{"id":2867,"date":"2022-05-09T15:09:02","date_gmt":"2022-05-09T19:09:02","guid":{"rendered":"https:\/\/openbooks.macewan.ca\/rcommander\/?post_type=chapter&p=2867"},"modified":"2024-02-08T14:38:13","modified_gmt":"2024-02-08T19:38:13","slug":"11-7-review-questions","status":"publish","type":"chapter","link":"https:\/\/openbooks.macewan.ca\/introstats\/chapter\/11-7-review-questions\/","title":{"raw":"11.7 Review Questions","rendered":"11.7 Review Questions"},"content":{"raw":"
    \r\n \t
  1. An American roulette wheel contains 18 red numbers, 18 black numbers, and 2 green numbers. The following table shows the frequency with which the ball landed on each color in 300 trials.\r\n
    \r\n\r\n\r\n\r\n\r\n\r\n
    Color<\/strong><\/th>\r\nRed<\/th>\r\nBlack<\/th>\r\nGreen<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    Frequency<\/strong><\/th>\r\n140<\/td>\r\n120<\/td>\r\n40<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\nAt the 5% significance level, do the data suggest that the wheel is out of balance?<\/li>\r\n \t
  2. A gambler thinks a die may be loaded and that the six numbers are not equally likely. To test his suspicion, he rolled the die 150 times and obtained the data in the following table.\r\n
    \r\n\r\n\r\n\r\n\r\n\r\n
    Number<\/strong><\/th>\r\n1<\/th>\r\n2<\/th>\r\n3<\/th>\r\n4<\/th>\r\n5<\/th>\r\n6<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    Frequency<\/strong><\/th>\r\n23<\/td>\r\n26<\/td>\r\n23<\/td>\r\n21<\/td>\r\n31<\/td>\r\n26<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\nDo the data provide sufficient evidence to conclude that the die is loaded? Perform the hypothesis test at the 5% significance level.<\/li>\r\n \t
  3. The following table reported the survey results on how members would prefer to receive ballots in annual elections. At the 5% significance level, do the data provide sufficient evidence to conclude that gender (column) and preference (row) are associated?\r\n
    \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    <\/td>\r\nMale<\/th>\r\nFemale<\/th>\r\nTotal<\/strong><\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    Mail<\/th>\r\n60<\/td>\r\n30<\/td>\r\n90<\/td>\r\n<\/tr>\r\n
    Email<\/th>\r\n150<\/td>\r\n90<\/td>\r\n240<\/td>\r\n<\/tr>\r\n
    Both<\/th>\r\n70<\/td>\r\n40<\/td>\r\n110<\/td>\r\n<\/tr>\r\n
    N\/A<\/th>\r\n80<\/td>\r\n50<\/td>\r\n130<\/td>\r\n<\/tr>\r\n
    Total<\/strong><\/th>\r\n360<\/td>\r\n210<\/td>\r\n570<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div><\/li>\r\n<\/ol>\r\n
    Show\/Hide Answer<\/summary>\r\n
      \r\n \t
    1. If the wheel is balanced, the chance of landing on a red number is [latex]p_1=\\frac{18}{18+18+2}[\/latex], on a black number is [latex]p_2=\\frac{18}{18+18+2}[\/latex], and on a green number is [latex]p_3=\\frac{2}{18+18+2}[\/latex]. If the wheel is out of balance, at least one proportion differs from the specified value. We can use the chi-square goodness of fit test to test this. The assumptions of a goodness-of-fit test are:\r\n
        \r\n \t
      1. All expected frequencies are at least 1.<\/li>\r\n \t
      2. At most 20% of the expected frequencies are less than 5.<\/li>\r\n \t
      3. Simple random sample (if you need to generalize the conclusion to a larger population)<\/li>\r\n<\/ol>\r\nThe expected frequencies are [latex]E_1=np_1=300\\times \\frac{18}{38}=142.1053[\/latex], [latex]E_2=np_2=300\\times \\frac{18}{38}=142.1053[\/latex], [latex]E_3=np_3=300\\times \\frac{2}{38}=15.78947[\/latex]. All the expected frequencies are greater than 5. We assume the trials were conducted randomly.\r\n\r\nSteps:\r\n
