{"id":302,"date":"2021-07-14T15:59:10","date_gmt":"2021-07-14T15:59:10","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/answers-to-selected-exercises-4\/"},"modified":"2023-12-05T09:44:30","modified_gmt":"2023-12-05T09:44:30","slug":"answers-to-selected-exercises-4","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/answers-to-selected-exercises-4\/","title":{"raw":"Answers to Selected Exercises","rendered":"Answers to Selected Exercises"},"content":{"raw":"

Facts About the Chi-Square Distribution - Practice<\/h2>\r\n1.\u00a0mean = 25 and standard deviation = 7.0711\r\n\r\n3.\u00a0when the number of degrees of freedom is greater than 90\r\n\r\n5.\u00a0df<\/em> = 2\r\n

Goodness-of-Fit Test - Practice<\/h2>\r\n7.\u00a0a goodness-of-fit test\r\n\r\n9. 3\r\n\r\n11.\u00a02.04\r\n\r\n13.\u00a0We decline to reject the null hypothesis. There is not enough evidence to suggest that the observed test scores are significantly different from the expected test scores.\r\n\r\n15.\u00a0H0<\/sub><\/em>: the distribution of AIDS cases follows the ethnicities of the general population of Santa Clara County.\r\n\r\n17. right-tailed\r\n\r\n19. 2016.136\r\n\r\n21.\u00a0Graph: Check student\u2019s solution.\u00a0<\/span>Decision: Reject the null hypothesis.\u00a0<\/span>Reason for the Decision: <\/span>p<\/em>-value < alpha\u00a0<\/span>Conclusion (write out in complete sentences): The make-up of AIDS cases does not fit the ethnicities of the general population of Santa Clara County.<\/span>\r\n

Test of Independence \u2013 Practice<\/h2>\r\n23.\u00a0a test of independence\r\n\r\n25.\u00a0a test of independence\r\n\r\n27. 8\r\n\r\n29. 6.6\r\n\r\n31.\u00a00.0435\r\n\r\n33.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Smoking Level Per Day<\/th>\r\nAfrican American<\/th>\r\nNative Hawaiian<\/th>\r\nLatino<\/th>\r\nJapanese Americans<\/th>\r\nWhite<\/th>\r\nTotals<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
1-10<\/td>\r\n9,886<\/td>\r\n2,745<\/td>\r\n12,831<\/td>\r\n8,378<\/td>\r\n7,650<\/td>\r\n41,490<\/td>\r\n<\/tr>\r\n
11-20<\/td>\r\n6,514<\/td>\r\n3,062<\/td>\r\n4,932<\/td>\r\n10,680<\/td>\r\n9,877<\/td>\r\n35,065<\/td>\r\n<\/tr>\r\n
21-30<\/td>\r\n1,671<\/td>\r\n1,419<\/td>\r\n1,406<\/td>\r\n4,715<\/td>\r\n6,062<\/td>\r\n15,273<\/td>\r\n<\/tr>\r\n
31+<\/td>\r\n759<\/td>\r\n788<\/td>\r\n800<\/td>\r\n2,305<\/td>\r\n3,970<\/td>\r\n8,622<\/td>\r\n<\/tr>\r\n
Totals<\/td>\r\n18,830<\/td>\r\n8,014<\/td>\r\n19,969<\/td>\r\n26,078<\/td>\r\n27,559<\/td>\r\n10,0450\r\n\r\n \r\n\r\n <\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n35.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Smoking Level Per Day<\/th>\r\nAfrican American<\/th>\r\nNative Hawaiian<\/th>\r\nLatino<\/th>\r\nJapanese Americans<\/th>\r\nWhite<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
1-10<\/td>\r\n7777.57<\/td>\r\n3310.11<\/td>\r\n8248.02<\/td>\r\n10771.29<\/td>\r\n11383.01<\/td>\r\n<\/tr>\r\n
11-20<\/td>\r\n6573.16<\/td>\r\n2797.52<\/td>\r\n6970.76<\/td>\r\n9103.29<\/td>\r\n9620.27<\/td>\r\n<\/tr>\r\n
21-30<\/td>\r\n2863.02<\/td>\r\n1218.49<\/td>\r\n3036.20<\/td>\r\n3965.05<\/td>\r\n4190.23<\/td>\r\n<\/tr>\r\n
31+<\/td>\r\n1616.25<\/td>\r\n687.87<\/td>\r\n1714.01<\/td>\r\n2238.37<\/td>\r\n2365.49<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n37.\u00a010,301.8\r\n\r\n39.\u00a0right\r\n\r\n41.\r\n\r\n
\r\n
    \r\n \t
  1. Reject the null hypothesis.<\/li>\r\n \t
  2. p<\/em>-value < alpha<\/li>\r\n \t
  3. There is sufficient evidence to conclude that smoking level is dependent on ethnic group.<\/li>\r\n<\/ol>\r\n<\/section>\r\n

    Test for Homogeneity \u2013 Practice<\/h2>\r\n
    \r\n\r\n43.\u00a0test for homogeneity\r\n\r\n45.\u00a0test for homogeneity\r\n\r\n47.\u00a0All values in the table must be greater than or equal to five.\r\n\r\n49. 3\r\n\r\n51.\u00a00.00005\r\n\r\n<\/div>\r\n
    \r\n

    Comparison of the Chi-Square Tests \u2013 Practice<\/h2>\r\n<\/div>\r\n
    \r\n\r\n53.\u00a0a goodness-of-fit test\r\n\r\n55.\u00a0a test for independence\r\n\r\n57.\u00a0Answers will vary. Sample answer: Tests of independence and tests for homogeneity both calculate the test statistic the same way, [latex]\\sum{ }_{(ij)}\\frac{{{({O}-{E})}^{2}}}{{E}}[\/latex] In addition, all values must be greater than or equal to five.\r\n\r\n<\/div>\r\n
    \r\n

