{"id":1017,"date":"2021-06-04T20:51:25","date_gmt":"2021-06-05T00:51:25","guid":{"rendered":"https:\/\/openbooks.macewan.ca\/rcommander\/?post_type=chapter&p=1017"},"modified":"2025-06-24T16:43:18","modified_gmt":"2025-06-24T20:43:18","slug":"9-2-two-sample-t-test","status":"publish","type":"chapter","link":"https:\/\/openbooks.macewan.ca\/introstats\/chapter\/9-2-two-sample-t-test\/","title":{"raw":"9.2 Two-Sample t Test and t Interval Based on Two Independent Samples","rendered":"9.2 Two-Sample t Test and t Interval Based on Two Independent Samples"},"content":{"raw":"Two-sample t-tests<\/i> are used to test hypotheses regarding the difference between two population means. Depending on whether the two population standard deviations ([latex]\\sigma_1[\/latex] and [latex]\\sigma_2[\/latex]) are equal or not, we have the non-pooled and pooled two-sample t-tests\u00a0and t<\/em> interval. Minor advantages of the pooled t-test are a slightly narrower confidence interval, a slightly more powerful test, and a simpler formula for the degrees of freedom. However, the pooled t-test is valid only when the two population standard deviations are close; otherwise,\u00a0it gives poor results. Therefore, we recommend using the non-pooled t-test<\/i>\u00a0unless we are quite confident that [latex]\\sigma_1[\/latex] = [latex]\\sigma_2[\/latex], which is very difficult to verify.\r\n

9.2.1 Non-Pooled Two-Sample t<\/em> Test and t<\/em> Interval<\/strong><\/h2>\r\n
\r\n\r\nAssumptions<\/strong>:\r\n
    \r\n \t
  1. Simple random samples<\/li>\r\n \t
  2. Two samples are independent<\/li>\r\n \t
  3. Normal populations or large sample sizes [latex](n_1 \\geq 30, n_2 \\geq 30)[\/latex]<\/li>\r\n<\/ol>\r\nSteps<\/strong>:\r\n
      \r\n \t
    1. Set up the hypotheses:\r\n
      \r\n\r\n\r\n\r\n\r\n\r\n\r\n
      \r\n

      Two-tailed test<\/strong><\/p>\r\n<\/td>\r\n

      		\r\n

      Right (upper)-tailed test<\/strong><\/p>\r\n<\/td>\r\n

      		\r\n

      Left (lower)-tailed test<\/strong><\/p>\r\n<\/td>\r\n<\/tr>\r\n<\/thead>\r\n

      \r\n
      [latex]H_0: \\mu_1 - \\mu_2 = \\Delta_0[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]H_0: \\mu_1 - \\mu_2 \\leq \\Delta_0[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]H_0: \\mu_1 - \\mu_2 \\geq \\Delta_0[\/latex]<\/div><\/td>\r\n<\/tr>\r\n
      \r\n
      [latex]H_a: \\mu_1 - \\mu_2 \\neq \\Delta_0[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]H_a: \\mu_1 - \\mu_2 \\: \\gt \\: \\Delta_0[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]H_a: \\mu_1 - \\mu_2 \\: \\lt \\: \\Delta_0[\/latex]<\/div><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\nNote that [latex]\\Delta_0[\/latex] can be zero or any value you want to test. In most cases, however, [latex]\\Delta_0=0[\/latex].<\/li>\r\n \t
    2. State the significance level [latex]\\alpha[\/latex].<\/li>\r\n \t
    3. Compute the value of the test statistic: [latex]t_o = \\frac{(\\bar{x}_1 - \\bar{x}_2) - (\\Delta_0)}{\\sqrt{ \\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2}}}[\/latex] with [latex]df = \\frac{ \\left( \\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2} \\right)^2}{\\frac{1}{n_1 - 1} \\left( \\frac{s_1^2}{n_1} \\right)^2 + \\frac{1}{n_2 - 1} \\left( \\frac{s_2^2}{n_2} \\right)^2 }[\/latex], rounded down<\/strong> to the nearest integer or [latex]\\min\\{n_1 -1, n_2 - 1 \\}[\/latex].<\/li>\r\n \t
    4. Use the t-score table (Table IV) to find the P-value or rejection region.\r\n
      \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      <\/td>\r\n\r\n
      Two-tailed<\/div><\/th>\r\n
      		\r\n
      Right-tailed<\/div><\/th>\r\n
      		\r\n
      Left-tailed<\/div><\/th>\r\n<\/tr>\r\n<\/thead>\r\n
      Null<\/th>\r\n\r\n
      [latex]H_0: \\mu_1 - \\mu_2 = \\Delta_0[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]H_0: \\mu_1 - \\mu_2 \\leq \\Delta_0[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]H_0: \\mu_1 - \\mu_2 \\geq \\Delta_0[\/latex]<\/div><\/td>\r\n<\/tr>\r\n
      Alternative<\/th>\r\n\r\n
      [latex]H_a: \\mu_1 - \\mu_2 \\neq \\Delta_0[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]H_a: \\mu_1 - \\mu_2 \\: \\gt \\: \\Delta_0[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]H_a: \\mu_1 - \\mu_2 \\: \\lt \\: \\Delta_0[\/latex]<\/div><\/td>\r\n<\/tr>\r\n
      P-value<\/th>\r\n\r\n
      [latex]2P(t \\geq |t_o|)[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]P(t \\geq t_o)[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]P(t \\leq t_o)[\/latex]<\/div><\/td>\r\n<\/tr>\r\n
      Rejection region<\/th>\r\n[latex]t \\geq t_{\\alpha \/ 2}[\/latex] or [latex]t \\leq - t_{\\alpha \/ 2}[\/latex]<\/td>\r\n\r\n
      [latex]t \\geq t_{\\alpha}[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]t \\leq - t_{\\alpha}[\/latex]<\/div><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div><\/li>\r\n \t
    5. Decision: Reject the null [latex]H_0[\/latex] if P-value [latex]\\leq \\alpha[\/latex] or [latex]t_o[\/latex] falls in the rejection region.<\/li>\r\n \t
    6. Conclusion.<\/li>\r\n<\/ol>\r\nA [latex](1 \u2013\\alpha) \\times 100\\%[\/latex]\u00a0<\/em>two-sample [latex]t[\/latex] confidence interval for [latex]\\mu_1[\/latex] \u2013 [latex]\\mu_2[\/latex] is\r\n
      \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      \r\n
      Two-tailed<\/div><\/th>\r\n
      		\r\n
      Right-tailed<\/div><\/th>\r\n
      		\r\n
      Left-tailed<\/div><\/th>\r\n<\/tr>\r\n<\/thead>\r\n
      \r\n
      [latex]H_0: \\mu_1 - \\mu_2 = \\Delta_0[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]H_0: \\mu_1 - \\mu_2 \\leq \\Delta_0[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]H_0: \\mu_1 - \\mu_2 \\geq \\Delta_0[\/latex]<\/div><\/td>\r\n<\/tr>\r\n
      \r\n
      [latex]H_a: \\mu_1 - \\mu_2 \\neq \\Delta_0[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]H_a: \\mu_1 - \\mu_2 \\: \\gt \\: \\Delta_0[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]H_a: \\mu_1 - \\mu_2 \\: \\lt \\: \\Delta_0[\/latex]<\/div><\/td>\r\n<\/tr>\r\n
      [latex](\\bar{x}_1 - \\bar{x}_2) \\pm t_{\\alpha \/ 2} \\sqrt{ \\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2}}[\/latex]<\/td>\r\n[latex]\\left((\\bar{x}_1 - \\bar{x}_2) - t_{\\alpha} \\sqrt{ \\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2}}, \\infty \\right)[\/latex]<\/td>\r\n[latex]\\left(- \\infty , (\\bar{x}_1 - \\bar{x}_2) + t_{\\alpha} \\sqrt{ \\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2}}\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<\/div>\r\n \r\n
      \r\n

