{"id":410,"date":"2016-04-21T22:43:39","date_gmt":"2016-04-21T22:43:39","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstats1xmaster\/?post_type=chapter&p=410"},"modified":"2021-06-15T17:53:14","modified_gmt":"2021-06-15T17:53:14","slug":"answers-to-selected-exercises-5","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/chapter\/answers-to-selected-exercises-5\/","title":{"raw":"Answers to Selected Exercises","rendered":"Answers to Selected Exercises"},"content":{"raw":"

Two Population Means with Unknown Standard Deviations<\/h2>\r\n1.\u00a0two proportions\r\n\r\n3.\u00a0matched or paired samples\r\n\r\n6. independent group means, population standard deviations and\/or variances unknown\r\n\r\n8.\u00a0two proportions\r\n\r\n10.\u00a0independent group means, population standard deviations and\/or variances unknown\r\n\r\n11.\u00a0independent group means, population standard deviations and\/or variances unknown\r\n\r\n13.\u00a0two proportions\r\n\r\n15.\u00a0The random variable is the difference between the mean amounts of sugar in the two soft drinks.\r\n\r\n17.\u00a0means\r\n\r\n19.\u00a0two-tailed\r\n\r\n21.\u00a0the difference between the mean life spans of whites and nonwhites\r\n\r\n23.\u00a0This is a comparison of two population means with unknown population standard deviations.\r\n\r\n27.\r\n\r\n
\r\n
    \r\n \t
  1. Reject the null hypothesis<\/li>\r\n \t
  2. p<\/em>-value < 0.05<\/li>\r\n \t
  3. There is not enough evidence at the 5% level of significance to support the claim that life expectancy in the 1900s is different between whites and nonwhites.<\/li>\r\n<\/ol>\r\n30.\r\n
      \r\n \t
    1. H0<\/sub><\/em>: \u03bc1<\/sub><\/em> \u2265 \u03bc2<\/sub><\/em><\/li>\r\n \t
    2. Ha<\/sub><\/em>: \u03bc1<\/sub><\/em> < \u03bc2<\/sub><\/em><\/li>\r\n \t
    3. [latex]\\overline{X}_{1}\u2013\\overline{X}_{2}[\/latex]is the difference between the mean enrollments of the two-year colleges and the four-year colleges.<\/li>\r\n \t
    4. Student\u2019s-t<\/em><\/li>\r\n \t
    5. test statistic: -0.2480<\/li>\r\n \t
    6. p<\/em>-value: 0.4019<\/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: Do not reject<\/li>\r\n \t
      3. Reason for Decision: p<\/em>-value > alpha<\/li>\r\n \t
      4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean enrollment at four-year colleges is higher than at two-year colleges.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n32.\r\n\r\n
        \r\n

        Subscripts: 1: mechanical engineering; 2: electrical engineering<\/p>\r\n\r\n

          \r\n \t
        1. H0<\/sub><\/em>: \u00b51<\/sub><\/em> \u2265 \u00b52<\/sub><\/em><\/li>\r\n \t
        2. Ha<\/sub><\/em>: \u00b51<\/sub><\/em> < \u00b52<\/sub><\/em><\/li>\r\n \t
        3. [latex]\\overline{X}_{1}\u2013\\overline{X}_{2}[\/latex] is the difference between the mean entry level salaries of mechanical engineers and electrical engineers.<\/li>\r\n \t
        4. t<\/em>108<\/sub><\/li>\r\n \t
        5. test statistic: t<\/em> = \u20130.82<\/li>\r\n \t
        6. p<\/em>-value: 0.2061<\/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: Do not reject the null hypothesis.<\/li>\r\n \t
          3. Reason for Decision: p<\/em>-value > alpha<\/li>\r\n \t
          4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean entry-level salaries of mechanical engineers is lower than that of electrical engineers.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<\/section>34.\r\n\r\n
            \r\n
              \r\n \t
            1. H0<\/sub><\/em>: \u00b51<\/sub><\/em> = \u00b52<\/sub><\/em><\/li>\r\n \t
            2. Ha<\/sub><\/em>: \u00b51<\/sub><\/em> \u2260 \u00b52<\/sub><\/em><\/li>\r\n \t
            3. [latex]\\overline{X}_{1}\u2013\\overline{X}_{2}[\/latex] is the difference between the mean times for completing a lap in races and in practices.<\/li>\r\n \t
            4. t<\/em>20.32<\/sub><\/li>\r\n \t
            5. test statistic: \u20134.70<\/li>\r\n \t
            6. p<\/em>-value: 0.0001<\/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: p<\/em>-value < alpha<\/li>\r\n \t
              4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n36.\r\n\r\n
                \r\n
                  \r\n \t
                1. H0<\/sub><\/em>: \u00b51<\/sub><\/em> = \u00b52<\/sub><\/em><\/li>\r\n \t
                2. Ha<\/sub><\/em>: \u00b51<\/sub><\/em> \u2260 \u00b52<\/sub><\/em><\/li>\r\n \t
                3. [latex]\\overline{X}_{1}\u2013\\overline{X}_{2}[\/latex] \u00a0is the difference between the mean times for completing a lap in races and in practices.<\/li>\r\n \t
                4. t<\/em>40.94<\/sub><\/li>\r\n \t
                5. test statistic: \u20135.08<\/li>\r\n \t
                6. p<\/em>-value: zero<\/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: p<\/em>-value < alpha<\/li>\r\n \t
                  4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n39.\u00a0There is insufficient evidence to conclude that the W<\/strong> teams score fewer goals, on average, than the E<\/strong> teams score.\r\n\r\n<\/section><\/section><\/section>41.Test: two independent sample means, population standard deviations unknown.Random variable:[latex]\\overline{X}_{1}\u2013\\overline{X}_{2}[\/latex]Distribution: H0<\/sub><\/em>: \u03bc<\/em>1<\/sub> = \u03bc<\/em>2<\/sub>Ha<\/sub><\/em>: \u03bc<\/em>1<\/sub> < \u03bc<\/em>2<\/sub> The mean age of entering prostitution in Canada is lower than the mean age in the United States.\r\n
                    \"This<\/span><\/figure>\r\n