          \r\n \t
        1. Hypotheses. [latex]H_0: p_1=\\frac{18}{38}, p_2=\\frac{18}{38}, p_3=\\frac{2}{38}[\/latex] versus [latex]H_a:[\/latex]\u00a0at least one proportion is different from the specified value.<\/li>\r\n \t
        2. Significance level [latex]\\alpha=0.05[\/latex]<\/li>\r\n \t
        3. Test statistic.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
          Color<\/th>\r\nObserved<\/th>\r\nExpected<\/th>\r\nContribution<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
          red<\/th>\r\n140<\/td>\r\n142.10526<\/td>\r\n0.0311891<\/td>\r\n<\/tr>\r\n
          black<\/th>\r\n120<\/td>\r\n142.10526<\/td>\r\n3.4385965<\/td>\r\n<\/tr>\r\n
          green<\/th>\r\n40<\/td>\r\n15.78947<\/td>\r\n37.1228070<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[latex]\\begin{aligned}\r\n\\chi_o^2 & =\\sum_{i=1}^3 \\frac{(O_i-E_i)^2}{E_i}=\\frac{(140- 142.10526)^2}{142.10526}+\\frac{(120-142.10526)^2}{142.10526}+\\frac{(40- 15.78947)^2}{15.78947} \\\\\r\n& =0.03118908+3.43859649+37.12280702 \\\\\r\n& =40.59259.\r\n\\end{aligned}[\/latex]\r\nwith [latex]df=k-1=3-1=2[\/latex].<\/li>\r\n \t
        4. P-value. The chi-square test is always right-tailed. P-value is the area to the right of the observed value [latex]\\chi^2_o[\/latex] under the chi-square density curve with degrees of freedom [latex]df=k-1=3-1=2[\/latex]. [latex]\\text{P-value}=P(\\chi^2 \\ge \\chi_o^2)=P(\\chi^2 \\ge 40.593) \\lt P(\\chi^2 \\ge 10.597)=0.005.[\/latex]\r\nThat is, P-value<0.005.\r\nIf the critical value approach is used, we need to find the rejection region. The chi-square test is always right-tailed. Make sure that the area of the rejection region is [latex]\\alpha[\/latex], so the critical value is [latex]\\chi^2_\\alpha=\\chi^2_{0.05}=5.991.[\/latex]<\/li>\r\n \t
        5. Decision. We reject [latex]H_0[\/latex] since P-value<0.005<0.05 ([latex]\\alpha[\/latex]). If the critical approach is used, we reject [latex]H_0[\/latex] since the observed test statistic [latex]\\chi_o^2=40.593>5.991[\/latex] falls in the rejection region.<\/li>\r\n \t
        6. Conclusion. At the 5% significance level, we have sufficient evidence that the wheel is out of balance.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
        7. If the die is balanced, the proportion of landing on each of 1, 2, [latex]\\cdots[\/latex], 6 should be the same, i.e., [latex]p_i=\\frac{1}{6}, i=1, 2, \\cdots, 6[\/latex]. All expected frequencies are [latex]E_i=np_i=150\\times \\frac{1}{6}=25>5[\/latex], the assumptions for a chi-square goodness-of-fit test are satisfied.\r\nSteps:\r\n
            \r\n \t
          1. Hypotheses. [latex]H_0: p_i=\\frac{1}{6}, i=1, 2, \\cdots 6[\/latex] versus [latex]H_a[\/latex]: at least one proportion is equal to [latex]\\frac{1}{6}[\/latex].<\/li>\r\n \t
          2. Significance level [latex]\\alpha=0.01[\/latex]<\/li>\r\n \t
          3. Test statistic. The following table is useful.\r\n[latex]\\chi^2_o=\\sum_{\\mbox{all cells}} \\frac{(O-E)^2}{E}=\\frac{(23-25)^2}{25}+\\frac{(26-25)^2}{25}+\\frac{(23-25)^2}{25}+\\frac{(21-25)^2}{25}+\\frac{(31-25)^2}{25}+\\frac{(26-25)^2}{25}=2.48[\/latex]\r\nwith the degrees of freedom of the test [latex]df=k-1=6-1=5.[\/latex]<\/span><\/li>\r\n \t