      Example: Two-Sample Non-Pooled t-Test and t Interval<\/p>\r\n\r\n<\/header>\r\n

      \r\n\r\nSome students attend class regularly, but some do not. An instructor wants to compare the class averages for those who attend lectures regularly ([latex]\\mu_1[\/latex]) with those who do not ([latex]\\mu_2[\/latex]). A simple random sample of size [latex]n_1 = 135[\/latex] is selected from the attendees and a simple random sample of size [latex]n_2=35[\/latex] is taken from the non-attendees. The sample mean and sample standard deviation for attendees are [latex]\\bar{x}_1 = 67, s_1 = 17[\/latex]; and for non-attendees are [latex]\\bar{x}_2 = 49, s_2 = 18[\/latex].\r\n
        \r\n \t
      1. Test at the 1% significance level whether those who attend lectures have a higher average<\/strong>, i.e., [latex]\\mu_1 \\: \\gt \\: \\mu_2[\/latex] or [latex]\\mu_1 - \\mu_2 \\: \\gt \\: 0[\/latex].<\/li>\r\n<\/ol>\r\n\u00a0 \u00a0 \u00a0 \u00a0Check the assumptions<\/strong>:<\/span>\r\n
          \r\n \t
        1. We have simple random samples from attendees and non-attendees.<\/li>\r\n \t
        2. The two samples are independent.<\/li>\r\n \t
        3. We do not have the data, so we cannot check whether two populations are normally distributed using normal probability plot (Q-Q plot); however, we have large sample sizes with [latex]n_1 = 135 \\: \\gt \\: 30, n_2 = 35 \\: \\gt \\: 30[\/latex].<\/li>\r\n<\/ol>\r\nTherefore, the assumptions are met.\r\n\r\nSteps:<\/strong>\r\n
            \r\n \t
          1. Set up the hypotheses: [latex]H_0: \\mu_1 - \\mu_2 \\leq 0[\/latex] versus [latex]H_a: \\mu_1 - \\mu_2 \\: \\gt \\: 0[\/latex].\r\nThis is a right-tailed test.<\/li>\r\n \t
          2. The significance level is [latex]\\alpha=0.01[\/latex].<\/li>\r\n \t
          3. Compute the value of the test statistic:\r\n

            [latex]t_o = \\frac{(\\bar{x}_1 - \\bar{x}_2) - \\Delta_0}{\\sqrt{ \\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2}}} = \\frac{(67 - 49) -0}{\\sqrt{ \\frac{17^2}{135} + \\frac{18^2}{35}}} = 5.332[\/latex] with<\/p>\r\n[latex]df = \\frac{ \\left( \\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2} \\right)^2}{\\frac{1}{n_1 - 1} \\left( \\frac{s_1^2}{n_1} \\right)^2 + \\frac{1}{n_2 - 1} \\left( \\frac{s_2^2}{n_2} \\right)^2 } = \\frac{ \\left( \\frac{17^2}{135} + \\frac{18^2}{35} \\right)^2}{\\frac{1}{135 - 1} \\left( \\frac{17^2}{135} \\right)^2 + \\frac{1}{35 - 1} \\left( \\frac{18^2}{35} \\right)^2 } = 50.85[\/latex], rounded down to [latex]df = 50[\/latex].<\/li>\r\n \t

          4. Find the P-value. For a right-tailed test with the observed test statistics [latex]t_o=5.332[\/latex], the P-value is the area to the right of [latex]t_o[\/latex], i.e., [latex]\\mbox{P-value} = P(t \\geq t_o) = P(t \\geq 5.332) < 0.0005, \\mbox{since} t_o=5.332 \\: \\gt \\: 3.496(t_{0.0005})[\/latex]<\/li>\r\n \t
          5. Decision: Since the P-value [latex] < 0.0005 < 0.01 (\\alpha)[\/latex], reject the null hypothesis [latex]H_0[\/latex].<\/li>\r\n \t
          6. Conclusion: At the 1% significance level, the data provide sufficient evidence that those who attend lectures have a higher average<\/strong>.<\/li>\r\n<\/ol>\r\nIf using the critical value approach, steps 1-3 are the same, steps 4-6 become:\r\n
              \r\n \t
            1. \r\n
                \r\n \t
              1. Rejection region:<\/a>\r\n
                \r\n\r\n\r\n\r\n
                [caption id=\"attachment_3592\" align=\"aligncenter\" width=\"300\"]\"A Figure 9.2<\/strong>: Rejection Region and Observed Value. [Image Description (See Appendix D Figure 9.2)<\/a>][\/caption]<\/td>\r\n[latex]\\alpha = 0.01 , t_{\\alpha} = t_{0.01} = 2.403[\/latex]\r\n\r\nFor a right-tailed test, the critical value is 2.403. The rejection region is to the right of 2.403.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div><\/li>\r\n \t
              2. Decision: Since the observed value [latex]t_o =5.332 \\: \\gt \\: 2.403[\/latex] falls in the rejection region, we reject the null hypothesis [latex]H_0[\/latex].<\/li>\r\n \t
              3. Conclusion: At the 1% significance level, the data provide sufficient evidence that those who attend lectures have a higher average<\/strong>.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
              4. Obtain a confidence interval for the difference between the class average for attendees and non-attendees [latex]\\mu_1 - \\mu_2[\/latex] corresponding to the test in part a).\r\nPart a) contains a right-tailed test at the 1% significance level. Therefore, we should obtain a 99% upper-tailed interval: [latex]\\left((\\bar{x}_1 - \\bar{x}_2) - t_{\\alpha} \\sqrt{ \\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2}}, \\infty \\right)[\/latex].\r\n[latex]\\alpha = 0.01, df = 50, t_{\\alpha} = t_{0.01} = 2.403[\/latex].\r\nThe lower bound for the upper-tailed interval is:\r\n