                    Graph: left-tailed<\/p>\r\n

                    p<\/em>-value : 0.0151<\/p>\r\n

                    Decision: Do not reject H<\/em>0<\/sub>.<\/p>\r\n

                    Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of entering prostitution in Canada is lower than the mean age in the United States.<\/p>\r\n43.\u00a0\u03bc<\/em>day<\/sub> \u2260 \u03bc<\/em>night<\/sub>\r\n

                    Two Population Means with Known Standard Deviations<\/h2>\r\n
                    44.\u00a0The difference in mean speeds of the fastball pitches of the two pitchers<\/div>\r\n
                    <\/div>\r\n
                    46.\u00a0\u20132.46<\/div>\r\n
                    <\/div>\r\n
                    48.\u00a0At the 1% significance level, we can reject the null hypothesis. There is sufficient data to conclude that the mean speed of Rodriguez\u2019s fastball is faster than Wesley\u2019s.<\/div>\r\n
                    <\/div>\r\n
                    \r\n\r\n50.\r\n\r\n
                    \r\n

                    Subscripts: 1 = Food, 2 = No Food<\/p>\r\n\r\n

                    <\/div>\r\nH0<\/sub><\/em>: \u03bc1<\/sub><\/em> \u2264 \u03bc2<\/sub><\/em>\r\n
                    <\/div>\r\nHa<\/sub><\/em>: \u03bc1<\/sub><\/em> > \u03bc2<\/sub><\/em><\/section><\/div>\r\n
                    <\/section>
                    52.<\/section>\r\n
                    \r\n\r\nthe graph of the p<\/em>-value.\r\n\r\n<\/div>\r\n
                    <\/div>\r\n
                    <\/div>\r\n
                    \r\n
                    \"This<\/span><\/figure>\r\n<\/section>54.\r\n

                    Subscripts: 1 = Gamma, 2 = Zeta<\/p>\r\n\r\n

                    <\/div>\r\nH0<\/sub><\/em>: \u03bc<\/em>1<\/sub> = \u03bc<\/em>2<\/sub>\r\n
                    <\/div>\r\nHa<\/sub><\/em>: \u03bc<\/em>1<\/sub> \u2260 \u03bc<\/em>2<\/sub>\r\n\r\n56.\u00a00.0062\r\n\r\n58.\u00a0There is sufficient evidence to reject the null hypothesis. The data support that the melting point for Alloy Zeta is different from the melting point of Alloy Gamma.\r\n\r\n60.\r\n\r\n
                    \r\n

                    Subscripts: 1 = boys, 2 = girls<\/p>\r\n\r\n

                      \r\n \t
                    1. H0<\/sub><\/em>: \u00b51<\/sub><\/em> \u2264 \u00b52<\/sub><\/em><\/li>\r\n \t
                    2. Ha<\/sub><\/em>: \u00b51<\/sub><\/em> > \u00b52<\/sub><\/em><\/li>\r\n \t
                    3. The random variable is the difference in the mean auto insurance costs for boys and girls.<\/li>\r\n \t
                    4. normal<\/li>\r\n \t
                    5. test statistic: z<\/em> = 2.50<\/li>\r\n \t
                    6. p<\/em>-value: 0.0062<\/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: p<\/em>-value < alpha<\/li>\r\n \t
                      4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean cost of auto insurance for teenage boys is greater than that for girls.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n62.\r\n\r\n
                        \r\n

                        Subscripts: 1 = non-hybrid sedans, 2 = hybrid sedans<\/p>\r\n\r\n

                          \r\n \t
                        1. H0<\/sub><\/em>: \u00b51<\/sub><\/em> \u2265 \u00b52<\/sub><\/em><\/li>\r\n \t
                        2. Ha<\/sub><\/em>: \u00b51<\/sub><\/em> < \u00b52<\/sub><\/em><\/li>\r\n \t
                        3. The random variable is the difference in the mean miles per gallon of non-hybrid sedans and hybrid sedans.<\/li>\r\n \t
                        4. normal<\/li>\r\n \t
                        5. test statistic: 6.36<\/li>\r\n \t
                        6. p<\/em>-value: 0<\/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: p<\/em>-value < alpha<\/li>\r\n \t
                          4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean miles per gallon of non-hybrid sedans is less than that of hybrid sedans.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n64.\r\n\r\n
                            \r\n
                              \r\n \t
                            1. H0<\/sub><\/em>: \u00b5d<\/sub><\/em> = 0<\/li>\r\n \t
                            2. Ha<\/sub><\/em>: \u00b5d<\/sub><\/em> < 0<\/li>\r\n \t
                            3. The random variable Xd<\/sub><\/em> is the average difference between husband\u2019s and wife\u2019s satisfaction level.<\/li>\r\n \t
                            4. t<\/em>9<\/sub><\/li>\r\n \t
                            5. test statistic: t<\/em> = \u20131.86<\/li>\r\n \t
                            6. p<\/em>-value: 0.0479<\/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, but run another test.<\/li>\r\n \t
                              3. Reason for Decision: p<\/em>-value < alpha<\/li>\r\n \t
                              4. Conclusion: This is a weak test because alpha and the p<\/em>-value are close. However, there is insufficient evidence to conclude that the mean difference is negative.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<\/section><\/section><\/section>\r\n

                                Comparing Two Independent Population Proportions<\/h2>\r\n66.\u00a0P<\/em>\u2032OS1<\/sub> \u2013 P<\/em>\u2032OS2<\/sub> = difference in the proportions of phones that had system failures within the first eight hours of operation with OS1<\/sub> and OS2<\/sub>.\r\n\r\n68.\u00a00.1018\r\n\r\n70.\u00a0proportions\r\n\r\n72.\u00a0right-tailed\r\n\r\n74.\u00a0The random variable is the difference in proportions (percents) of the populations that are of two or more races in Nevada and North Dakota.\r\n\r\n76.\u00a0Our sample sizes are much greater than five each, so we use the normal for two proportions distribution for this hypothesis test.\r\n\r\n80.\r\n\r\n
                                Reject the null hypothesis.p<\/em>-value < alphaAt the 5% significance level, there is sufficient evidence to conclude that the proportion (percent) of the population that is of two or more races in Nevada is statistically higher than that in North Dakota.83.
                                H0<\/sub><\/em>: PW<\/sub><\/em> = PB<\/sub><\/em>Ha<\/sub><\/em>: PW<\/sub><\/em> \u2260 PB<\/sub><\/em>The random variable is the difference in the proportions of white and black suicide victims, aged 15 to 24.normal for two proportionstest statistic: \u20130.1944\r\n\r\np<\/em>-value: 0.8458\r\n\r\nCheck 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: p<\/em>-value > alpha<\/li>\r\n \t
                                4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the proportions of white and black female suicide victims, aged 15 to 24, are different.<\/li>\r\n<\/ol>\r\n85.\r\n