    Test of a Single Variance \u2013 Practice<\/h2>\r\n
    \r\n
    \r\n\r\n59.\u00a0a test of a single variance\r\n\r\n61.\u00a0a left-tailed test\r\n\r\n63.\r\n
      \r\n \t
    • H0<\/sub><\/em>: \u03c3<\/em>2<\/sup> = 0.812<\/sup>;<\/li>\r\n \t
    • Ha<\/sub><\/em>: <\/span>\u03c3<\/em>2<\/sup> > 0.81<\/span>2<\/sup><\/li>\r\n<\/ul>\r\n65. a test of a single variance\r\n\r\n67. 0.0542\r\n

      Facts About the Chi-Square Distribution \u2013 Homework<\/h2>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n69.\u00a0true\r\n\r\n71.\u00a0false\r\n

      Goodness-of-Fit Test - Homework<\/h2>\r\n
      73.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      Marital Status<\/th>\r\nPercent<\/th>\r\nExpected Frequency<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
      never married<\/td>\r\n31.3<\/td>\r\n125.2<\/td>\r\n<\/tr>\r\n
      married<\/td>\r\n56.1<\/td>\r\n224.4<\/td>\r\n<\/tr>\r\n
      widowed<\/td>\r\n2.5<\/td>\r\n10<\/td>\r\n<\/tr>\r\n
      divorced\/separated<\/td>\r\n10.1<\/td>\r\n40.4<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
        \r\n \t
      1. The data fit the distribution.<\/li>\r\n \t
      2. The data does not fit the distribution.<\/li>\r\n \t
      3. 3<\/li>\r\n \t
      4. chi-square distribution with df<\/em> = 3<\/li>\r\n \t
      5. 19.27<\/li>\r\n \t
      6. 0.0002<\/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 null<\/li>\r\n \t
        3. Reason for decision: p<\/em>-value < alpha<\/li>\r\n \t
        4. Conclusion: Data does not fit the distribution.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n75.\r\nH0<\/sub><\/em>: The local results follow the distribution of the U.S. AP examinee population\r\nHa<\/sub><\/em>: The local results do not follow the distribution of the U.S. AP examinee population\r\ndf<\/em> = 5\r\nchi-square distribution with df<\/em> = 5\r\nchi-square test statistic = 13.4\r\np<\/em>-value = 0.0199\r\nCheck student\u2019s solution.\r\n
          \r\n
          Alpha = 0.05<\/div>\r\n
          Decision: Reject null when a<\/em> = 0.05<\/div>\r\n
          Reason for Decision: p<\/em>-value < alpha<\/div>\r\n
          Conclusion: Local data do not fit the AP Examinee Distribution.<\/div>\r\n
          Decision: Do not reject null when a<\/em> = 0.01<\/div>\r\n
          Conclusion: There is insufficient evidence to conclude that local data do not follow the distribution of the U.S. AP examinee distribution.<\/div>\r\n<\/div>\r\n
          \r\n\r\n77.\r\n\r\n
          \r\n
            \r\n \t
          1. H0<\/sub><\/em>: The actual college majors of graduating females fit the distribution of their expected majors<\/li>\r\n \t
          2. Ha<\/sub><\/em>: The actual college majors of graduating females do not fit the distribution of their expected majors<\/li>\r\n \t
          3. df<\/em> = 10<\/li>\r\n \t
          4. chi-square distribution with df<\/em> = 10<\/li>\r\n \t
          5. test statistic = 11.48<\/li>\r\n \t
          6. p<\/em>-value = 0.3211<\/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 null when a<\/em> = 0.05 and a<\/em> = 0.01<\/li>\r\n \t
            3. Reason for decision: p<\/em>-value > alpha<\/li>\r\n \t
            4. Conclusion: There is insufficient evidence to conclude that the distribution of actual college majors of graduating females fits the distribution of their expected majors.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n79. true\r\n\r\n81. true\r\n\r\n83. false\r\n\r\n85.\r\n\r\n
              \r\n
                \r\n \t
              1. H0<\/sub><\/em>: Surveyed obese fit the distribution of expected obese<\/li>\r\n \t
              2. Ha<\/sub><\/em>: Surveyed obese do not fit the distribution of expected obese<\/li>\r\n \t
              3. df<\/em> = 4<\/li>\r\n \t
              4. chi-square distribution with df<\/em> = 4<\/li>\r\n \t
              5. test statistic = 54.01<\/li>\r\n \t
              6. p<\/em>-value = 0<\/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: p<\/em>-value < alpha<\/li>\r\n \t
                4. Conclusion: At the 5% level of significance, from the data, there is sufficient evidence to conclude that the surveyed obese do not fit the distribution of expected obese.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n
                  \r\n

