\/li>\n
                                                                                                                                            5. \u20131.70<\/li>\n
                                                                                                                                            6. p<\/em>-value = 0.0448<\/li>\n
                                                                                                                                            7. Check student\u2019s solution.<\/li>\n
                                                                                                                                            8. \n
                                                                                                                                                \n
                                                                                                                                              1. Alpha: 0.05<\/li>\n
                                                                                                                                              2. Decision: Reject the null hypothesis<\/li>\n
                                                                                                                                              3. Reason for decision: The\u00a0p<\/em>-value is less than 0.05. (However, they are very close.)<\/li>\n
                                                                                                                                              4. Conclusion: There is sufficient evidence to support the claim that less than 50% of his friends have pierced ears.<\/li>\n<\/ol>\n<\/li>\n
                                                                                                                                              5. Confidence Interval: (0.245, 0.515): The \u201cplus-4s\u201d confidence interval is (0.259, 0.519).<\/li>\n<\/ol>\n
                                                                                                                                                \n
                                                                                                                                                100.<\/p>\n
                                                                                                                                                \n
                                                                                                                                                  \n
                                                                                                                                                1. H0<\/sub><\/em>: p<\/em> = 0.40<\/li>\n
                                                                                                                                                2. Ha<\/sub><\/em>: p<\/em> < 0.40<\/li>\n
                                                                                                                                                3. Let P\u2032<\/em> = the proportion of schoolmates who fear public speaking.<\/li>\n
                                                                                                                                                4. normal for a single proportion<\/li>\n
                                                                                                                                                5. \u20131.01<\/li>\n
                                                                                                                                                6. p<\/em>-value = 0.1563<\/li>\n
                                                                                                                                                7. Check student\u2019s solution.\n
                                                                                                                                                    \n
                                                                                                                                                  1. Alpha: 0.05<\/li>\n
                                                                                                                                                  2. Decision: Do not reject the null hypothesis.<\/li>\n
                                                                                                                                                  3. Reason for decision: The p<\/em>-value is greater than 0.05.<\/li>\n
                                                                                                                                                  4. Conclusion: There is insufficient evidence to support the claim that less than 40% of students at the school fear public speaking.<\/li>\n<\/ol>\n<\/li>\n
                                                                                                                                                  5. Confidence Interval: (0.3241, 0.4240): The \u201cplus-4s\u201d confidence interval is (0.3257, 0.4250).<\/li>\n<\/ol>\n

                                                                                                                                                    102.<\/p>\n

                                                                                                                                                    \n
                                                                                                                                                      \n
                                                                                                                                                    1. H0<\/sub><\/em>: p<\/em> = 0.14<\/li>\n
                                                                                                                                                    2. Ha<\/sub><\/em>: p<\/em> < 0.14<\/li>\n
                                                                                                                                                    3. Let P\u2032<\/em> = the proportion of NYC residents that smoke.<\/li>\n
                                                                                                                                                    4. normal for a single proportion<\/li>\n
                                                                                                                                                    5. \u20130.2756<\/li>\n
                                                                                                                                                    6. p<\/em>-value = 0.3914<\/li>\n
                                                                                                                                                    7. Check student\u2019s solution.\n
                                                                                                                                                        \n
                                                                                                                                                      1. alpha: 0.05<\/li>\n
                                                                                                                                                      2. Decision: Do not reject the null hypothesis.<\/li>\n
                                                                                                                                                      3. Reason for decision: The p<\/em>-value is greater than 0.05.<\/li>\n
                                                                                                                                                      4. At the 5% significance level, there is insufficient evidence to conclude that the proportion of NYC residents who smoke is less than 0.14.<\/li>\n<\/ol>\n<\/li>\n
                                                                                                                                                      5. Confidence Interval: (0.0502, 0.2070): The \u201cplus-4s\u201d confidence interval (see chapter 8) is (0.0676, 0.2297).<\/li>\n<\/ol>\n

                                                                                                                                                        104.<\/p>\n

                                                                                                                                                        \n
                                                                                                                                                          \n
                                                                                                                                                        1. H0<\/sub><\/em>: \u03bc<\/em> = 69,110<\/li>\n
                                                                                                                                                        2. Ha<\/sub><\/em>: \u03bc<\/em> > 69,110<\/li>\n
                                                                                                                                                        3. Let [latex]\\overline{X}[\/latex]\u00a0= the mean salary in dollars for California registered nurses.<\/li>\n
                                                                                                                                                        4. Student’s t<\/em>-distribution<\/li>\n
                                                                                                                                                        5. t<\/em> = 1.719<\/li>\n
                                                                                                                                                        6. p<\/em>-value: 0.0466<\/li>\n
                                                                                                                                                        7. Check student\u2019s solution.\n
                                                                                                                                                            \n
                                                                                                                                                          1. Alpha: 0.05<\/li>\n
                                                                                                                                                          2. Decision: Reject the null hypothesis.<\/li>\n
                                                                                                                                                          3. Reason for decision: The p<\/em>-value is less than 0.05.<\/li>\n
                                                                                                                                                          4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean salary of California registered nurses exceeds $69,110.<\/li>\n<\/ol>\n<\/li>\n
                                                                                                                                                          5. ($68,757, $73,485)<\/li>\n<\/ol>\n