                                  Subscripts: 1 = Cabrillo College, 2 = Lake Tahoe College<\/p>\r\n\r\n

                                    \r\n \t
                                  1. H0<\/sub><\/em>: p1<\/sub><\/em> = p2<\/sub><\/em><\/li>\r\n \t
                                  2. Ha<\/sub><\/em>: p1<\/sub><\/em> \u2260 p2<\/sub><\/em><\/li>\r\n \t
                                  3. The random variable is the difference between the proportions of Hispanic students at Cabrillo College and Lake Tahoe College.<\/li>\r\n \t
                                  4. normal for two proportions<\/li>\r\n \t
                                  5. test statistic: 4.29<\/li>\r\n \t
                                  6. p<\/em>-value: 0.00002<\/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: p<\/em>-value < alpha<\/li>\r\n \t
                                    4. Conclusion: There is sufficient evidence to conclude that the proportions of Hispanic students at Cabrillo College and Lake Tahoe College are different.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n87. \u00a0\u00a0p2011<\/sub><\/em> \u2264 p2010<\/sub><\/em>\r\n\r\n89.\r\n\r\n
                                      \r\n

                                      Test: two independent sample proportions.<\/p>\r\n

                                      Random variable: p<\/em>\u20321<\/sub> - p<\/em>\u20322<\/sub><\/p>\r\n

                                      Distribution:<\/p>\r\n\r\n

                                      <\/div>\r\nH0<\/sub><\/em>: p1<\/sub><\/em> = p2<\/sub><\/em>\r\n
                                      <\/div>\r\nHa<\/sub><\/em>: p1<\/sub><\/em> \u2260 p2<\/sub><\/em>\r\n

                                      The proportion of eReader users is different for the 16- to 29-year-old users from that of the 30 and older users.<\/p>\r\n

                                      Graph: two-tailed<\/p>\r\n\r\n

                                      \"This<\/span><\/figure>\r\n

                                      p<\/em>-value : 0.0033<\/p>\r\n

                                      Decision: Reject the null hypothesis.<\/p>\r\n

                                      Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that the proportion of eReader users 16 to 29 years old is different from the proportion of eReader users 30 and older.<\/p>\r\n\r\n<\/section>91.\r\n\r\n

                                      \r\n

                                      Test: two independent sample proportions<\/p>\r\n

                                      Random variable: p\u20321<\/sub><\/em> \u2212 p\u20322<\/sub><\/em><\/p>\r\n

                                      Distribution:<\/p>\r\n

                                      H0<\/sub><\/em>: p1<\/sub><\/em> = p2<\/sub><\/em><\/p>\r\n\r\n

                                      <\/div>\r\nHa<\/sub><\/em>: p1<\/sub><\/em> > p2<\/sub><\/em>\r\n

                                      A higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older.<\/p>\r\n

                                      Graph: right-tailed<\/p>\r\n\r\n

                                      \"This<\/span><\/figure>\r\np<\/em>-value: 0.2354\r\n

                                      Decision: Do not reject the H0<\/sub><\/em>.<\/p>\r\n

                                      Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older.<\/p>\r\n\r\n<\/section>93.\r\n\r\n

                                      \r\n

                                      Subscripts: 1: men; 2: women<\/p>\r\n\r\n

                                        \r\n \t
                                      1. H0<\/sub><\/em>: p1<\/sub><\/em> \u2264 p2<\/sub><\/em><\/li>\r\n \t
                                      2. Ha<\/sub><\/em>: p1<\/sub><\/em> > p2<\/sub><\/em><\/li>\r\n \t
                                      3. P<\/span>\u2032<\/span><\/span>1<\/span><\/span>\u2212<\/span>P<\/span>\u2032<\/span><\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> is the difference between the proportions of men and women who enjoy shopping for electronic equipment.<\/li>\r\n \t
                                      4. normal for two proportions<\/li>\r\n \t
                                      5. test statistic: 0.22<\/li>\r\n \t
                                      6. p<\/em>-value: 0.4133<\/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: Do not reject the null hypothesis.<\/li>\r\n \t
                                        3. Reason for Decision: p<\/em>-value > alpha<\/li>\r\n \t
                                        4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the proportion of men who enjoy shopping for electronic equipment is more than the proportion of women.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n95.\r\n\r\n
                                          \r\n
                                            \r\n \t
                                          1. H0<\/sub><\/em>: p1<\/sub><\/em> = p2<\/sub><\/em><\/li>\r\n \t
                                          2. Ha<\/sub><\/em>: p1<\/sub><\/em> \u2260 p2<\/sub><\/em><\/li>\r\n \t
                                          3. P<\/span>\u2032<\/span><\/span>1<\/span><\/span>\u2212<\/span>P<\/span>\u2032<\/span><\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> is the difference between the proportions of men and women that have at least one pierced ear.<\/li>\r\n \t
                                          4. normal for two proportions<\/li>\r\n \t
                                          5. test statistic: \u20134.82<\/li>\r\n \t
                                          6. p<\/em>-value: zero<\/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: p<\/em>-value < alpha<\/li>\r\n \t
                                            4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportions of males and females with at least one pierced ear is different.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n97.\r\n\r\n
                                              \r\n
                                                \r\n \t
                                              1. H0<\/sub><\/em>: \u00b5d<\/sub><\/em> = 0<\/li>\r\n \t
                                              2. Ha<\/sub><\/em>: \u00b5d<\/sub><\/em> > 0<\/li>\r\n \t
                                              3. The random variable Xd<\/sub><\/em> is the mean difference in work times on days when eating breakfast and on days when not eating breakfast.<\/li>\r\n \t
                                              4. t<\/em>9<\/sub><\/li>\r\n \t
                                              5. test statistic: 4.8963<\/li>\r\n \t
                                              6. p<\/em>-value: 0.0004<\/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: p<\/em>-value < alpha<\/li>\r\n \t
                                                4. Conclusion: At the 5% level of significance, there is sufficient evidence to conclude that the mean difference in work times on days when eating breakfast and on days when not eating breakfast has increased.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n98.\u00a0the mean difference of the system failures\r\n\r\n100.\u00a00.0067\r\n\r\n102.\u00a0With a p<\/em>-value 0.0067, we can reject the null hypothesis. There is enough evidence to support that the software patch is effective in reducing the number of system failures.\r\n\r\n104.\u00a00.0021\r\n\r\n106.\r\n
                                                  \r\n\r\nthe graph of the p<\/em>-value.\r\n\r\n<\/div>\r\n
                                                  <\/div>\r\n
                                                  <\/div>\r\n
                                                  \r\n
                                                  \"This<\/span><\/figure>\r\n108.\r\n\r\n
                                                  \r\n