                  Test of Independence \u2013 Homework<\/h2>\r\n87.\r\n\r\n
                  \r\n
                    \r\n \t
                  1. H0<\/sub><\/em>: Car size is independent of family size.<\/li>\r\n \t
                  2. Ha<\/sub><\/em>: Car size is dependent on family size.<\/li>\r\n \t
                  3. df<\/em> = 9<\/li>\r\n \t
                  4. chi-square distribution with df<\/em> = 9<\/li>\r\n \t
                  5. test statistic = 15.8284<\/li>\r\n \t
                  6. p<\/em>-value = 0.0706<\/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: p<\/em>-value > alpha<\/li>\r\n \t
                    4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that car size and family size are dependent.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n89.\r\n\r\n
                      \r\n
                        \r\n \t
                      1. H0<\/sub><\/em>: Honeymoon locations are independent of the bride\u2019s age.<\/li>\r\n \t
                      2. Ha<\/sub><\/em>: Honeymoon locations are dependent on the bride\u2019s age.<\/li>\r\n \t
                      3. df<\/em> = 9<\/li>\r\n \t
                      4. chi-square distribution with df<\/em> = 9<\/li>\r\n \t
                      5. test statistic = 15.7027<\/li>\r\n \t
                      6. p<\/em>-value = 0.0734<\/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: p<\/em>-value > alpha<\/li>\r\n \t
                        4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that honeymoon location and bride age are dependent.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n91.\r\n\r\n
                          \r\n
                            \r\n \t
                          1. H0<\/sub><\/em>: The types of fries sold are independent of the location.<\/li>\r\n \t
                          2. Ha<\/sub><\/em>: The types of fries sold are dependent on the location.<\/li>\r\n \t
                          3. df<\/em> = 6<\/li>\r\n \t
                          4. chi-square distribution with df<\/em> = 6<\/li>\r\n \t
                          5. test statistic =18.8369<\/li>\r\n \t
                          6. p<\/em>-value = 0.0044<\/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: p<\/em>-value < alpha<\/li>\r\n \t
                            4. Conclusion: At the 5% significance level, there is sufficient evidence that types of fries and location are dependent.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n93.\r\n\r\n
                              \r\n
                                \r\n \t
                              1. H0<\/sub><\/em>: Salary is independent of level of education.<\/li>\r\n \t
                              2. Ha<\/sub><\/em>: Salary is dependent on level of education.<\/li>\r\n \t
                              3. df<\/em> = 12<\/li>\r\n \t
                              4. chi-square distribution with df<\/em> = 12<\/li>\r\n \t
                              5. test statistic = 255.7704<\/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

                                Alpha: 0.05<\/p>\r\n

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

                                Reason for decision: p<\/em>-value < alpha<\/p>\r\n

                                Conclusion: At the 5% significance level, there is sufficient evidence to conclude that salary and level of education are dependent.<\/p>\r\n<\/li>\r\n<\/ol>\r\n95. true\r\n\r\n97. true\r\n\r\n99.\r\n\r\n

                                \r\n
                                  \r\n \t
                                1. H0<\/sub><\/em>: Age is independent of the youngest online entrepreneurs\u2019 net worth.<\/li>\r\n \t
                                2. Ha<\/sub><\/em>: Age is dependent on the net worth of the youngest online entrepreneurs.<\/li>\r\n \t
                                3. df<\/em> = 2<\/li>\r\n \t
                                4. chi-square distribution with df<\/em> = 2<\/li>\r\n \t
                                5. test statistic = 1.76<\/li>\r\n \t
                                6. p<\/em>-value 0.4144<\/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: p<\/em>-value > alpha<\/li>\r\n \t
                                  4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the age and net worth of the youngest online entrepreneurs are dependent.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n

                                    Test for Homogeneity - Homework<\/h2>\r\n101.\r\n\r\n
                                    \r\n
                                      \r\n \t
                                    1. H0<\/sub><\/em>: The distribution for personality types is the same for both majors<\/li>\r\n \t
                                    2. Ha<\/sub><\/em>: The distribution for personality types is not the same for both majors<\/li>\r\n \t
                                    3. df<\/em> = 4<\/li>\r\n \t
                                    4. chi-square with df<\/em> = 4<\/li>\r\n \t
                                    5. test statistic = 3.01<\/li>\r\n \t
                                    6. p<\/em>-value = 0.5568<\/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: p<\/em>-value > alpha<\/li>\r\n \t
                                      4. Conclusion: There is insufficient evidence to conclude that the distribution of personality types is different for business and social science majors.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n103.\r\n\r\n
                                        \r\n
                                          \r\n \t
                                        1. H0<\/sub><\/em>: The distribution of fish caught is the same in Green Valley Lake and in Echo Lake.<\/li>\r\n \t
                                        2. Ha<\/sub><\/em>: The distribution of fish caught is not the same in Green Valley Lake and in Echo Lake.<\/li>\r\n \t
                                        3. 3<\/li>\r\n \t
                                        4. chi-square with df<\/em> = 3<\/li>\r\n \t
                                        5. 11.75<\/li>\r\n \t
                                        6. p<\/em>-value = 0.0083<\/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: p<\/em>-value < alpha<\/li>\r\n \t
                                          4. Conclusion: There is evidence to conclude that the distribution of fish caught is different in Green Valley Lake and in Echo Lake<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n105.\r\n\r\n
                                            \r\n
                                              \r\n \t
                                            1. H0<\/sub><\/em>: The distribution of average energy use in the USA is the same as in Europe between 2005 and 2010.<\/li>\r\n \t
                                            2. Ha<\/sub><\/em>: The distribution of average energy use in the USA is not the same as in Europe between 2005 and 2010.<\/li>\r\n \t
                                            3. df<\/em> = 4<\/li>\r\n \t
                                            4. chi-square with df<\/em> = 4<\/li>\r\n \t
                                            5. test statistic = 2.7434<\/li>\r\n \t
                                            6. p<\/em>-value = 0.7395<\/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: p<\/em>-value > alpha<\/li>\r\n \t
                                              4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the average energy use values in the US and EU are not derived from different distributions for the period from 2005 to 2010.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n

                                                Comparison of the Chi-Square Tests - Homework<\/h2>\r\n<\/section>107.<\/span>\r\n\r\n<\/section>
                                                \r\n
                                                  \r\n \t
                                                1. H0<\/sub><\/em>: The distribution for technology use is the same for community college students and university students.<\/li>\r\n \t
                                                2. Ha<\/sub><\/em>: The distribution for technology use is not the same for community college students and university students.<\/li>\r\n \t
                                                3. 2<\/li>\r\n \t
                                                4. chi-square with df<\/em> = 2<\/li>\r\n \t
                                                5. 7.05<\/li>\r\n \t
                                                6. p<\/em>-value = 0.0294<\/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: p<\/em>-value < alpha<\/li>\r\n \t
                                                  4. Conclusion: There is sufficient evidence to conclude that the distribution of technology use for statistics homework is not the same for statistics students at community colleges and at universities.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n
                                                    \r\n