          4. P-value. The chi-square test is always right-tailed. P-value is the area to the right of the observed value [latex]\\chi^2_o[\/latex] under the chi-square density curve with degrees of freedom [latex]df=k-1=6-1=5[\/latex]. P-value=[latex]P(\\chi^2\\ge \\chi_o^2)=P(\\chi^2\\ge 2.48)>P(\\chi^2\\ge 9.236)=0.1[\/latex], i.e., P-value>0.1.\r\nOr\r\nRejection region. The chi-square test is always right-tailed. Make sure that the area of the rejection region is [latex]\\alpha[\/latex], so the critical value is [latex]\\chi^2_{\\alpha}=\\chi^2_{0.01}=15.086[\/latex].<\/li>\r\n \t
          5. Decision: We cannot reject [latex]H_0[\/latex] since P-value>0.1>0.01 ([latex]\\alpha)[\/latex].\r\nOr\r\nWe cannot reject [latex]H_0[\/latex] since [latex]\\chi_o^2=2.48<15.086[\/latex]\u00a0falls in the non-rejection region.<\/li>\r\n \t
          6. Conclusion: At the 5% significance level, we do not have sufficient evidence that the die is loaded.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
          7. We should use the chi-square independence test. The expected frequencies are given in the following table. All the expected frequencies are greater than 5, the assumptions for a chi-square independence test are satisfied.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
            Preference<\/th>\r\nMale<\/th>\r\nFemale<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
            Mail<\/th>\r\n56.84211<\/td>\r\n33.15789<\/td>\r\n<\/tr>\r\n
            Email<\/th>\r\n151.57895<\/td>\r\n88.42105<\/td>\r\n<\/tr>\r\n
            Both<\/th>\r\n69.47368<\/td>\r\n40.52632<\/td>\r\n<\/tr>\r\n
            N\/A<\/th>\r\n82.10526<\/td>\r\n47.89474<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSteps:\r\n
              \r\n \t
            1. Hypotheses. [latex]H_0:[\/latex] gender and preference are independent. versus [latex]H_a[\/latex]: gender and preference are associated.<\/li>\r\n \t
            2. Significance level [latex]\\alpha=0.05[\/latex].<\/li>\r\n \t
            3. Test statistic.\r\n[latex]\\begin{aligned}\r\n\\chi^2_o &=\\sum_\\text{all cells} \\frac{(O-E)^2}{E}\\\\&=\\frac{(60-56.842)^2}{56.842}+\\frac{(30-33.158)^2}{33.158}+\\frac{(150-151.579)^2}{151.579}\r\n+\\frac{(90-88.421)^2}{88.421}\\\\&+\\frac{(70-69.474)^2}{69.474}+\\frac{(40-40.526)^2}{40.526}+\\frac{(80-82.105)^2}{82.105}+\\frac{(50-47.895)^2}{47.895} \\\\\r\n& =0.678\r\n\\end{aligned}[\/latex]\r\nwith the degrees of freedom of the test [latex]df=(r-1)\\times (c-1)=(4-1)\\times (2-1)=3.[\/latex]<\/li>\r\n \t
            4. P-value. The chi-square test is always right-tailed. P-value is the area to the right of the observed value [latex]\\chi^2_o[\/latex] under the chi-square density curve with degrees of freedom [latex]df=3[\/latex]. P-value=[latex]P(\\chi^2\\ge \\chi_o^2)=P(\\chi^2\\ge 0.678)>P(\\chi^2\\ge 6.251)=0.1[\/latex], i.e., P-value >0.10.\r\nOr\r\nRejection region. The chi-square test is always right-tailed. Make sure that the area of the rejection region is [latex]\\alpha[\/latex], so the critical value is [latex]\\chi^2_{\\alpha}=\\chi^2_{0.05}=7.815[\/latex].<\/li>\r\n \t
            5. Decision: Don\u2019t reject [latex]H_0[\/latex], since [latex]P\\text{-value} > 0.1 > 0.05(\\alpha).[\/latex]\r\nOr\r\nWe cannot reject [latex]H_0[\/latex] since [latex]\\chi_o^2=0.678<7.815[\/latex]\u00a0falls in the non-rejection region.<\/li>\r\n \t
            6. Conclusion: At the 5% significance level, we do not have sufficient evidence that gender and preference are associated.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n \r\n\r\n<\/details>","rendered":"