                                                  H0<\/sub><\/em>: \u03bcd<\/sub><\/em> \u2265 0<\/p>\r\n

                                                  Ha<\/sub><\/em>: \u03bcd<\/sub><\/em> < 0<\/p>\r\n110.\u00a00.0699\r\n\r\n112.\u00a0We decline to reject the null hypothesis. There is not sufficient evidence to support that the medication is effective.\r\n\r\n113.\r\n\r\n

                                                  \r\n

                                                  p<\/em>-value = 0.1494<\/p>\r\n

                                                  At the 5% significance level, there is insufficient evidence to conclude that the medication lowered cholesterol levels after 12 weeks.<\/p>\r\n115.\u00a0There is sufficient evidence to conclude that the method reduces the proportion of HIV positive patients who develop AIDS after four years.\r\n\r\n117.\u00a00.0155; There is sufficient evidence to conclude that the blood pressure decreased after the training.\r\n\r\n119.\r\n\r\n

                                                  \r\n

                                                  Test: two matched pairs or paired samples (t<\/em>-test)<\/p>\r\n

                                                  Random variable:\u00a0[latex]\\overline{X}_{d}[\/latex]<\/p>\r\n

                                                  Distribution: t<\/em>12<\/sub><\/p>\r\n

                                                  H0<\/sub><\/em>: \u03bc<\/em>d<\/sub><\/em> = 0 Ha<\/sub><\/em>: \u03bc<\/em>d<\/sub><\/em> > 0<\/p>\r\n

                                                  The mean of the differences of new female breast cancer cases in the south between 2013 and 2012 is greater than zero. The estimate for new female breast cancer cases in the south is higher in 2013 than in 2012.<\/p>\r\n

                                                  Graph: right-tailed<\/p>\r\n

                                                  p<\/em>-value: 0.0004<\/p>\r\n\r\n

                                                  \"This<\/span><\/figure>\r\n

                                                  Decision: Reject H0<\/sub><\/em><\/p>\r\n

                                                  Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that there was a higher estimate of new female breast cancer cases in 2013 than in 2012.<\/p>\r\n\r\n<\/section>121.\r\n\r\n

                                                  \r\n

                                                  Test: matched or paired samples (t<\/em>-test)<\/p>\r\n

                                                  Difference data: {\u20130.9, \u20133.7, \u20133.2, \u20130.5, 0.6, \u20131.9, \u20130.5, 0.2, 0.6, 0.4, 1.7, \u20132.4, 1.8}<\/p>\r\n

                                                  Random Variable:[latex]\\overline{X}_{d}[\/latex]<\/p>\r\n

                                                  Distribution: H0<\/sub><\/em>: \u03bcd<\/sub><\/em> = 0 Ha<\/sub><\/em>: \u03bcd<\/sub><\/em> < 0<\/p>\r\n

                                                  The mean of the differences of the rate of underemployment in the northeastern states between 2012 and 2011 is less than zero. The underemployment rate went down from 2011 to 2012.<\/p>\r\n

                                                  Graph: left-tailed.<\/p>\r\n\r\n

                                                  \"This<\/span><\/figure>\r\n

                                                  p<\/em>-value: 0.1207<\/p>\r\n

                                                  Decision: Do not reject H0<\/sub><\/em>.<\/p>\r\n

                                                  Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that there was a decrease in the underemployment rates of the northeastern states from 2011 to 2012.<\/p>\r\n123.\u00a0two proportions\r\n\r\n125.\u00a0single mean\r\n\r\n127.\u00a0single proportion\r\n\r\n129.\u00a0two proportions\r\n\r\n131.\u00a0single proportion\r\n\r\n133.\u00a0a test of two independent means.\r\n\r\n<\/section><\/section><\/section><\/section><\/section><\/section><\/section><\/section><\/section>\r\n

                                                  <\/h2>\r\n
                                                  <\/div>","rendered":"

                                                  Two Population Means with Unknown Standard Deviations<\/h2>\n

                                                  1.\u00a0two proportions<\/p>\n

                                                  3.\u00a0matched or paired samples<\/p>\n

                                                  6. independent group means, population standard deviations and\/or variances unknown<\/p>\n

                                                  8.\u00a0two proportions<\/p>\n

                                                  10.\u00a0independent group means, population standard deviations and\/or variances unknown<\/p>\n

                                                  11.\u00a0independent group means, population standard deviations and\/or variances unknown<\/p>\n

                                                  13.\u00a0two proportions<\/p>\n

                                                  15.\u00a0The random variable is the difference between the mean amounts of sugar in the two soft drinks.<\/p>\n

                                                  17.\u00a0means<\/p>\n

                                                  19.\u00a0two-tailed<\/p>\n

                                                  21.\u00a0the difference between the mean life spans of whites and nonwhites<\/p>\n

                                                  23.\u00a0This is a comparison of two population means with unknown population standard deviations.<\/p>\n