                                                    Test of a Single Variance - Homework<\/h2>\r\n110.\u00a0225\r\n\r\n112.\u00a0H0<\/sub><\/em>: \u03c3<\/em>2<\/sup> \u2264 150\r\n\r\n114. 36\r\n\r\n118.\u00a0The claim is that the variance is no more than 150 minutes.\r\n\r\n120.\u00a0a Student's t<\/em>- or normal distribution\r\n\r\n122.\r\n\r\n
                                                    \r\n
                                                      \r\n \t
                                                    1. H0<\/sub><\/em>: \u03c3<\/em> = 15<\/li>\r\n \t
                                                    2. Ha<\/sub><\/em>: \u03c3<\/em> > 15<\/li>\r\n \t
                                                    3. df<\/em> = 42<\/li>\r\n \t
                                                    4. chi-square with df<\/em> = 42<\/li>\r\n \t
                                                    5. test statistic = 26.88<\/li>\r\n \t
                                                    6. p<\/em>-value = 0.9663<\/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 null hypothesis.<\/li>\r\n \t
                                                      3. Reason for decision: p<\/em>-value > alpha<\/li>\r\n \t
                                                      4. Conclusion: There is insufficient evidence to conclude that the standard deviation is greater than 15.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n124.\r\n\r\n
                                                        \r\n
                                                          \r\n \t
                                                        1. H0<\/sub><\/em>: \u03c3<\/em> \u2264 3<\/li>\r\n \t
                                                        2. Ha<\/sub><\/em>: \u03c3<\/em> > 3<\/li>\r\n \t
                                                        3. df<\/em> = 17<\/li>\r\n \t
                                                        4. chi-square distribution with df<\/em> = 17<\/li>\r\n \t
                                                        5. test statistic = 28.73<\/li>\r\n \t
                                                        6. p<\/em>-value = 0.0371<\/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: p<\/em>-value < alpha<\/li>\r\n \t
                                                          4. Conclusion: There is sufficient evidence to conclude that the standard deviation is greater than three.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n126.\r\n\r\n
                                                            \r\n
                                                              \r\n \t
                                                            1. H0<\/sub><\/em>: \u03c3<\/em> = 2<\/li>\r\n \t
                                                            2. Ha<\/sub><\/em>: \u03c3<\/em> \u2260 2<\/li>\r\n \t
                                                            3. df<\/em> = 14<\/li>\r\n \t
                                                            4. chi-square distiribution with df<\/em> = 14<\/li>\r\n \t
                                                            5. chi-square test statistic = 5.2094<\/li>\r\n \t
                                                            6. p<\/em>-value = 0.0346<\/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: p<\/em>-value < alpha<\/li>\r\n \t
                                                              4. Conclusion: There is sufficient evidence to conclude that the standard deviation is different than 2.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n128.\r\n

                                                                The sample standard deviation is $34.29.<\/p>\r\n

                                                                H0<\/sub><\/em> : \u03c3<\/em>2<\/sup> = 252<\/sup><\/p>\r\nHa<\/sub><\/em> : \u03c3<\/em>2<\/sup> > 252<\/sup>\r\n\r\ndf<\/em> = n<\/em> \u2013 1 = 7.\r\n\r\ntest statistic:[latex]{x}^{2}={x}^{2}_{7}=\\frac{{{(n-1)}{s}^{2}}}{{{25}^{2}}}=\\frac{{{(8-1)}{34.29}^{2}}}{{{25}^{2}}}={13.169}[\/latex]\r\n\r\np<\/em>-value: P<\/span>(<\/span>x<\/span>2<\/span>7<\/span><\/span>><\/span>13.169<\/span><\/span>)<\/span><\/span>=<\/span>1<\/span>\u2013<\/span>P<\/span>(<\/span>x<\/span>2<\/span>7<\/span><\/span>\u00a0<\/span>\u2264<\/span>13.169<\/span><\/span>)<\/span><\/span>=<\/span>0.0681<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\r\n\r\nAlpha: 0.05\r\n\r\nDecision: Do not reject the null hypothesis.\r\n\r\nReason for decision: p<\/em>-value > alpha\r\n\r\nConclusion: At the 5% level, there is insufficient evidence to conclude that the variance is more than 625.\r\n\r\n<\/section>\r\n

                                                                Bringing It Together: Homework<\/h2>\r\n130.\r\n\r\n \r\n
                                                                  \r\n \t
                                                                1. The test statistic is always positive and if the expected and observed values are not close together, the test statistic is large and the null hypothesis will be rejected.<\/li>\r\n \t
                                                                2. Testing to see if the data fits the distribution \u201ctoo well\u201d or is too perfect.<\/li>\r\n<\/ol>\r\n \r\n\r\n<\/section><\/section><\/section><\/section><\/section><\/section><\/section><\/section><\/section><\/section><\/section><\/section><\/section><\/div>\r\n<\/section>","rendered":"

                                                                  Facts About the Chi-Square Distribution – Practice<\/h2>\n

                                                                  1.\u00a0mean = 25 and standard deviation = 7.0711<\/p>\n

                                                                  3.\u00a0when the number of degrees of freedom is greater than 90<\/p>\n

                                                                  5.\u00a0df<\/em> = 2<\/p>\n

                                                                  Goodness-of-Fit Test – Practice<\/h2>\n

                                                                  7.\u00a0a goodness-of-fit test<\/p>\n

                                                                  9. 3<\/p>\n

                                                                  11.\u00a02.04<\/p>\n

                                                                  13.\u00a0We decline to reject the null hypothesis. There is not enough evidence to suggest that the observed test scores are significantly different from the expected test scores.<\/p>\n

                                                                  15.\u00a0H0<\/sub><\/em>: the distribution of AIDS cases follows the ethnicities of the general population of Santa Clara County.<\/p>\n