                \n
              1. An American roulette wheel contains 18 red numbers, 18 black numbers, and 2 green numbers. The following table shows the frequency with which the ball landed on each color in 300 trials.\n
                \n\n\n\n\n\n
                Color<\/strong><\/th>\nRed<\/th>\nBlack<\/th>\nGreen<\/th>\n<\/tr>\n<\/thead>\n
                Frequency<\/strong><\/th>\n140<\/td>\n120<\/td>\n40<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

                At the 5% significance level, do the data suggest that the wheel is out of balance?<\/li>\n

              2. A gambler thinks a die may be loaded and that the six numbers are not equally likely. To test his suspicion, he rolled the die 150 times and obtained the data in the following table.\n
                \n\n\n\n\n\n
                Number<\/strong><\/th>\n1<\/th>\n2<\/th>\n3<\/th>\n4<\/th>\n5<\/th>\n6<\/th>\n<\/tr>\n<\/thead>\n
                Frequency<\/strong><\/th>\n23<\/td>\n26<\/td>\n23<\/td>\n21<\/td>\n31<\/td>\n26<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

                Do the data provide sufficient evidence to conclude that the die is loaded? Perform the hypothesis test at the 5% significance level.<\/li>\n

              3. The following table reported the survey results on how members would prefer to receive ballots in annual elections. At the 5% significance level, do the data provide sufficient evidence to conclude that gender (column) and preference (row) are associated?\n
                \n\n\n\n\n\n\n\n\n\n
                <\/td>\nMale<\/th>\nFemale<\/th>\nTotal<\/strong><\/th>\n<\/tr>\n<\/thead>\n
                Mail<\/th>\n60<\/td>\n30<\/td>\n90<\/td>\n<\/tr>\n
                Email<\/th>\n150<\/td>\n90<\/td>\n240<\/td>\n<\/tr>\n
                Both<\/th>\n70<\/td>\n40<\/td>\n110<\/td>\n<\/tr>\n
                N\/A<\/th>\n80<\/td>\n50<\/td>\n130<\/td>\n<\/tr>\n
                Total<\/strong><\/th>\n360<\/td>\n210<\/td>\n570<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/li>\n<\/ol>\n
                \nShow\/Hide Answer<\/summary>\n
                  \n
                1. If the wheel is balanced, the chance of landing on a red number is [latex]p_1=\\frac{18}{18+18+2}[\/latex], on a black number is [latex]p_2=\\frac{18}{18+18+2}[\/latex], and on a green number is [latex]p_3=\\frac{2}{18+18+2}[\/latex]. If the wheel is out of balance, at least one proportion differs from the specified value. We can use the chi-square goodness of fit test to test this. The assumptions of a goodness-of-fit test are:\n
                    \n
                  1. All expected frequencies are at least 1.<\/li>\n
                  2. At most 20% of the expected frequencies are less than 5.<\/li>\n
                  3. Simple random sample (if you need to generalize the conclusion to a larger population)<\/li>\n<\/ol>\n

                    The expected frequencies are [latex]E_1=np_1=300\\times \\frac{18}{38}=142.1053[\/latex], [latex]E_2=np_2=300\\times \\frac{18}{38}=142.1053[\/latex], [latex]E_3=np_3=300\\times \\frac{2}{38}=15.78947[\/latex]. All the expected frequencies are greater than 5. We assume the trials were conducted randomly.<\/p>\n

                    Steps:<\/p>\n

                      \n
                    1. Hypotheses. [latex]H_0: p_1=\\frac{18}{38}, p_2=\\frac{18}{38}, p_3=\\frac{2}{38}[\/latex] versus [latex]H_a:[\/latex]\u00a0at least one proportion is different from the specified value.<\/li>\n