                                                  27.<\/p>\n

                                                  \n
                                                    \n
                                                  1. Reject the null hypothesis<\/li>\n
                                                  2. p<\/em>-value < 0.05<\/li>\n
                                                  3. There is not enough evidence at the 5% level of significance to support the claim that life expectancy in the 1900s is different between whites and nonwhites.<\/li>\n<\/ol>\n

                                                    30.<\/p>\n

                                                      \n
                                                    1. H0<\/sub><\/em>: \u03bc1<\/sub><\/em> \u2265 \u03bc2<\/sub><\/em><\/li>\n
                                                    2. Ha<\/sub><\/em>: \u03bc1<\/sub><\/em> < \u03bc2<\/sub><\/em><\/li>\n
                                                    3. [latex]\\overline{X}_{1}\u2013\\overline{X}_{2}[\/latex]is the difference between the mean enrollments of the two-year colleges and the four-year colleges.<\/li>\n
                                                    4. Student\u2019s-t<\/em><\/li>\n
                                                    5. test statistic: -0.2480<\/li>\n
                                                    6. p<\/em>-value: 0.4019<\/li>\n
                                                    7. Check student\u2019s solution.<\/li>\n
                                                    8. \n
                                                        \n
                                                      1. Alpha: 0.05<\/li>\n
                                                      2. Decision: Do not reject<\/li>\n
                                                      3. Reason for Decision: p<\/em>-value > alpha<\/li>\n
                                                      4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean enrollment at four-year colleges is higher than at two-year colleges.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                                                        32.<\/p>\n

                                                        \n

                                                        Subscripts: 1: mechanical engineering; 2: electrical engineering<\/p>\n

                                                          \n
                                                        1. H0<\/sub><\/em>: \u00b51<\/sub><\/em> \u2265 \u00b52<\/sub><\/em><\/li>\n
                                                        2. Ha<\/sub><\/em>: \u00b51<\/sub><\/em> < \u00b52<\/sub><\/em><\/li>\n
                                                        3. [latex]\\overline{X}_{1}\u2013\\overline{X}_{2}[\/latex] is the difference between the mean entry level salaries of mechanical engineers and electrical engineers.<\/li>\n
                                                        4. t<\/em>108<\/sub><\/li>\n
                                                        5. test statistic: t<\/em> = \u20130.82<\/li>\n
                                                        6. p<\/em>-value: 0.2061<\/li>\n
                                                        7. Check student\u2019s solution.<\/li>\n
                                                        8. \n
                                                            \n
                                                          1. Alpha: 0.05<\/li>\n
                                                          2. Decision: Do not reject the null hypothesis.<\/li>\n
                                                          3. Reason for Decision: p<\/em>-value > alpha<\/li>\n
                                                          4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean entry-level salaries of mechanical engineers is lower than that of electrical engineers.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/section>\n

                                                            34.<\/p>\n

                                                            \n
                                                              \n
                                                            1. H0<\/sub><\/em>: \u00b51<\/sub><\/em> = \u00b52<\/sub><\/em><\/li>\n
                                                            2. Ha<\/sub><\/em>: \u00b51<\/sub><\/em> \u2260 \u00b52<\/sub><\/em><\/li>\n
                                                            3. [latex]\\overline{X}_{1}\u2013\\overline{X}_{2}[\/latex] is the difference between the mean times for completing a lap in races and in practices.<\/li>\n
                                                            4. t<\/em>20.32<\/sub><\/li>\n
                                                            5. test statistic: \u20134.70<\/li>\n
                                                            6. p<\/em>-value: 0.0001<\/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: p<\/em>-value < alpha<\/li>\n
                                                              4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                                                                36.<\/p>\n

                                                                \n
                                                                  \n
                                                                1. H0<\/sub><\/em>: \u00b51<\/sub><\/em> = \u00b52<\/sub><\/em><\/li>\n
                                                                2. Ha<\/sub><\/em>: \u00b51<\/sub><\/em> \u2260 \u00b52<\/sub><\/em><\/li>\n
                                                                3. [latex]\\overline{X}_{1}\u2013\\overline{X}_{2}[\/latex] \u00a0is the difference between the mean times for completing a lap in races and in practices.<\/li>\n
                                                                4. t<\/em>40.94<\/sub><\/li>\n
                                                                5. test statistic: \u20135.08<\/li>\n
                                                                6. p<\/em>-value: zero<\/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: p<\/em>-value < alpha<\/li>\n
                                                                  4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                                                                    39.\u00a0There is insufficient evidence to conclude that the W<\/strong> teams score fewer goals, on average, than the E<\/strong> teams score.<\/p>\n<\/section>\n<\/section>\n<\/section>\n

                                                                    41.Test: two independent sample means, population standard deviations unknown.Random variable:[latex]\\overline{X}_{1}\u2013\\overline{X}_{2}[\/latex]Distribution: H0<\/sub><\/em>: \u03bc<\/em>1<\/sub> = \u03bc<\/em>2<\/sub>Ha<\/sub><\/em>: \u03bc<\/em>1<\/sub> < \u03bc<\/em>2<\/sub> The mean age of entering prostitution in Canada is lower than the mean age in the United States.<\/p>\n

                                                                    \"This<\/span><\/figure>\n

                                                                    Graph: left-tailed<\/p>\n

                                                                    p<\/em>-value : 0.0151<\/p>\n

                                                                    Decision: Do not reject H<\/em>0<\/sub>.<\/p>\n

                                                                    Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of entering prostitution in Canada is lower than the mean age in the United States.<\/p>\n

                                                                    43.\u00a0\u03bc<\/em>day<\/sub> \u2260 \u03bc<\/em>night<\/sub><\/p>\n

                                                                    Two Population Means with Known Standard Deviations<\/h2>\n
                                                                    44.\u00a0The difference in mean speeds of the fastball pitches of the two pitchers<\/div>\n
                                                                    <\/div>\n
                                                                    46.\u00a0\u20132.46<\/div>\n
                                                                    <\/div>\n
                                                                    48.\u00a0At the 1% significance level, we can reject the null hypothesis. There is sufficient data to conclude that the mean speed of Rodriguez\u2019s fastball is faster than Wesley\u2019s.<\/div>\n
                                                                    <\/div>\n
                                                                    \n