                [latex](\\bar{x}_1 - \\bar{x}_2) - t_{\\alpha} \\sqrt{ \\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2}} = (67 - 49) - 2.403 \\times \\sqrt{ \\frac{17^2}{135} + \\frac{18^2}{35}} = 9.887[\/latex].<\/p>\r\nThus, the corresponding 99% confidence interval for [latex]\\mu_1 - \\mu_2[\/latex] is [latex](9.887, \\infty)[\/latex].\r\nInterpretation: we are 99% confident that the difference in average grades is at least 9.887 between attendees and non-attendees.<\/li>\r\n \t

              5. Does the interval in part (b) support the conclusion in part a)?\r\nIn part a), we reject [latex]H_0[\/latex] at the 1% significance level and claim that [latex]\\mu_1 - \\mu_2 \\: \\gt \\: 0[\/latex].\r\nIn part b), since the entire interval is above 0, we can claim that [latex]\\mu_1 - \\mu_2 \\: \\gt \\: 0[\/latex] with 99% confidence, which supports the results obtained in part a).<\/li>\r\n \t
              6. Based on the interval obtained in part b), can we claim that the class average of attendees is at least 5 marks higher than that of the non-attendees? How about 10 marks higher?\r\nWe can claim that the class average of attendees is at least 5 marks higher than that of the non-attendees since the entire interval is above 5. However, we cannot claim that the class average of attendees is at least 10 marks higher than that of the non-attendees since the interval contains 10.<\/a>[caption id=\"attachment_1021\" align=\"aligncenter\" width=\"432\"]\"A Figure 9.3<\/strong>: Confidence Interval of difference in Class Average. [Image Description (See Appendix D Figure 9.3)<\/a>][\/caption]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