                    2. Significance level [latex]\\alpha=0.05[\/latex]<\/li>\n
                    3. Test statistic.
                      \n\n\n\n\n\n\n\n
                      Color<\/th>\nObserved<\/th>\nExpected<\/th>\nContribution<\/th>\n<\/tr>\n<\/thead>\n
                      red<\/th>\n140<\/td>\n142.10526<\/td>\n0.0311891<\/td>\n<\/tr>\n
                      black<\/th>\n120<\/td>\n142.10526<\/td>\n3.4385965<\/td>\n<\/tr>\n
                      green<\/th>\n40<\/td>\n15.78947<\/td>\n37.1228070<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                      [latex]\\begin{aligned} \\chi_o^2 & =\\sum_{i=1}^3 \\frac{(O_i-E_i)^2}{E_i}=\\frac{(140- 142.10526)^2}{142.10526}+\\frac{(120-142.10526)^2}{142.10526}+\\frac{(40- 15.78947)^2}{15.78947} \\\\ & =0.03118908+3.43859649+37.12280702 \\\\ & =40.59259. \\end{aligned}[\/latex]
                      \nwith [latex]df=k-1=3-1=2[\/latex].<\/li>\n

                    4. P-value. The chi-square test is always right-tailed. P-value is the area to the right of the observed value [latex]\\chi^2_o[\/latex] under the chi-square density curve with degrees of freedom [latex]df=k-1=3-1=2[\/latex]. [latex]\\text{P-value}=P(\\chi^2 \\ge \\chi_o^2)=P(\\chi^2 \\ge 40.593) \\lt P(\\chi^2 \\ge 10.597)=0.005.[\/latex]
                      \nThat is, P-value<0.005.
                      \nIf the critical value approach is used, we need to find the rejection region. The chi-square test is always right-tailed. Make sure that the area of the rejection region is [latex]\\alpha[\/latex], so the critical value is [latex]\\chi^2_\\alpha=\\chi^2_{0.05}=5.991.[\/latex]<\/li>\n
                    5. Decision. We reject [latex]H_0[\/latex] since P-value<0.005<0.05 ([latex]\\alpha[\/latex]). If the critical approach is used, we reject [latex]H_0[\/latex] since the observed test statistic [latex]\\chi_o^2=40.593>5.991[\/latex] falls in the rejection region.<\/li>\n
                    6. Conclusion. At the 5% significance level, we have sufficient evidence that the wheel is out of balance.<\/li>\n<\/ol>\n<\/li>\n
                    7. If the die is balanced, the proportion of landing on each of 1, 2, [latex]\\cdots[\/latex], 6 should be the same, i.e., [latex]p_i=\\frac{1}{6}, i=1, 2, \\cdots, 6[\/latex]. All expected frequencies are [latex]E_i=np_i=150\\times \\frac{1}{6}=25>5[\/latex], the assumptions for a chi-square goodness-of-fit test are satisfied.
                      \nSteps:<\/p>\n
                        \n
                      1. Hypotheses. [latex]H_0: p_i=\\frac{1}{6}, i=1, 2, \\cdots 6[\/latex] versus [latex]H_a[\/latex]: at least one proportion is equal to [latex]\\frac{1}{6}[\/latex].<\/li>\n
                      2. Significance level [latex]\\alpha=0.01[\/latex]<\/li>\n
                      3. Test statistic. The following table is useful.
                        \n[latex]\\chi^2_o=\\sum_{\\mbox{all cells}} \\frac{(O-E)^2}{E}=\\frac{(23-25)^2}{25}+\\frac{(26-25)^2}{25}+\\frac{(23-25)^2}{25}+\\frac{(21-25)^2}{25}+\\frac{(31-25)^2}{25}+\\frac{(26-25)^2}{25}=2.48[\/latex]
                        \nwith the degrees of freedom of the test [latex]df=k-1=6-1=5.[\/latex]<\/span><\/li>\n
                      4. P-value. The chi-square test is always right-tailed. P-value is the area to the right of the observed value [latex]\\chi^2_o[\/latex] under the chi-square density curve with degrees of freedom [latex]df=k-1=6-1=5[\/latex]. P-value=[latex]P(\\chi^2\\ge \\chi_o^2)=P(\\chi^2\\ge 2.48)>P(\\chi^2\\ge 9.236)=0.1[\/latex], i.e., P-value>0.1.