                                                                  17. right-tailed<\/p>\n

                                                                  19. 2016.136<\/p>\n

                                                                  21.\u00a0Graph: Check student\u2019s solution.\u00a0<\/span>Decision: Reject the null hypothesis.\u00a0<\/span>Reason for the Decision: <\/span>p<\/em>-value < alpha\u00a0<\/span>Conclusion (write out in complete sentences): The make-up of AIDS cases does not fit the ethnicities of the general population of Santa Clara County.<\/span><\/p>\n

                                                                  Test of Independence \u2013 Practice<\/h2>\n

                                                                  23.\u00a0a test of independence<\/p>\n

                                                                  25.\u00a0a test of independence<\/p>\n

                                                                  27. 8<\/p>\n

                                                                  29. 6.6<\/p>\n

                                                                  31.\u00a00.0435<\/p>\n

                                                                  33.<\/p>\n\n\n\n\n\n\n\n\n\n
                                                                  Smoking Level Per Day<\/th>\nAfrican American<\/th>\nNative Hawaiian<\/th>\nLatino<\/th>\nJapanese Americans<\/th>\nWhite<\/th>\nTotals<\/th>\n<\/tr>\n<\/thead>\n
                                                                  1-10<\/td>\n9,886<\/td>\n2,745<\/td>\n12,831<\/td>\n8,378<\/td>\n7,650<\/td>\n41,490<\/td>\n<\/tr>\n
                                                                  11-20<\/td>\n6,514<\/td>\n3,062<\/td>\n4,932<\/td>\n10,680<\/td>\n9,877<\/td>\n35,065<\/td>\n<\/tr>\n
                                                                  21-30<\/td>\n1,671<\/td>\n1,419<\/td>\n1,406<\/td>\n4,715<\/td>\n6,062<\/td>\n15,273<\/td>\n<\/tr>\n
                                                                  31+<\/td>\n759<\/td>\n788<\/td>\n800<\/td>\n2,305<\/td>\n3,970<\/td>\n8,622<\/td>\n<\/tr>\n
                                                                  Totals<\/td>\n18,830<\/td>\n8,014<\/td>\n19,969<\/td>\n26,078<\/td>\n27,559<\/td>\n10,0450<\/p>\n

                                                                   <\/p>\n

                                                                   <\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                                                  35.<\/p>\n\n\n\n\n\n\n\n\n
                                                                  Smoking Level Per Day<\/th>\nAfrican American<\/th>\nNative Hawaiian<\/th>\nLatino<\/th>\nJapanese Americans<\/th>\nWhite<\/th>\n<\/tr>\n<\/thead>\n
                                                                  1-10<\/td>\n7777.57<\/td>\n3310.11<\/td>\n8248.02<\/td>\n10771.29<\/td>\n11383.01<\/td>\n<\/tr>\n
                                                                  11-20<\/td>\n6573.16<\/td>\n2797.52<\/td>\n6970.76<\/td>\n9103.29<\/td>\n9620.27<\/td>\n<\/tr>\n
                                                                  21-30<\/td>\n2863.02<\/td>\n1218.49<\/td>\n3036.20<\/td>\n3965.05<\/td>\n4190.23<\/td>\n<\/tr>\n
                                                                  31+<\/td>\n1616.25<\/td>\n687.87<\/td>\n1714.01<\/td>\n2238.37<\/td>\n2365.49<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                                                  37.\u00a010,301.8<\/p>\n

                                                                  39.\u00a0right<\/p>\n

                                                                  41.<\/p>\n

                                                                  \n
                                                                    \n
                                                                  1. Reject the null hypothesis.<\/li>\n
                                                                  2. p<\/em>-value < alpha<\/li>\n
                                                                  3. There is sufficient evidence to conclude that smoking level is dependent on ethnic group.<\/li>\n<\/ol>\n<\/section>\n

                                                                    Test for Homogeneity \u2013 Practice<\/h2>\n
                                                                    \n

                                                                    43.\u00a0test for homogeneity<\/p>\n

                                                                    45.\u00a0test for homogeneity<\/p>\n

                                                                    47.\u00a0All values in the table must be greater than or equal to five.<\/p>\n

                                                                    49. 3<\/p>\n

                                                                    51.\u00a00.00005<\/p>\n<\/div>\n

                                                                    \n

                                                                    Comparison of the Chi-Square Tests \u2013 Practice<\/h2>\n<\/div>\n
                                                                    \n

                                                                    53.\u00a0a goodness-of-fit test<\/p>\n

                                                                    55.\u00a0a test for independence<\/p>\n

                                                                    57.\u00a0Answers will vary. Sample answer: Tests of independence and tests for homogeneity both calculate the test statistic the same way, [latex]\\sum{ }_{(ij)}\\frac{{{({O}-{E})}^{2}}}{{E}}[\/latex] In addition, all values must be greater than or equal to five.<\/p>\n<\/div>\n

                                                                    \n

                                                                    Test of a Single Variance \u2013 Practice<\/h2>\n
                                                                    \n
                                                                    \n
                                                                    \n

                                                                    59.\u00a0a test of a single variance<\/p>\n

                                                                    61.\u00a0a left-tailed test<\/p>\n

                                                                    63.<\/p>\n

                                                                      \n
                                                                    • H0<\/sub><\/em>: \u03c3<\/em>2<\/sup> = 0.812<\/sup>;<\/li>\n
                                                                    • Ha<\/sub><\/em>: <\/span>\u03c3<\/em>2<\/sup> > 0.81<\/span>2<\/sup><\/li>\n<\/ul>\n

                                                                      65. a test of a single variance<\/p>\n

                                                                      67. 0.0542<\/p>\n

                                                                      Facts About the Chi-Square Distribution \u2013 Homework<\/h2>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n