                9.2.2 Pooled Two-Sample t<\/em> Test and t<\/em> Interval<\/strong><\/h2>\r\nIf the two population standard deviations are equal, i.e., [latex]\\sigma_1 = \\sigma_2 = \\sigma[\/latex], we can pool the two samples together to get a better estimate of the common standard deviation [latex]\\sigma[\/latex]\r\n\r\n[latex]\\hat{\\sigma} = s_p = \\sqrt{ \\frac{(n_1 - 1)s_1^2 + (n_2 -1)s_2^2 } {(n_1 - 1) + (n_2 -1)}}[\/latex]\r\n\r\nwhere the term [latex](n_1 - 1)s_1^2 = \\sum_{\\text{sample 1}}(x - \\bar{x}_1)^2 [\/latex] is the variation of the data within sample 1, and [latex](n_2 - 1)s_2^2 = \\sum_{\\text{sample 2}}(x - \\bar{x}_2)^2 [\/latex] is the variation of the data within sample 2.\r\n\r\nRecall that the standard deviation of [latex]\\bar{X}_1 - \\bar{X}_2[\/latex] is [latex]\\sigma_{\\scriptsize \\bar{X}_1 - \\bar{X}_2} = \\sqrt{ \\frac{\\sigma_1^2}{n_1} + \\frac{\\sigma_2^2}{n_2} }[\/latex]. Thus, if [latex]\\sigma_1 = \\sigma_2 = \\sigma[\/latex], then [latex]\\sigma_{\\scriptsize \\bar{X}_1 - \\bar{X}_2}[\/latex] reduces to [latex]\\sqrt{ \\frac{\\sigma^2}{n_1} + \\frac{\\sigma^2}{n_2} } = \\sigma \\sqrt{ \\frac{1}{n_1} + \\frac{1}{n_2} }[\/latex]. Estimating [latex]\\sigma[\/latex] with [latex]s_p[\/latex] leads to the pooled test statistic:\r\n\r\n[latex] t = \\frac{(\\bar{X}_1 - \\bar{X}_2) - (\\mu_1 - \\mu_2)}{s_p \\sqrt{\\frac{1}{n_1} + \\frac{1}{n_2} } } \\sim \\text{$t$ distribution}[\/latex]\r\n\r\nwith [latex]df = (n_1 - 1) + (n_2 -1) = n_1 + n_2 -2[\/latex].\r\n\r\nThe assumption [latex]\\sigma_1 = \\sigma_2[\/latex] is very difficult to verify. Some textbooks suggest a rule of thumb:\r\n\r\nIf the ratio of the larger to the smaller sample standard deviation is less than 2, then the assumption is considered to be reasonable, i.e., [latex]\\frac{\\max \\{ s_1, s_2\\}}{\\min \\{ s_1, s_2\\}} < 2[\/latex]<\/strong>.\r\n
                \r\n\r\nAssumptions<\/strong>:\r\n
                  \r\n \t
                1. Simple random samples.<\/li>\r\n \t
                2. Independent samples.<\/li>\r\n \t
                3. Normal populations or large sample sizes [latex](n_1 \\geq 30, n_2 \\geq 30)[\/latex].<\/li>\r\n \t
                4. Equal population standard deviations. This assumption is reasonable if [latex]\\frac{\\max \\{ s_1, s_2\\}}{\\min \\{ s_1, s_2\\}} < 2[\/latex].<\/li>\r\n<\/ol>\r\nSteps<\/strong>:\r\n
                    \r\n \t
                  1. Set up the hypotheses:\r\n
                    \r\n\r\n\r\n\r\n\r\n\r\n
                    Two-tailed test<\/strong><\/td>\r\nRight (upper)-tailed test<\/strong><\/td>\r\nLeft (lower)-tailed test<\/strong><\/td>\r\n<\/tr>\r\n
                    \r\n
                    [latex]H_0: \\mu_1 - \\mu_2 = \\Delta_0[\/latex]<\/div><\/td>\r\n
                    		\r\n
                    [latex]H_0: \\mu_1 - \\mu_2 \\leq \\Delta_0[\/latex]<\/div><\/td>\r\n
                    		\r\n
                    [latex]H_0: \\mu_1 - \\mu_2 \\geq \\Delta_0[\/latex]<\/div><\/td>\r\n<\/tr>\r\n
                    \r\n
                    [latex]H_a: \\mu_1 - \\mu_2 \\neq \\Delta_0[\/latex]<\/div><\/td>\r\n
                    		\r\n
                    [latex]H_a: \\mu_1 - \\mu_2 \\: \\gt \\: \\Delta_0[\/latex]<\/div><\/td>\r\n
                    		\r\n
                    [latex]H_a: \\mu_1 - \\mu_2 \\: \\lt \\: \\Delta_0[\/latex]<\/div><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\nNote that [latex]\\Delta_0[\/latex] can be zero or any value you want to test.<\/li>\r\n \t
                  2. State the significance level [latex]\\alpha[\/latex].<\/li>\r\n \t
                  3. Compute the value of the test statistic: [latex]t_o = \\frac{(\\bar{x}_1 - \\bar{x}_2) - \\Delta_0}{s_p \\sqrt{ \\frac{1}{n_1} + \\frac{1}{n_2}}}[\/latex], with [latex]df = n_1 + n_2 \u2013 2[\/latex] and [latex]s_p = \\sqrt{ \\frac{(n_1 - 1)s_1^2 + (n_2 -1)s_2^2 } {(n_1 - 1) + (n_2 -1)}}[\/latex].<\/li>\r\n \t
                  4. Use the t-score table (Table IV) to find the P-value or rejection region.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
                    <\/td>\r\n\r\n
                    Two-tailed<\/div><\/th>\r\n
                    		\r\n
                    Right-tailed<\/div><\/th>\r\n
                    		\r\n
                    Left-tailed<\/div><\/th>\r\n<\/tr>\r\n<\/thead>\r\n
                    Null<\/th>\r\n\r\n
                    [latex]H_0: \\mu_1 - \\mu_2 = \\Delta_0[\/latex]<\/div><\/td>\r\n
                    		\r\n
                    [latex]H_0: \\mu_1 - \\mu_2 \\leq \\Delta_0[\/latex]<\/div><\/td>\r\n
                    		\r\n
                    [latex]H_0: \\mu_1 - \\mu_2 \\geq \\Delta_0[\/latex]<\/div><\/td>\r\n<\/tr>\r\n
                    Alternative<\/th>\r\n\r\n
                    [latex]H_a: \\mu_1 - \\mu_2 \\neq \\Delta_0[\/latex]<\/div><\/td>\r\n
                    		\r\n
                    [latex]H_a: \\mu_1 - \\mu_2 \\: \\gt \\: \\Delta_0[\/latex]<\/div><\/td>\r\n
                    		\r\n
                    [latex]H_a: \\mu_1 - \\mu_2 \\: \\lt \\: \\Delta_0[\/latex]<\/div><\/td>\r\n<\/tr>\r\n
                    P-value<\/th>\r\n\r\n
                    [latex]2P(t \\geq |t_o|)[\/latex]<\/div><\/td>\r\n
                    		\r\n
                    [latex]P(t \\geq t_o)[\/latex]<\/div><\/td>\r\n
                    		\r\n
                    [latex]P(t \\leq t_o)[\/latex]<\/div><\/td>\r\n<\/tr>\r\n
                    Rejection region<\/th>\r\n[latex]t \\geq t_{\\alpha \/ 2}[\/latex] or [latex]t \\leq - t_{\\alpha \/ 2}[\/latex]<\/td>\r\n\r\n
                    [latex]t \\geq t_{\\alpha}[\/latex]<\/div><\/td>\r\n
                    		\r\n
                    [latex]t \\leq - t_{\\alpha}[\/latex]<\/div><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n \t
                  5. Decision: Reject the null [latex]H_0[\/latex] if P-value [latex]\\leq \\alpha[\/latex] or [latex]t_o[\/latex] falls in the rejection region.<\/li>\r\n \t
                  6. Conclusion.<\/li>\r\n<\/ol>\r\nA [latex](1 - \\alpha) \\times 100%[\/latex] two-sample t<\/em> confidence interval for [latex]\\mu_1 - \\mu_2[\/latex] is\r\n
                    \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
                    \r\n
                    Two-tailed<\/div><\/th>\r\n
                    		\r\n
                    Right-tailed<\/div><\/th>\r\n
                    		\r\n
                    Left-tailed<\/div><\/th>\r\n<\/tr>\r\n<\/thead>\r\n
                    \r\n
                    [latex]H_0: \\mu_1 - \\mu_2 = \\Delta_0[\/latex]<\/div><\/td>\r\n
                    		\r\n
                    [latex]H_0: \\mu_1 - \\mu_2 \\leq \\Delta_0[\/latex]<\/div><\/td>\r\n
                    		\r\n
                    [latex]H_0: \\mu_1 - \\mu_2 \\geq \\Delta_0[\/latex]<\/div><\/td>\r\n<\/tr>\r\n
                    \r\n
                    [latex]H_a: \\mu_1 - \\mu_2 \\neq \\Delta_0[\/latex]<\/div><\/td>\r\n
                    		\r\n
                    [latex]H_a: \\mu_1 - \\mu_2 \\: \\gt \\: \\Delta_0[\/latex]<\/div><\/td>\r\n
                    		\r\n
                    [latex]H_a: \\mu_1 - \\mu_2 \\: \\lt \\: \\Delta_0[\/latex]<\/div><\/td>\r\n<\/tr>\r\n
                    [latex] \\small{(\\bar{x}_1 - \\bar{x}_2) \\pm t_{\\alpha \/ 2} s_p \\sqrt{ \\frac{1}{n_1} + \\frac{1}{n_2}}}[\/latex]<\/td>\r\n[latex] \\small{\\left((\\bar{x}_1 - \\bar{x}_2) - t_{\\alpha} s_p \\sqrt{ \\frac{1}{n_1} + \\frac{1}{n_2}}, \\infty \\right)}[\/latex]<\/td>\r\n[latex] \\small{\\left(- \\infty , (\\bar{x}_1 - \\bar{x}_2) + t_{\\alpha} \u00a0s_p \\sqrt{ \\frac{1}{n_1} + \\frac{1}{n_2}}\\right)}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<\/div>\r\n \r\n
                    \r\n

                    Example: Pooled Two-Sample t Test and Interval<\/p>\r\n\r\n<\/header>\r\n

                    \r\n\r\nSome students attend class regularly, but some do not. An instructor wants to compare the class averages for those who attend lectures regularly ([latex]\\mu_1[\/latex]) with those who do not ([latex]\\mu_2[\/latex]). A simple random sample of size [latex]n_1=135[\/latex] is selected from the attendees, and a simple random sample of size [latex]n_2=35[\/latex] is taken from the non-attendees. The sample mean and sample standard deviation for attendees are [latex]\\bar{x}_1 = 67, s_1 = 17[\/latex]; and for non-attendees are [latex]\\bar{x}_2 = 49, s_2 = 18[\/latex].\r\n
                      \r\n \t
                    1. Is it reasonable to conduct a pooled two-sample t-test<\/i>\u00a0to test whether those who attend lectures have a higher average<\/strong>? If yes, run the test at the 1% significance level.<\/li>\r\n<\/ol>\r\n