                        \nOr
                        \nRejection region. The chi-square test is always right-tailed. Make sure that the area of the rejection region is [latex]\\alpha[\/latex], so the critical value is [latex]\\chi^2_{\\alpha}=\\chi^2_{0.01}=15.086[\/latex].<\/li>\n
                      5. Decision: We cannot reject [latex]H_0[\/latex] since P-value>0.1>0.01 ([latex]\\alpha)[\/latex].
                        \nOr
                        \nWe cannot reject [latex]H_0[\/latex] since [latex]\\chi_o^2=2.48<15.086[\/latex]\u00a0falls in the non-rejection region.<\/li>\n
                      6. Conclusion: At the 5% significance level, we do not have sufficient evidence that the die is loaded.<\/li>\n<\/ol>\n<\/li>\n
                      7. We should use the chi-square independence test. The expected frequencies are given in the following table. All the expected frequencies are greater than 5, the assumptions for a chi-square independence test are satisfied.
                        \n\n\n\n\n\n\n\n\n
                        Preference<\/th>\nMale<\/th>\nFemale<\/th>\n<\/tr>\n<\/thead>\n
                        Mail<\/th>\n56.84211<\/td>\n33.15789<\/td>\n<\/tr>\n
                        Email<\/th>\n151.57895<\/td>\n88.42105<\/td>\n<\/tr>\n
                        Both<\/th>\n69.47368<\/td>\n40.52632<\/td>\n<\/tr>\n
                        N\/A<\/th>\n82.10526<\/td>\n47.89474<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                        Steps:<\/p>\n

                          \n
                        1. Hypotheses. [latex]H_0:[\/latex] gender and preference are independent. versus [latex]H_a[\/latex]: gender and preference are associated.<\/li>\n
                        2. Significance level [latex]\\alpha=0.05[\/latex].<\/li>\n
                        3. Test statistic.
                          \n[latex]\\begin{aligned} \\chi^2_o &=\\sum_\\text{all cells} \\frac{(O-E)^2}{E}\\\\&=\\frac{(60-56.842)^2}{56.842}+\\frac{(30-33.158)^2}{33.158}+\\frac{(150-151.579)^2}{151.579} +\\frac{(90-88.421)^2}{88.421}\\\\&+\\frac{(70-69.474)^2}{69.474}+\\frac{(40-40.526)^2}{40.526}+\\frac{(80-82.105)^2}{82.105}+\\frac{(50-47.895)^2}{47.895} \\\\ & =0.678 \\end{aligned}[\/latex]
                          \nwith the degrees of freedom of the test [latex]df=(r-1)\\times (c-1)=(4-1)\\times (2-1)=3.[\/latex]<\/li>\n
                        4. P-value. The chi-square test is always right-tailed. P-value is the area to the right of the observed value [latex]\\chi^2_o[\/latex] under the chi-square density curve with degrees of freedom [latex]df=3[\/latex]. P-value=[latex]P(\\chi^2\\ge \\chi_o^2)=P(\\chi^2\\ge 0.678)>P(\\chi^2\\ge 6.251)=0.1[\/latex], i.e., P-value >0.10.
                          \nOr
                          \nRejection region. The chi-square test is always right-tailed. Make sure that the area of the rejection region is [latex]\\alpha[\/latex], so the critical value is [latex]\\chi^2_{\\alpha}=\\chi^2_{0.05}=7.815[\/latex].<\/li>\n
                        5. Decision: Don\u2019t reject [latex]H_0[\/latex], since [latex]P\\text{-value} > 0.1 > 0.05(\\alpha).[\/latex]
                          \nOr
                          \nWe cannot reject [latex]H_0[\/latex] since [latex]\\chi_o^2=0.678<7.815[\/latex]\u00a0falls in the non-rejection region.<\/li>\n
                        6. Conclusion: At the 5% significance level, we do not have sufficient evidence that gender and preference are associated.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

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