                                                                      69.\u00a0true<\/p>\n

                                                                      71.\u00a0false<\/p>\n

                                                                      Goodness-of-Fit Test – Homework<\/h2>\n
                                                                      73.<\/p>\n\n\n\n\n\n\n\n\n
                                                                      Marital Status<\/th>\nPercent<\/th>\nExpected Frequency<\/th>\n<\/tr>\n<\/thead>\n
                                                                      never married<\/td>\n31.3<\/td>\n125.2<\/td>\n<\/tr>\n
                                                                      married<\/td>\n56.1<\/td>\n224.4<\/td>\n<\/tr>\n
                                                                      widowed<\/td>\n2.5<\/td>\n10<\/td>\n<\/tr>\n
                                                                      divorced\/separated<\/td>\n10.1<\/td>\n40.4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
                                                                        \n
                                                                      1. The data fit the distribution.<\/li>\n
                                                                      2. The data does not fit the distribution.<\/li>\n
                                                                      3. 3<\/li>\n
                                                                      4. chi-square distribution with df<\/em> = 3<\/li>\n
                                                                      5. 19.27<\/li>\n
                                                                      6. 0.0002<\/li>\n
                                                                      7. Check student\u2019s solution.\n
                                                                          \n
                                                                        1. Alpha = 0.05<\/li>\n
                                                                        2. Decision: Reject null<\/li>\n
                                                                        3. Reason for decision: p<\/em>-value < alpha<\/li>\n
                                                                        4. Conclusion: Data does not fit the distribution.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                                                                          75.
                                                                          \nH0<\/sub><\/em>: The local results follow the distribution of the U.S. AP examinee population
                                                                          \nHa<\/sub><\/em>: The local results do not follow the distribution of the U.S. AP examinee population
                                                                          \ndf<\/em> = 5
                                                                          \nchi-square distribution with df<\/em> = 5
                                                                          \nchi-square test statistic = 13.4
                                                                          \np<\/em>-value = 0.0199
                                                                          \nCheck student\u2019s solution.<\/p>\n

                                                                          \n
                                                                          Alpha = 0.05<\/div>\n
                                                                          Decision: Reject null when a<\/em> = 0.05<\/div>\n
                                                                          Reason for Decision: p<\/em>-value < alpha<\/div>\n
                                                                          Conclusion: Local data do not fit the AP Examinee Distribution.<\/div>\n
                                                                          Decision: Do not reject null when a<\/em> = 0.01<\/div>\n
                                                                          Conclusion: There is insufficient evidence to conclude that local data do not follow the distribution of the U.S. AP examinee distribution.<\/div>\n<\/div>\n
                                                                          \n

                                                                          77.<\/p>\n

                                                                          \n
                                                                            \n
                                                                          1. H0<\/sub><\/em>: The actual college majors of graduating females fit the distribution of their expected majors<\/li>\n
                                                                          2. Ha<\/sub><\/em>: The actual college majors of graduating females do not fit the distribution of their expected majors<\/li>\n
                                                                          3. df<\/em> = 10<\/li>\n
                                                                          4. chi-square distribution with df<\/em> = 10<\/li>\n
                                                                          5. test statistic = 11.48<\/li>\n
                                                                          6. p<\/em>-value = 0.3211<\/li>\n
                                                                          7. Check student\u2019s solution.\n
                                                                              \n
                                                                            1. Alpha = 0.05<\/li>\n
                                                                            2. Decision: Do not reject null when a<\/em> = 0.05 and a<\/em> = 0.01<\/li>\n
                                                                            3. Reason for decision: p<\/em>-value > alpha<\/li>\n
                                                                            4. Conclusion: There is insufficient evidence to conclude that the distribution of actual college majors of graduating females fits the distribution of their expected majors.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                                                                              79. true<\/p>\n

                                                                              81. true<\/p>\n

                                                                              83. false<\/p>\n

                                                                              85.<\/p>\n

                                                                              \n
                                                                                \n
                                                                              1. H0<\/sub><\/em>: Surveyed obese fit the distribution of expected obese<\/li>\n
                                                                              2. Ha<\/sub><\/em>: Surveyed obese do not fit the distribution of expected obese<\/li>\n
                                                                              3. df<\/em> = 4<\/li>\n
                                                                              4. chi-square distribution with df<\/em> = 4<\/li>\n
                                                                              5. test statistic = 54.01<\/li>\n
                                                                              6. p<\/em>-value = 0<\/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: p<\/em>-value < alpha<\/li>\n
                                                                                4. Conclusion: At the 5% level of significance, from the data, there is sufficient evidence to conclude that the surveyed obese do not fit the distribution of expected obese.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n
                                                                                  \n

                                                                                  Test of Independence \u2013 Homework<\/h2>\n

                                                                                  87.<\/p>\n

                                                                                  \n
                                                                                    \n
                                                                                  1. H0<\/sub><\/em>: Car size is independent of family size.<\/li>\n
                                                                                  2. Ha<\/sub><\/em>: Car size is dependent on family size.<\/li>\n
                                                                                  3. df<\/em> = 9<\/li>\n
                                                                                  4. chi-square distribution with df<\/em> = 9<\/li>\n
                                                                                  5. test statistic = 15.8284<\/li>\n
                                                                                  6. p<\/em>-value = 0.0706<\/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: p<\/em>-value > alpha<\/li>\n
                                                                                    4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that car size and family size are dependent.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                                                                                      89.<\/p>\n

                                                                                      \n
                                                                                        \n
                                                                                      1. H0<\/sub><\/em>: Honeymoon locations are independent of the bride\u2019s age.<\/li>\n
                                                                                      2. Ha<\/sub><\/em>: Honeymoon locations are dependent on the bride\u2019s age.<\/li>\n
                                                                                      3. df<\/em> = 9<\/li>\n
                                                                                      4. chi-square distribution with df<\/em> = 9<\/li>\n
                                                                                      5. test statistic = 15.7027<\/li>\n
                                                                                      6. p<\/em>-value = 0.0734<\/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: p<\/em>-value > alpha<\/li>\n
                                                                                        4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that honeymoon location and bride age are dependent.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                                                                                          91.<\/p>\n