                      Check the assumptions<\/strong>:<\/p>\r\n\r\n

                        \r\n \t
                      1. We have simple random samples.<\/li>\r\n \t
                      2. The two samples are independent.<\/li>\r\n \t
                      3. We have large sample sizes [latex](n_1 = 135 > 30, n_2 =35 > 30)[\/latex].<\/li>\r\n \t
                      4. Equal standard deviation [latex]\\frac{\\max \\{ s_1, s_2 \\} }{\\min \\{ s_1, s_2 \\}} = \\frac{\\max \\{ 17, 18 \\}}{\\min \\{ 17, 18 \\}} = \\frac{18}{17} < 2[\/latex].<\/li>\r\n<\/ol>\r\nIt is reasonable to conduct a pooled two-sample t-test since all the assumptions for pooled two-sample t-test\u00a0are met.\r\n\r\nSteps<\/strong>:\r\n
                          \r\n \t
                        1. \r\n
                            \r\n \t
                          1. Set up the hypotheses: [latex]H_0: \\mu_1 - \\mu_2 \\leq 0[\/latex] versus [latex]H_a: \\mu_1 - \\mu_2 \\: \\gt \\: 0[\/latex]. This is a right-tailed test.<\/li>\r\n \t
                          2. The significance level is [latex]\\alpha=0.01[\/latex].<\/li>\r\n \t
                          3. Compute the value of the test statistic:\r\n[latex]t_o = \\frac{(\\bar{x}_1 - \\bar{x}_2) - \\Delta_0}{s_p \\sqrt{ \\frac{1}{n_1} + \\frac{1}{n_2}}} = \\frac{(67-49) - 0}{17.207 \\sqrt{ \\frac{1}{135} + \\frac{1}{35}}} = 5.515[\/latex] with [latex]df = n_1 + n_2 \u2013 2 = 135+35\u20132=168[\/latex] (not given in Table IV, use df=100), and with\r\n[latex]s_p = \\sqrt{ \\frac{(n_1 - 1)s_1^2 + (n_2 -1)s_2^2 } {n_1 + n_2 -2}} = \u00a0\\sqrt{ \\frac{(135 - 1)17^2 + (35 -1)18^2 } {135 + 35 - 2}} = 17.207.[\/latex]<\/li>\r\n \t
                          4. Find the P-value. For a right-tailed test with the observed test statistics [latex]t_o=5.515[\/latex], the P-value is the area to the right of [latex]t_o[\/latex] i.e., p-value [latex]=P(t \\geq t_o) = P(t \\geq 5.515) < 0.0005[\/latex], since [latex]t_o=5.515 \\: \\gt \\: 3.390 (t_{0.0005})[\/latex] with [latex]df=100[\/latex].<\/li>\r\n \t
                          5. Decision: Since the P-value [latex]<0.0005< 0.01(\\alpha)[\/latex] reject the null hypothesis [latex]H_0[\/latex]<\/li>\r\n \t
                          6. Conclusion: At the 1% significance level, the data provide sufficient evidence that those who attend lectures have a higher average<\/strong>.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                          7. Obtain a confidence interval for the difference between the class average for attendees and non-attendees, [latex]\\mu_1 - \\mu_2[\/latex], corresponding to the test in part a).\r\nPart a) contains a right-tailed test at the 1% significance level. Therefore, we should obtain a 99% upper-tailed interval [latex]((\\bar{x}_1 - \\bar{x}_2) - t_{\\alpha} s_p \\sqrt{\\frac{1}{n_1} + \\frac{1}{n_2} }, \\infty)[\/latex], with [latex]\\alpha=0.01, df=100[\/latex], and [latex]t_{0.01}=2.364[\/latex]. The lower bound for the upper-tailed interval is [latex]\\begin{align*} (\\bar{x}_1 - \\bar{x}_2) - t_{\\alpha} s_p \\sqrt{\\frac{1}{n_1} + \\frac{1}{n_2} } &= (67 - 49) - 2.364 \\times 17.207 \\times \\sqrt{\\frac{1}{135} + \\frac{1}{35}} \\\\ &= 10.284. \\end{align*}[\/latex] Thus, the corresponding 99% confidence interval for [latex]\\mu_1 - \\mu_2[\/latex] is \u00a0[latex](10.284, \\infty)[\/latex].\r\nInterpretation<\/strong>: we are 99% confident that the difference in average grades is at least 10.284 between attendees and non-attendees.<\/li>\r\n \t
                          8. Based on the confidence interval in part b), can we claim that the class average of attendees is at least 10 marks higher than that of the non-attendees?\r\nYes, since the entire interval is above 10, we can claim that [latex]\\mu_1 - \\mu_2 \\: \\gt \\: 10[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n \r\n
                            \"\"<\/div>\r\n
                            <\/div>\r\n
                            \r\n
                            \r\n

                            Exercise: Two-Sample Test<\/p>\r\n\r\n<\/header>\r\n

                            \r\n\r\nThe following table summarizes the operative times of neurosurgeries conducted by a dynamic system (Z-plate) and a static system (ALPS plate).\r\n