                                                                                                                                                            106. Do not reject H0<\/sub><\/em>: There is not sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.<\/p>\n<\/section>\n

                                                                                                                                                            108.\u00a0There is not enough evidence to conclude that the mean number of hours is more than 4.5<\/p>\n

                                                                                                                                                            110.<\/p>\n

                                                                                                                                                            \n
                                                                                                                                                              \n
                                                                                                                                                            1. H0<\/sub><\/em>: p<\/em> = 0.488 Ha<\/sub><\/em>: p<\/em> \u2260 0.488<\/li>\n
                                                                                                                                                            2. p<\/em>-value = 0.0114<\/li>\n
                                                                                                                                                            3. alpha = 0.05<\/li>\n
                                                                                                                                                            4. Reject the null hypothesis.<\/li>\n
                                                                                                                                                            5. At the 5% level of significance, there is enough evidence to conclude that 48.8% of families own stocks.<\/li>\n
                                                                                                                                                            6. The survey does not appear to be accurate.<\/li>\n<\/ol>\n

                                                                                                                                                              112.<\/p>\n

                                                                                                                                                              \n
                                                                                                                                                                \n
                                                                                                                                                              1. H0<\/sub><\/em>: p<\/em> = 0.517 Ha<\/sub><\/em>: p<\/em> \u2260 0.517<\/li>\n
                                                                                                                                                              2. p<\/em>-value = 0.9203.<\/li>\n
                                                                                                                                                              3. alpha = 0.05.<\/li>\n
                                                                                                                                                              4. Do not reject the null hypothesis.<\/li>\n
                                                                                                                                                              5. At the 5% significance level, there is not enough evidence to conclude that the proportion of homes in Kentucky that are heated by natural gas is 0.517.<\/li>\n
                                                                                                                                                              6. However, we cannot generalize this result to the entire nation. First, the sample\u2019s population is only the state of Kentucky. Second, it is reasonable to assume that homes in the extreme north and south will have extreme high usage and low usage, respectively. We would need to expand our sample base to include these possibilities if we wanted to generalize this claim to the entire nation.<\/li>\n<\/ol>\n

                                                                                                                                                                114.<\/p>\n

                                                                                                                                                                \n
                                                                                                                                                                  \n
                                                                                                                                                                1. H0<\/sub><\/em>: \u00b5<\/em> \u2265 11.52 Ha<\/sub><\/em>: \u00b5<\/em> < 11.52<\/li>\n
                                                                                                                                                                2. p<\/em>-value = 0.000002 which is almost 0.<\/li>\n
                                                                                                                                                                3. alpha = 0.05.<\/li>\n
                                                                                                                                                                4. Reject the null hypothesis.<\/li>\n
                                                                                                                                                                5. At the 5% significance level, there is enough evidence to conclude that the mean amount of summer rain in the northeastern US is less than 11.52 inches, on average.<\/li>\n
                                                                                                                                                                6. We would make the same conclusion if alpha was 1% because the p<\/em>-value is almost 0.<\/li>\n<\/ol>\n

                                                                                                                                                                  116.<\/p>\n

                                                                                                                                                                  \n
                                                                                                                                                                    \n
                                                                                                                                                                  1. H0<\/sub><\/em>: \u00b5<\/em> \u2264 5.8 Ha<\/sub><\/em>: \u00b5<\/em> > 5.8<\/li>\n
                                                                                                                                                                  2. p<\/em>-value = 0.9987<\/li>\n
                                                                                                                                                                  3. alpha = 0.05<\/li>\n
                                                                                                                                                                  4. Do not reject the null hypothesis.<\/li>\n
                                                                                                                                                                  5. At the 5% level of significance, there is not enough evidence to conclude that a woman visits her doctor, on average, more than 5.8 times a year.<\/li>\n<\/ol>\n

                                                                                                                                                                    118.<\/p>\n

                                                                                                                                                                    \n
                                                                                                                                                                      \n
                                                                                                                                                                    1. H0<\/sub><\/em>: \u00b5<\/em> \u2265 150 Ha<\/sub><\/em>: \u00b5<\/em> < 150<\/li>\n
                                                                                                                                                                    2. p<\/em>-value = 0.0622<\/li>\n
                                                                                                                                                                    3. alpha = 0.01<\/li>\n
                                                                                                                                                                    4. Do not reject the null hypothesis.<\/li>\n
                                                                                                                                                                    5. At the 1% significance level, there is not enough evidence to conclude that freshmen students study less than 2.5 hours per day, on average.<\/li>\n
                                                                                                                                                                    6. The student academic group\u2019s claim appears to be correct.<\/li>\n<\/ol>\n<\/section>\n<\/section>\n<\/section>\n<\/section>\n<\/section>\n<\/section>\n<\/section>\n<\/section>\n<\/section>\n<\/section>\n<\/section>\n<\/section>\n\n\t\t\t
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                                                                                                                                                                      Candela Citations<\/h3>\n\t\t\t\t\t
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