                                                                    50.<\/p>\n

                                                                    \n

                                                                    Subscripts: 1 = Food, 2 = No Food<\/p>\n

                                                                    <\/div>\n

                                                                    H0<\/sub><\/em>: \u03bc1<\/sub><\/em> \u2264 \u03bc2<\/sub><\/em><\/p>\n

                                                                    <\/div>\n

                                                                    Ha<\/sub><\/em>: \u03bc1<\/sub><\/em> > \u03bc2<\/sub><\/em><\/section>\n<\/div>\n

                                                                    <\/section>\n
                                                                    52.<\/section>\n
                                                                    \n

                                                                    the graph of the p<\/em>-value.<\/p>\n<\/div>\n

                                                                    <\/div>\n
                                                                    <\/div>\n
                                                                    \n
                                                                    \"This<\/span><\/figure>\n<\/section>\n

                                                                    54.<\/p>\n

                                                                    Subscripts: 1 = Gamma, 2 = Zeta<\/p>\n

                                                                    <\/div>\n

                                                                    H0<\/sub><\/em>: \u03bc<\/em>1<\/sub> = \u03bc<\/em>2<\/sub><\/p>\n

                                                                    <\/div>\n

                                                                    Ha<\/sub><\/em>: \u03bc<\/em>1<\/sub> \u2260 \u03bc<\/em>2<\/sub><\/p>\n

                                                                    56.\u00a00.0062<\/p>\n

                                                                    58.\u00a0There is sufficient evidence to reject the null hypothesis. The data support that the melting point for Alloy Zeta is different from the melting point of Alloy Gamma.<\/p>\n

                                                                    60.<\/p>\n

                                                                    \n

                                                                    Subscripts: 1 = boys, 2 = girls<\/p>\n

                                                                      \n
                                                                    1. H0<\/sub><\/em>: \u00b51<\/sub><\/em> \u2264 \u00b52<\/sub><\/em><\/li>\n
                                                                    2. Ha<\/sub><\/em>: \u00b51<\/sub><\/em> > \u00b52<\/sub><\/em><\/li>\n
                                                                    3. The random variable is the difference in the mean auto insurance costs for boys and girls.<\/li>\n
                                                                    4. normal<\/li>\n
                                                                    5. test statistic: z<\/em> = 2.50<\/li>\n
                                                                    6. p<\/em>-value: 0.0062<\/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: p<\/em>-value < alpha<\/li>\n
                                                                      4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean cost of auto insurance for teenage boys is greater than that for girls.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                                                                        62.<\/p>\n

                                                                        \n

                                                                        Subscripts: 1 = non-hybrid sedans, 2 = hybrid sedans<\/p>\n

                                                                          \n
                                                                        1. H0<\/sub><\/em>: \u00b51<\/sub><\/em> \u2265 \u00b52<\/sub><\/em><\/li>\n
                                                                        2. Ha<\/sub><\/em>: \u00b51<\/sub><\/em> < \u00b52<\/sub><\/em><\/li>\n
                                                                        3. The random variable is the difference in the mean miles per gallon of non-hybrid sedans and hybrid sedans.<\/li>\n
                                                                        4. normal<\/li>\n
                                                                        5. test statistic: 6.36<\/li>\n
                                                                        6. p<\/em>-value: 0<\/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: p<\/em>-value < alpha<\/li>\n
                                                                          4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean miles per gallon of non-hybrid sedans is less than that of hybrid sedans.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                                                                            64.<\/p>\n

                                                                            \n
                                                                              \n
                                                                            1. H0<\/sub><\/em>: \u00b5d<\/sub><\/em> = 0<\/li>\n
                                                                            2. Ha<\/sub><\/em>: \u00b5d<\/sub><\/em> < 0<\/li>\n
                                                                            3. The random variable Xd<\/sub><\/em> is the average difference between husband\u2019s and wife\u2019s satisfaction level.<\/li>\n
                                                                            4. t<\/em>9<\/sub><\/li>\n
                                                                            5. test statistic: t<\/em> = \u20131.86<\/li>\n
                                                                            6. p<\/em>-value: 0.0479<\/li>\n
                                                                            7. Check student\u2019s solution<\/li>\n
                                                                            8. \n
                                                                                \n
                                                                              1. Alpha: 0.05<\/li>\n
                                                                              2. Decision: Reject the null hypothesis, but run another test.<\/li>\n
                                                                              3. Reason for Decision: p<\/em>-value < alpha<\/li>\n
                                                                              4. Conclusion: This is a weak test because alpha and the p<\/em>-value are close. However, there is insufficient evidence to conclude that the mean difference is negative.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/section>\n<\/section>\n<\/section>\n

                                                                                Comparing Two Independent Population Proportions<\/h2>\n

                                                                                66.\u00a0P<\/em>\u2032OS1<\/sub> \u2013 P<\/em>\u2032OS2<\/sub> = difference in the proportions of phones that had system failures within the first eight hours of operation with OS1<\/sub> and OS2<\/sub>.<\/p>\n

                                                                                68.\u00a00.1018<\/p>\n

                                                                                70.\u00a0proportions<\/p>\n

                                                                                72.\u00a0right-tailed<\/p>\n

                                                                                74.\u00a0The random variable is the difference in proportions (percents) of the populations that are of two or more races in Nevada and North Dakota.<\/p>\n

                                                                                76.\u00a0Our sample sizes are much greater than five each, so we use the normal for two proportions distribution for this hypothesis test.<\/p>\n

                                                                                80.<\/p>\n

                                                                                Reject the null hypothesis.p<\/em>-value < alphaAt the 5% significance level, there is sufficient evidence to conclude that the proportion (percent) of the population that is of two or more races in Nevada is statistically higher than that in North Dakota.83.<\/p>\n
                                                                                H0<\/sub><\/em>: PW<\/sub><\/em> = PB<\/sub><\/em>Ha<\/sub><\/em>: PW<\/sub><\/em> \u2260 PB<\/sub><\/em>The random variable is the difference in the proportions of white and black suicide victims, aged 15 to 24.normal for two proportionstest statistic: \u20130.1944<\/p>\n