                            Table 9.1<\/strong>: Operating Time of Dynamic and Static System<\/p>\r\n\r\n

                            \r\n\r\n\r\n\r\n\r\n\r\n\r\n
                            \r\n
                            Dynamic<\/strong><\/div><\/td>\r\n
                            		\r\n
                            Static<\/strong><\/div><\/td>\r\n<\/tr>\r\n
                            \r\n
                            [latex]\\bar{x}_1 = 400[\/latex]<\/div><\/td>\r\n
                            		\r\n
                            [latex]\\bar{x}_2 = 480[\/latex]<\/div><\/td>\r\n<\/tr>\r\n
                            \r\n
                            [latex]s_1 = 85[\/latex]<\/div><\/td>\r\n
                            		\r\n
                            [latex]s_2 = 40[\/latex]<\/div><\/td>\r\n<\/tr>\r\n
                            \r\n
                            [latex]n_1 = 60[\/latex]<\/div><\/td>\r\n
                            		\r\n
                            [latex]n_2 = 30[\/latex]<\/div><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n
                              \r\n \t
                            1. Test at the 5% significance level whether the dynamic system (Z-plate) has a lower mean operative time than the static system (ALPS plate).<\/li>\r\n \t
                            2. Obtain a confidence interval for the difference in mean operative time between the dynamic and the static systems, [latex]\\mu_1 - \\mu_2[\/latex], corresponding to the test in part a).<\/li>\r\n<\/ol>\r\n
                              Show\/Hide Answer<\/summary>\r\n
                                \r\n \t
                              1. Check the assumptions<\/strong>:\r\n
                                  \r\n \t
                                1. We have simple random samples.<\/li>\r\n \t
                                2. The two samples are independent.<\/li>\r\n \t
                                3. We have large sample sizes [latex]n_1 = 60 > 30, n_2 = 30 \\geq 30[\/latex].<\/li>\r\n \t
                                4. Equal standard deviations [latex]\\frac{\\max \\{ s_1, s_2 \\} }{\\min \\{ s_1, s_2 \\}} = \\frac{\\max \\{ 85, 40 \\}}{\\min \\{ 85, 40 \\}} = \\frac{85}{40} \\: \\gt \\: 2[\/latex].<\/li>\r\n<\/ol>\r\nSince the equal standard deviation assumption is violated, we should use the non-pooled two-sample t-test<\/i>.\r\n\r\nSteps<\/strong>:\r\n
                                    \r\n \t
                                  1. Set up the hypotheses: [latex]H_0: \\mu_1 - \\mu_2 \\geq 0[\/latex] versus [latex]H_a: \\mu_1 - \\mu_2 < 0[\/latex].\r\nThis is a left-tailed test.<\/li>\r\n \t
                                  2. The significance level is [latex]\\alpha = 0.05[\/latex].<\/li>\r\n \t
                                  3. Compute the value of the test statistic:[latex]t_o = \\frac{(\\bar{x}_1 - \\bar{x}_2) - \\Delta_0}{\\sqrt{ \\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2}}} = \\frac{(400 - 480) - 0}{\\sqrt{ \\frac{85^2}{60} + \\frac{40^2}{30}}} = -6.069[\/latex] with[latex]df = \\frac{ \\left( \\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2} \\right)^2}{\\frac{1}{n_1 - 1} \\left( \\frac{s_1^2}{n_1} \\right)^2 + \\frac{1}{n_2 - 1} \\left( \\frac{s_2^2}{n_2} \\right)^2 } = \\frac{ \\left( \\frac{85^2}{60} + \\frac{40^2}{30} \\right)^2}{\\frac{1}{60 - 1} \\left( \\frac{85^2}{60} \\right)^2 + \\frac{1}{30 - 1} \\left( \\frac{40^2}{30} \\right)^2 } = 87.797,[\/latex] rounded down to [latex]df = 87[\/latex].<\/li>\r\n \t
                                  4. Find the P-value. For a left-tailed test with the observed test statistics [latex]t_o = \u2013 6.069[\/latex], the P-value is the area to the left of [latex]t_o[\/latex], i.e., [latex]\\mbox{P-value} = P(t \\leq t_o) = P(t \\leq -6.069) = P(t \\geq 6.069) < 0.0005,[\/latex] since [latex]6.069 \\: \\gt \\: 3.406(t_{0.0005})[\/latex].<\/li>\r\n \t
                                  5. Decision: Since the P-value [latex] < 0.0005 < 0.05 (\\alpha)[\/latex], reject the null hypothesis [latex]H_0[\/latex].<\/li>\r\n \t
                                  6. Conclusion: At the 5% significance level, the data provide sufficient evidence that the dynamic system (Z-plate) has a lower mean operative time than the static system (ALPS plate).<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                  7. For a left-tailed test at the 5% significance level, the corresponding confidence interval is a 95% lower-tailed interval [latex]( - \\infty, (\\bar{x}_1 - \\bar{x}_2) + t_{\\alpha} \\sqrt{\\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2} })[\/latex] with the upper confidence bound [latex] (\\bar{x}_1 - \\bar{x}_2) + t_{\\alpha} \\sqrt{\\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2} } = (400 - 480) + 1.663 \\times \\sqrt{ \\frac{85^2}{60} + \\frac{40^2}{30}} = -58.079.[\/latex] Note that for [latex]df=87[\/latex], [latex]t_{0.05}=1.663[\/latex]. Therefore, the 95% lower-tailed interval is [latex]( - \\infty, (\\bar{x}_1 - \\bar{x}_2) + t_{\\alpha} \\sqrt{\\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2} }) = (- \\infty , -58.079) [\/latex].\r\nInterpretation<\/strong>: we are 95% confident that the difference in mean operative time between the dynamic and the static systems is below -58.097. Since the entire interval is below 0, we can claim that [latex]\\mu_1 - \\mu_2 < 0[\/latex], which supports the conclusion of the hypothesis test in part a).<\/li>\r\n<\/ol>\r\n<\/details><\/div>\r\n<\/div>\r\n<\/div>\r\n
                                    \"\"<\/div>\r\nIt is safer to use the non-pooled two-sample [latex]t[\/latex] test if we are not sure whether the two population standard deviations are equal. Use the pooled two-sample [latex]t[\/latex] test only if we have evidence that the population standard deviations are equal. For example, we can use the pooled two-sample [latex]t[\/latex] test when we compare two independent groups in a one-way ANOVA (analysis of variance) analysis since equal standard deviation is one of the assumptions of the one-way ANOVA F test which will be covered in Chapter 13.","rendered":"

                                    Two-sample t-tests<\/i> are used to test hypotheses regarding the difference between two population means. Depending on whether the two population standard deviations ([latex]\\sigma_1[\/latex] and [latex]\\sigma_2[\/latex]) are equal or not, we have the non-pooled and pooled two-sample t-tests\u00a0and t<\/em> interval. Minor advantages of the pooled t-test are a slightly narrower confidence interval, a slightly more powerful test, and a simpler formula for the degrees of freedom. However, the pooled t-test is valid only when the two population standard deviations are close; otherwise,\u00a0it gives poor results. Therefore, we recommend using the non-pooled t-test<\/i>\u00a0unless we are quite confident that [latex]\\sigma_1[\/latex] = [latex]\\sigma_2[\/latex], which is very difficult to verify.<\/p>\n

                                    9.2.1 Non-Pooled Two-Sample t<\/em> Test and t<\/em> Interval<\/strong><\/h2>\n
                                    \n

                                    Assumptions<\/strong>:<\/p>\n

                                      \n
                                    1. Simple random samples<\/li>\n
                                    2. Two samples are independent<\/li>\n
                                    3. Normal populations or large sample sizes [latex](n_1 \\geq 30, n_2 \\geq 30)[\/latex]<\/li>\n<\/ol>\n