                                                                                          \n
                                                                                            \n
                                                                                          1. H0<\/sub><\/em>: The types of fries sold are independent of the location.<\/li>\n
                                                                                          2. Ha<\/sub><\/em>: The types of fries sold are dependent on the location.<\/li>\n
                                                                                          3. df<\/em> = 6<\/li>\n
                                                                                          4. chi-square distribution with df<\/em> = 6<\/li>\n
                                                                                          5. test statistic =18.8369<\/li>\n
                                                                                          6. p<\/em>-value = 0.0044<\/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: p<\/em>-value < alpha<\/li>\n
                                                                                            4. Conclusion: At the 5% significance level, there is sufficient evidence that types of fries and location are dependent.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                                                                                              93.<\/p>\n

                                                                                              \n
                                                                                                \n
                                                                                              1. H0<\/sub><\/em>: Salary is independent of level of education.<\/li>\n
                                                                                              2. Ha<\/sub><\/em>: Salary is dependent on level of education.<\/li>\n
                                                                                              3. df<\/em> = 12<\/li>\n
                                                                                              4. chi-square distribution with df<\/em> = 12<\/li>\n
                                                                                              5. test statistic = 255.7704<\/li>\n
                                                                                              6. p<\/em>-value = 0<\/li>\n
                                                                                              7. Check student\u2019s solution.<\/li>\n
                                                                                              8. \n

                                                                                                Alpha: 0.05<\/p>\n

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

                                                                                                Reason for decision: p<\/em>-value < alpha<\/p>\n

                                                                                                Conclusion: At the 5% significance level, there is sufficient evidence to conclude that salary and level of education are dependent.<\/p>\n<\/li>\n<\/ol>\n

                                                                                                95. true<\/p>\n

                                                                                                97. true<\/p>\n

                                                                                                99.<\/p>\n

                                                                                                \n
                                                                                                  \n
                                                                                                1. H0<\/sub><\/em>: Age is independent of the youngest online entrepreneurs\u2019 net worth.<\/li>\n
                                                                                                2. Ha<\/sub><\/em>: Age is dependent on the net worth of the youngest online entrepreneurs.<\/li>\n
                                                                                                3. df<\/em> = 2<\/li>\n
                                                                                                4. chi-square distribution with df<\/em> = 2<\/li>\n
                                                                                                5. test statistic = 1.76<\/li>\n
                                                                                                6. p<\/em>-value 0.4144<\/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: p<\/em>-value > alpha<\/li>\n
                                                                                                  4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the age and net worth of the youngest online entrepreneurs are dependent.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                                                                                                    Test for Homogeneity – Homework<\/h2>\n

                                                                                                    101.<\/p>\n

                                                                                                    \n
                                                                                                      \n
                                                                                                    1. H0<\/sub><\/em>: The distribution for personality types is the same for both majors<\/li>\n
                                                                                                    2. Ha<\/sub><\/em>: The distribution for personality types is not the same for both majors<\/li>\n
                                                                                                    3. df<\/em> = 4<\/li>\n
                                                                                                    4. chi-square with df<\/em> = 4<\/li>\n
                                                                                                    5. test statistic = 3.01<\/li>\n
                                                                                                    6. p<\/em>-value = 0.5568<\/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: p<\/em>-value > alpha<\/li>\n
                                                                                                      4. Conclusion: There is insufficient evidence to conclude that the distribution of personality types is different for business and social science majors.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                                                                                                        103.<\/p>\n

                                                                                                        \n
                                                                                                          \n
                                                                                                        1. H0<\/sub><\/em>: The distribution of fish caught is the same in Green Valley Lake and in Echo Lake.<\/li>\n
                                                                                                        2. Ha<\/sub><\/em>: The distribution of fish caught is not the same in Green Valley Lake and in Echo Lake.<\/li>\n
                                                                                                        3. 3<\/li>\n
                                                                                                        4. chi-square with df<\/em> = 3<\/li>\n
                                                                                                        5. 11.75<\/li>\n
                                                                                                        6. p<\/em>-value = 0.0083<\/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: p<\/em>-value < alpha<\/li>\n
                                                                                                          4. Conclusion: There is evidence to conclude that the distribution of fish caught is different in Green Valley Lake and in Echo Lake<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                                                                                                            105.<\/p>\n

                                                                                                            \n
                                                                                                              \n
                                                                                                            1. H0<\/sub><\/em>: The distribution of average energy use in the USA is the same as in Europe between 2005 and 2010.<\/li>\n
                                                                                                            2. Ha<\/sub><\/em>: The distribution of average energy use in the USA is not the same as in Europe between 2005 and 2010.<\/li>\n
                                                                                                            3. df<\/em> = 4<\/li>\n
                                                                                                            4. chi-square with df<\/em> = 4<\/li>\n
                                                                                                            5. test statistic = 2.7434<\/li>\n
                                                                                                            6. p<\/em>-value = 0.7395<\/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: p<\/em>-value > alpha<\/li>\n
                                                                                                              4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the average energy use values in the US and EU are not derived from different distributions for the period from 2005 to 2010.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                                                                                                                Comparison of the Chi-Square Tests – Homework<\/h2>\n<\/section>\n

                                                                                                                107.<\/span><\/p>\n<\/section>\n

                                                                                                                \n
                                                                                                                  \n
                                                                                                                1. H0<\/sub><\/em>: The distribution for technology use is the same for community college students and university students.<\/li>\n
                                                                                                                2. Ha<\/sub><\/em>: The distribution for technology use is not the same for community college students and university students.<\/li>\n
                                                                                                                3. 2<\/li>\n
                                                                                                                4. chi-square with df<\/em> = 2<\/li>\n
                                                                                                                5. 7.05<\/li>\n
                                                                                                                6. p<\/em>-value = 0.0294<\/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: p<\/em>-value < alpha<\/li>\n
                                                                                                                  4. Conclusion: There is sufficient evidence to conclude that the distribution of technology use for statistics homework is not the same for statistics students at community colleges and at universities.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n
                                                                                                                    \n