                                                                                p<\/em>-value: 0.8458<\/p>\n

                                                                                Check student\u2019s solution.<\/p>\n

                                                                                  \n
                                                                                1. Alpha: 0.05<\/li>\n
                                                                                2. Decision: Reject the null hypothesis.<\/li>\n
                                                                                3. Reason for decision: p<\/em>-value > alpha<\/li>\n
                                                                                4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the proportions of white and black female suicide victims, aged 15 to 24, are different.<\/li>\n<\/ol>\n

                                                                                  85.<\/p>\n

                                                                                  Subscripts: 1 = Cabrillo College, 2 = Lake Tahoe College<\/p>\n

                                                                                    \n
                                                                                  1. H0<\/sub><\/em>: p1<\/sub><\/em> = p2<\/sub><\/em><\/li>\n
                                                                                  2. Ha<\/sub><\/em>: p1<\/sub><\/em> \u2260 p2<\/sub><\/em><\/li>\n
                                                                                  3. The random variable is the difference between the proportions of Hispanic students at Cabrillo College and Lake Tahoe College.<\/li>\n
                                                                                  4. normal for two proportions<\/li>\n
                                                                                  5. test statistic: 4.29<\/li>\n
                                                                                  6. p<\/em>-value: 0.00002<\/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: p<\/em>-value < alpha<\/li>\n
                                                                                    4. Conclusion: There is sufficient evidence to conclude that the proportions of Hispanic students at Cabrillo College and Lake Tahoe College are different.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                                                                                      87. \u00a0\u00a0p2011<\/sub><\/em> \u2264 p2010<\/sub><\/em><\/p>\n

                                                                                      89.<\/p>\n

                                                                                      \n

                                                                                      Test: two independent sample proportions.<\/p>\n

                                                                                      Random variable: p<\/em>\u20321<\/sub> – p<\/em>\u20322<\/sub><\/p>\n

                                                                                      Distribution:<\/p>\n

                                                                                      <\/div>\n

                                                                                      H0<\/sub><\/em>: p1<\/sub><\/em> = p2<\/sub><\/em><\/p>\n

                                                                                      <\/div>\n

                                                                                      Ha<\/sub><\/em>: p1<\/sub><\/em> \u2260 p2<\/sub><\/em><\/p>\n

                                                                                      The proportion of eReader users is different for the 16- to 29-year-old users from that of the 30 and older users.<\/p>\n

                                                                                      Graph: two-tailed<\/p>\n

                                                                                      \"This<\/span><\/figure>\n

                                                                                      p<\/em>-value : 0.0033<\/p>\n

                                                                                      Decision: Reject the null hypothesis.<\/p>\n

                                                                                      Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that the proportion of eReader users 16 to 29 years old is different from the proportion of eReader users 30 and older.<\/p>\n<\/section>\n

                                                                                      91.<\/p>\n

                                                                                      \n

                                                                                      Test: two independent sample proportions<\/p>\n

                                                                                      Random variable: p\u20321<\/sub><\/em> \u2212 p\u20322<\/sub><\/em><\/p>\n

                                                                                      Distribution:<\/p>\n

                                                                                      H0<\/sub><\/em>: p1<\/sub><\/em> = p2<\/sub><\/em><\/p>\n

                                                                                      <\/div>\n

                                                                                      Ha<\/sub><\/em>: p1<\/sub><\/em> > p2<\/sub><\/em><\/p>\n

                                                                                      A higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older.<\/p>\n

                                                                                      Graph: right-tailed<\/p>\n

                                                                                      \"This<\/span><\/figure>\n

                                                                                      p<\/em>-value: 0.2354<\/p>\n

                                                                                      Decision: Do not reject the H0<\/sub><\/em>.<\/p>\n

                                                                                      Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older.<\/p>\n<\/section>\n

                                                                                      93.<\/p>\n

                                                                                      \n

                                                                                      Subscripts: 1: men; 2: women<\/p>\n

                                                                                        \n
                                                                                      1. H0<\/sub><\/em>: p1<\/sub><\/em> \u2264 p2<\/sub><\/em><\/li>\n
                                                                                      2. Ha<\/sub><\/em>: p1<\/sub><\/em> > p2<\/sub><\/em><\/li>\n
                                                                                      3. P<\/span>\u2032<\/span><\/span>1<\/span><\/span>\u2212<\/span>P<\/span>\u2032<\/span><\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> is the difference between the proportions of men and women who enjoy shopping for electronic equipment.<\/li>\n
                                                                                      4. normal for two proportions<\/li>\n
                                                                                      5. test statistic: 0.22<\/li>\n
                                                                                      6. p<\/em>-value: 0.4133<\/li>\n
                                                                                      7. Check student\u2019s solution.<\/li>\n
                                                                                      8. \n
                                                                                          \n
                                                                                        1. Alpha: 0.05<\/li>\n
                                                                                        2. Decision: Do not reject the null hypothesis.<\/li>\n
                                                                                        3. Reason for Decision: p<\/em>-value > alpha<\/li>\n
                                                                                        4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the proportion of men who enjoy shopping for electronic equipment is more than the proportion of women.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                                                                                          95.<\/p>\n

                                                                                          \n
                                                                                            \n
                                                                                          1. H0<\/sub><\/em>: p1<\/sub><\/em> = p2<\/sub><\/em><\/li>\n
                                                                                          2. Ha<\/sub><\/em>: p1<\/sub><\/em> \u2260 p2<\/sub><\/em><\/li>\n
                                                                                          3. P<\/span>\u2032<\/span><\/span>1<\/span><\/span>\u2212<\/span>P<\/span>\u2032<\/span><\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> is the difference between the proportions of men and women that have at least one pierced ear.<\/li>\n
                                                                                          4. normal for two proportions<\/li>\n
                                                                                          5. test statistic: \u20134.82<\/li>\n
                                                                                          6. p<\/em>-value: zero<\/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: p<\/em>-value < alpha<\/li>\n
                                                                                            4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportions of males and females with at least one pierced ear is different.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                                                                                              97.<\/p>\n