                                      Steps<\/strong>:<\/p>\n

                                        \n
                                      1. Set up the hypotheses:\n
                                        \n\n\n\n\n\n\n
                                        \n

                                        Two-tailed test<\/strong><\/p>\n<\/td>\n

                                        		\n

                                        Right (upper)-tailed test<\/strong><\/p>\n<\/td>\n

                                        		\n

                                        Left (lower)-tailed test<\/strong><\/p>\n<\/td>\n<\/tr>\n<\/thead>\n

                                        \n
                                        [latex]H_0: \\mu_1 - \\mu_2 = \\Delta_0[\/latex]<\/div>\n<\/td>\n
                                        		\n
                                        [latex]H_0: \\mu_1 - \\mu_2 \\leq \\Delta_0[\/latex]<\/div>\n<\/td>\n
                                        		\n
                                        [latex]H_0: \\mu_1 - \\mu_2 \\geq \\Delta_0[\/latex]<\/div>\n<\/td>\n<\/tr>\n
                                        \n
                                        [latex]H_a: \\mu_1 - \\mu_2 \\neq \\Delta_0[\/latex]<\/div>\n<\/td>\n
                                        		\n
                                        [latex]H_a: \\mu_1 - \\mu_2 \\: \\gt \\: \\Delta_0[\/latex]<\/div>\n<\/td>\n
                                        		\n
                                        [latex]H_a: \\mu_1 - \\mu_2 \\: \\lt \\: \\Delta_0[\/latex]<\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

                                        Note that [latex]\\Delta_0[\/latex] can be zero or any value you want to test. In most cases, however, [latex]\\Delta_0=0[\/latex].<\/li>\n

                                      2. State the significance level [latex]\\alpha[\/latex].<\/li>\n
                                      3. Compute the value of the test statistic: [latex]t_o = \\frac{(\\bar{x}_1 - \\bar{x}_2) - (\\Delta_0)}{\\sqrt{ \\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2}}}[\/latex] with [latex]df = \\frac{ \\left( \\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2} \\right)^2}{\\frac{1}{n_1 - 1} \\left( \\frac{s_1^2}{n_1} \\right)^2 + \\frac{1}{n_2 - 1} \\left( \\frac{s_2^2}{n_2} \\right)^2 }[\/latex], rounded down<\/strong> to the nearest integer or [latex]\\min\\{n_1 -1, n_2 - 1 \\}[\/latex].<\/li>\n
                                      4. Use the t-score table (Table IV) to find the P-value or rejection region.\n
                                        \n\n\n\n\n\n\n\n\n
                                        <\/td>\n\n
                                        Two-tailed<\/div>\n<\/th>\n
                                        		\n
                                        Right-tailed<\/div>\n<\/th>\n
                                        		\n
                                        Left-tailed<\/div>\n<\/th>\n<\/tr>\n<\/thead>\n
                                        Null<\/th>\n\n
                                        [latex]H_0: \\mu_1 - \\mu_2 = \\Delta_0[\/latex]<\/div>\n<\/td>\n
                                        		\n
                                        [latex]H_0: \\mu_1 - \\mu_2 \\leq \\Delta_0[\/latex]<\/div>\n<\/td>\n
                                        		\n
                                        [latex]H_0: \\mu_1 - \\mu_2 \\geq \\Delta_0[\/latex]<\/div>\n<\/td>\n<\/tr>\n
                                        Alternative<\/th>\n\n
                                        [latex]H_a: \\mu_1 - \\mu_2 \\neq \\Delta_0[\/latex]<\/div>\n<\/td>\n
                                        		\n
                                        [latex]H_a: \\mu_1 - \\mu_2 \\: \\gt \\: \\Delta_0[\/latex]<\/div>\n<\/td>\n
                                        		\n
                                        [latex]H_a: \\mu_1 - \\mu_2 \\: \\lt \\: \\Delta_0[\/latex]<\/div>\n<\/td>\n<\/tr>\n
                                        P-value<\/th>\n\n
                                        [latex]2P(t \\geq |t_o|)[\/latex]<\/div>\n<\/td>\n
                                        		\n
                                        [latex]P(t \\geq t_o)[\/latex]<\/div>\n<\/td>\n
                                        		\n
                                        [latex]P(t \\leq t_o)[\/latex]<\/div>\n<\/td>\n<\/tr>\n
                                        Rejection region<\/th>\n[latex]t \\geq t_{\\alpha \/ 2}[\/latex] or [latex]t \\leq - t_{\\alpha \/ 2}[\/latex]<\/td>\n\n
                                        [latex]t \\geq t_{\\alpha}[\/latex]<\/div>\n<\/td>\n
                                        		\n
                                        [latex]t \\leq - t_{\\alpha}[\/latex]<\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/li>\n
                                      5. Decision: Reject the null [latex]H_0[\/latex] if P-value [latex]\\leq \\alpha[\/latex] or [latex]t_o[\/latex] falls in the rejection region.<\/li>\n
                                      6. Conclusion.<\/li>\n<\/ol>\n

                                        A [latex](1 \u2013\\alpha) \\times 100\\%[\/latex]\u00a0<\/em>two-sample [latex]t[\/latex] confidence interval for [latex]\\mu_1[\/latex] \u2013 [latex]\\mu_2[\/latex] is<\/p>\n

                                        \n\n\n\n\n\n\n\n
                                        \n
                                        Two-tailed<\/div>\n<\/th>\n
                                        		\n
                                        Right-tailed<\/div>\n<\/th>\n
                                        		\n
                                        Left-tailed<\/div>\n<\/th>\n<\/tr>\n<\/thead>\n
                                        \n
                                        [latex]H_0: \\mu_1 - \\mu_2 = \\Delta_0[\/latex]<\/div>\n<\/td>\n
                                        		\n
                                        [latex]H_0: \\mu_1 - \\mu_2 \\leq \\Delta_0[\/latex]<\/div>\n<\/td>\n
                                        		\n
                                        [latex]H_0: \\mu_1 - \\mu_2 \\geq \\Delta_0[\/latex]<\/div>\n<\/td>\n<\/tr>\n
                                        \n
                                        [latex]H_a: \\mu_1 - \\mu_2 \\neq \\Delta_0[\/latex]<\/div>\n<\/td>\n
                                        		\n
                                        [latex]H_a: \\mu_1 - \\mu_2 \\: \\gt \\: \\Delta_0[\/latex]<\/div>\n<\/td>\n
                                        		\n
                                        [latex]H_a: \\mu_1 - \\mu_2 \\: \\lt \\: \\Delta_0[\/latex]<\/div>\n<\/td>\n<\/tr>\n
                                        [latex](\\bar{x}_1 - \\bar{x}_2) \\pm t_{\\alpha \/ 2} \\sqrt{ \\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2}}[\/latex]<\/td>\n[latex]\\left((\\bar{x}_1 - \\bar{x}_2) - t_{\\alpha} \\sqrt{ \\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2}}, \\infty \\right)[\/latex]<\/td>\n[latex]\\left(- \\infty , (\\bar{x}_1 - \\bar{x}_2) + t_{\\alpha} \\sqrt{ \\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2}}\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n