                                                                                                                    Test of a Single Variance – Homework<\/h2>\n

                                                                                                                    110.\u00a0225<\/p>\n

                                                                                                                    112.\u00a0H0<\/sub><\/em>: \u03c3<\/em>2<\/sup> \u2264 150<\/p>\n

                                                                                                                    114. 36<\/p>\n

                                                                                                                    118.\u00a0The claim is that the variance is no more than 150 minutes.<\/p>\n

                                                                                                                    120.\u00a0a Student’s t<\/em>– or normal distribution<\/p>\n

                                                                                                                    122.<\/p>\n

                                                                                                                    \n
                                                                                                                      \n
                                                                                                                    1. H0<\/sub><\/em>: \u03c3<\/em> = 15<\/li>\n
                                                                                                                    2. Ha<\/sub><\/em>: \u03c3<\/em> > 15<\/li>\n
                                                                                                                    3. df<\/em> = 42<\/li>\n
                                                                                                                    4. chi-square with df<\/em> = 42<\/li>\n
                                                                                                                    5. test statistic = 26.88<\/li>\n
                                                                                                                    6. p<\/em>-value = 0.9663<\/li>\n
                                                                                                                    7. Check student\u2019s solution.\n
                                                                                                                        \n
                                                                                                                      1. Alpha = 0.05<\/li>\n
                                                                                                                      2. Decision: Do not reject null hypothesis.<\/li>\n
                                                                                                                      3. Reason for decision: p<\/em>-value > alpha<\/li>\n
                                                                                                                      4. Conclusion: There is insufficient evidence to conclude that the standard deviation is greater than 15.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                                                                                                                        124.<\/p>\n

                                                                                                                        \n
                                                                                                                          \n
                                                                                                                        1. H0<\/sub><\/em>: \u03c3<\/em> \u2264 3<\/li>\n
                                                                                                                        2. Ha<\/sub><\/em>: \u03c3<\/em> > 3<\/li>\n
                                                                                                                        3. df<\/em> = 17<\/li>\n
                                                                                                                        4. chi-square distribution with df<\/em> = 17<\/li>\n
                                                                                                                        5. test statistic = 28.73<\/li>\n
                                                                                                                        6. p<\/em>-value = 0.0371<\/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: p<\/em>-value < alpha<\/li>\n
                                                                                                                          4. Conclusion: There is sufficient evidence to conclude that the standard deviation is greater than three.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                                                                                                                            126.<\/p>\n

                                                                                                                            \n
                                                                                                                              \n
                                                                                                                            1. H0<\/sub><\/em>: \u03c3<\/em> = 2<\/li>\n
                                                                                                                            2. Ha<\/sub><\/em>: \u03c3<\/em> \u2260 2<\/li>\n
                                                                                                                            3. df<\/em> = 14<\/li>\n
                                                                                                                            4. chi-square distiribution with df<\/em> = 14<\/li>\n
                                                                                                                            5. chi-square test statistic = 5.2094<\/li>\n
                                                                                                                            6. p<\/em>-value = 0.0346<\/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: p<\/em>-value < alpha<\/li>\n
                                                                                                                              4. Conclusion: There is sufficient evidence to conclude that the standard deviation is different than 2.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                                                                                                                                128.<\/p>\n

                                                                                                                                The sample standard deviation is $34.29.<\/p>\n

                                                                                                                                H0<\/sub><\/em> : \u03c3<\/em>2<\/sup> = 252<\/sup><\/p>\n

                                                                                                                                Ha<\/sub><\/em> : \u03c3<\/em>2<\/sup> > 252<\/sup><\/p>\n

                                                                                                                                df<\/em> = n<\/em> \u2013 1 = 7.<\/p>\n

                                                                                                                                test statistic:[latex]{x}^{2}={x}^{2}_{7}=\\frac{{{(n-1)}{s}^{2}}}{{{25}^{2}}}=\\frac{{{(8-1)}{34.29}^{2}}}{{{25}^{2}}}={13.169}[\/latex]<\/p>\n

                                                                                                                                p<\/em>-value: P<\/span>(<\/span>x<\/span>2<\/span>7<\/span><\/span>><\/span>13.169<\/span><\/span>)<\/span><\/span>=<\/span>1<\/span>\u2013<\/span>P<\/span>(<\/span>x<\/span>2<\/span>7<\/span><\/span>\u00a0<\/span>\u2264<\/span>13.169<\/span><\/span>)<\/span><\/span>=<\/span>0.0681<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

                                                                                                                                Alpha: 0.05<\/p>\n

                                                                                                                                Decision: Do not reject the null hypothesis.<\/p>\n

                                                                                                                                Reason for decision: p<\/em>-value > alpha<\/p>\n

                                                                                                                                Conclusion: At the 5% level, there is insufficient evidence to conclude that the variance is more than 625.<\/p>\n<\/section>\n

                                                                                                                                Bringing It Together: Homework<\/h2>\n

                                                                                                                                130.<\/p>\n

                                                                                                                                 <\/p>\n

                                                                                                                                  \n
                                                                                                                                1. The test statistic is always positive and if the expected and observed values are not close together, the test statistic is large and the null hypothesis will be rejected.<\/li>\n
                                                                                                                                2. Testing to see if the data fits the distribution \u201ctoo well\u201d or is too perfect.<\/li>\n<\/ol>\n

                                                                                                                                   <\/p>\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<\/section>\n<\/div>\n<\/section>\n\n\t\t\t

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                                                                                                                                  Candela Citations<\/h3>\n\t\t\t\t\t
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