                                                                                              \n
                                                                                                \n
                                                                                              1. H0<\/sub><\/em>: \u00b5d<\/sub><\/em> = 0<\/li>\n
                                                                                              2. Ha<\/sub><\/em>: \u00b5d<\/sub><\/em> > 0<\/li>\n
                                                                                              3. The random variable Xd<\/sub><\/em> is the mean difference in work times on days when eating breakfast and on days when not eating breakfast.<\/li>\n
                                                                                              4. t<\/em>9<\/sub><\/li>\n
                                                                                              5. test statistic: 4.8963<\/li>\n
                                                                                              6. p<\/em>-value: 0.0004<\/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: p<\/em>-value < alpha<\/li>\n
                                                                                                4. Conclusion: At the 5% level of significance, there is sufficient evidence to conclude that the mean difference in work times on days when eating breakfast and on days when not eating breakfast has increased.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                                                                                                  98.\u00a0the mean difference of the system failures<\/p>\n

                                                                                                  100.\u00a00.0067<\/p>\n

                                                                                                  102.\u00a0With a p<\/em>-value 0.0067, we can reject the null hypothesis. There is enough evidence to support that the software patch is effective in reducing the number of system failures.<\/p>\n

                                                                                                  104.\u00a00.0021<\/p>\n

                                                                                                  106.<\/p>\n

                                                                                                  \n

                                                                                                  the graph of the p<\/em>-value.<\/p>\n<\/div>\n

                                                                                                  <\/div>\n
                                                                                                  <\/div>\n
                                                                                                  \n
                                                                                                  \"This<\/span><\/figure>\n

                                                                                                  108.<\/p>\n

                                                                                                  \n

                                                                                                  H0<\/sub><\/em>: \u03bcd<\/sub><\/em> \u2265 0<\/p>\n

                                                                                                  Ha<\/sub><\/em>: \u03bcd<\/sub><\/em> < 0<\/p>\n

                                                                                                  110.\u00a00.0699<\/p>\n

                                                                                                  112.\u00a0We decline to reject the null hypothesis. There is not sufficient evidence to support that the medication is effective.<\/p>\n

                                                                                                  113.<\/p>\n

                                                                                                  \n

                                                                                                  p<\/em>-value = 0.1494<\/p>\n

                                                                                                  At the 5% significance level, there is insufficient evidence to conclude that the medication lowered cholesterol levels after 12 weeks.<\/p>\n

                                                                                                  115.\u00a0There is sufficient evidence to conclude that the method reduces the proportion of HIV positive patients who develop AIDS after four years.<\/p>\n

                                                                                                  117.\u00a00.0155; There is sufficient evidence to conclude that the blood pressure decreased after the training.<\/p>\n

                                                                                                  119.<\/p>\n

                                                                                                  \n

                                                                                                  Test: two matched pairs or paired samples (t<\/em>-test)<\/p>\n

                                                                                                  Random variable:\u00a0[latex]\\overline{X}_{d}[\/latex]<\/p>\n

                                                                                                  Distribution: t<\/em>12<\/sub><\/p>\n

                                                                                                  H0<\/sub><\/em>: \u03bc<\/em>d<\/sub><\/em> = 0 Ha<\/sub><\/em>: \u03bc<\/em>d<\/sub><\/em> > 0<\/p>\n

                                                                                                  The mean of the differences of new female breast cancer cases in the south between 2013 and 2012 is greater than zero. The estimate for new female breast cancer cases in the south is higher in 2013 than in 2012.<\/p>\n

                                                                                                  Graph: right-tailed<\/p>\n

                                                                                                  p<\/em>-value: 0.0004<\/p>\n

                                                                                                  \"This<\/span><\/figure>\n

                                                                                                  Decision: Reject H0<\/sub><\/em><\/p>\n

                                                                                                  Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that there was a higher estimate of new female breast cancer cases in 2013 than in 2012.<\/p>\n<\/section>\n

                                                                                                  121.<\/p>\n

                                                                                                  \n

                                                                                                  Test: matched or paired samples (t<\/em>-test)<\/p>\n

                                                                                                  Difference data: {\u20130.9, \u20133.7, \u20133.2, \u20130.5, 0.6, \u20131.9, \u20130.5, 0.2, 0.6, 0.4, 1.7, \u20132.4, 1.8}<\/p>\n

                                                                                                  Random Variable:[latex]\\overline{X}_{d}[\/latex]<\/p>\n

                                                                                                  Distribution: H0<\/sub><\/em>: \u03bcd<\/sub><\/em> = 0 Ha<\/sub><\/em>: \u03bcd<\/sub><\/em> < 0<\/p>\n

                                                                                                  The mean of the differences of the rate of underemployment in the northeastern states between 2012 and 2011 is less than zero. The underemployment rate went down from 2011 to 2012.<\/p>\n

                                                                                                  Graph: left-tailed.<\/p>\n

                                                                                                  \"This<\/span><\/figure>\n

                                                                                                  p<\/em>-value: 0.1207<\/p>\n

                                                                                                  Decision: Do not reject H0<\/sub><\/em>.<\/p>\n

                                                                                                  Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that there was a decrease in the underemployment rates of the northeastern states from 2011 to 2012.<\/p>\n

                                                                                                  123.\u00a0two proportions<\/p>\n

                                                                                                  125.\u00a0single mean<\/p>\n

                                                                                                  127.\u00a0single proportion<\/p>\n

                                                                                                  129.\u00a0two proportions<\/p>\n

                                                                                                  131.\u00a0single proportion<\/p>\n

                                                                                                  133.\u00a0a test of two independent means.<\/p>\n<\/section>\n<\/section>\n<\/section>\n<\/section>\n<\/section>\n<\/section>\n<\/section>\n<\/section>\n<\/section>\n

                                                                                                  <\/h2>\n
                                                                                                  <\/div>\n\n\t\t\t
                                                                                                  \n\t\t\t

                                                                                                  Candela Citations<\/h3>\n\t\t\t\t\t
                                                                                                  \n\t\t\t\t\t\t
                                                                                                  \n\t\t\t\t\t\t\t
                                                                                                  CC licensed content, Shared previously<\/div>