                                         <\/p>\n

                                        \n
                                        \n

                                        Example: Two-Sample Non-Pooled t-Test and t Interval<\/p>\n<\/header>\n

                                        \n

                                        Some students attend class regularly, but some do not. An instructor wants to compare the class averages for those who attend lectures regularly ([latex]\\mu_1[\/latex]) with those who do not ([latex]\\mu_2[\/latex]). A simple random sample of size [latex]n_1 = 135[\/latex] is selected from the attendees and a simple random sample of size [latex]n_2=35[\/latex] is taken from the non-attendees. The sample mean and sample standard deviation for attendees are [latex]\\bar{x}_1 = 67, s_1 = 17[\/latex]; and for non-attendees are [latex]\\bar{x}_2 = 49, s_2 = 18[\/latex].<\/p>\n

                                          \n
                                        1. Test at the 1% significance level whether those who attend lectures have a higher average<\/strong>, i.e., [latex]\\mu_1 \\: \\gt \\: \\mu_2[\/latex] or [latex]\\mu_1 - \\mu_2 \\: \\gt \\: 0[\/latex].<\/li>\n<\/ol>\n

                                          \u00a0 \u00a0 \u00a0 \u00a0Check the assumptions<\/strong>:<\/span><\/p>\n

                                            \n
                                          1. We have simple random samples from attendees and non-attendees.<\/li>\n
                                          2. The two samples are independent.<\/li>\n
                                          3. We do not have the data, so we cannot check whether two populations are normally distributed using normal probability plot (Q-Q plot); however, we have large sample sizes with [latex]n_1 = 135 \\: \\gt \\: 30, n_2 = 35 \\: \\gt \\: 30[\/latex].<\/li>\n<\/ol>\n

                                            Therefore, the assumptions are met.
                                            \n
                                            \nSteps:<\/strong><\/p>\n

                                              \n
                                            1. Set up the hypotheses: [latex]H_0: \\mu_1 - \\mu_2 \\leq 0[\/latex] versus [latex]H_a: \\mu_1 - \\mu_2 \\: \\gt \\: 0[\/latex].
                                              \nThis is a right-tailed test.<\/li>\n
                                            2. The significance level is [latex]\\alpha=0.01[\/latex].<\/li>\n
                                            3. Compute the value of the test statistic:\n

                                              [latex]t_o = \\frac{(\\bar{x}_1 - \\bar{x}_2) - \\Delta_0}{\\sqrt{ \\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2}}} = \\frac{(67 - 49) -0}{\\sqrt{ \\frac{17^2}{135} + \\frac{18^2}{35}}} = 5.332[\/latex] with<\/p>\n

                                              [latex]df = \\frac{ \\left( \\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2} \\right)^2}{\\frac{1}{n_1 - 1} \\left( \\frac{s_1^2}{n_1} \\right)^2 + \\frac{1}{n_2 - 1} \\left( \\frac{s_2^2}{n_2} \\right)^2 } = \\frac{ \\left( \\frac{17^2}{135} + \\frac{18^2}{35} \\right)^2}{\\frac{1}{135 - 1} \\left( \\frac{17^2}{135} \\right)^2 + \\frac{1}{35 - 1} \\left( \\frac{18^2}{35} \\right)^2 } = 50.85[\/latex], rounded down to [latex]df = 50[\/latex].<\/li>\n

                                            4. Find the P-value. For a right-tailed test with the observed test statistics [latex]t_o=5.332[\/latex], the P-value is the area to the right of [latex]t_o[\/latex], i.e., [latex]\\mbox{P-value} = P(t \\geq t_o) = P(t \\geq 5.332) < 0.0005, \\mbox{since} t_o=5.332 \\: \\gt \\: 3.496(t_{0.0005})[\/latex]<\/li>\n
                                            5. Decision: Since the P-value [latex]< 0.0005 < 0.01 (\\alpha)[\/latex], reject the null hypothesis [latex]H_0[\/latex].<\/li>\n
                                            6. Conclusion: At the 1% significance level, the data provide sufficient evidence that those who attend lectures have a higher average<\/strong>.<\/li>\n<\/ol>\n

                                              If using the critical value approach, steps 1-3 are the same, steps 4-6 become:<\/p>\n

                                                \n
                                              1. \n
                                                  \n
                                                1. Rejection region:<\/a>\n
                                                  \n\n\n\n
                                                  \n
                                                  \"A
                                                  Figure 9.2<\/strong>: Rejection Region and Observed Value. [Image Description (See Appendix D Figure 9.2)<\/a>]<\/figcaption><\/figure>\n<\/td>\n
                                                  		[latex]\\alpha = 0.01 , t_{\\alpha} = t_{0.01} = 2.403[\/latex]<\/p>\n

                                                  For a right-tailed test, the critical value is 2.403. The rejection region is to the right of 2.403.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/li>\n

                                                2. Decision: Since the observed value [latex]t_o =5.332 \\: \\gt \\: 2.403[\/latex] falls in the rejection region, we reject the null hypothesis [latex]H_0[\/latex].<\/li>\n
                                                3. Conclusion: At the 1% significance level, the data provide sufficient evidence that those who attend lectures have a higher average<\/strong>.<\/li>\n<\/ol>\n<\/li>\n
                                                4. Obtain a confidence interval for the difference between the class average for attendees and non-attendees [latex]\\mu_1 - \\mu_2[\/latex] corresponding to the test in part a).
                                                  \nPart a) contains a right-tailed test at the 1% significance level. Therefore, we should obtain a 99% upper-tailed interval: [latex]\\left((\\bar{x}_1 - \\bar{x}_2) - t_{\\alpha} \\sqrt{ \\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2}}, \\infty \\right)[\/latex].
                                                  \n[latex]\\alpha = 0.01, df = 50, t_{\\alpha} = t_{0.01} = 2.403[\/latex].
                                                  \nThe lower bound for the upper-tailed interval is:<\/p>\n

                                                  [latex](\\bar{x}_1 - \\bar{x}_2) - t_{\\alpha} \\sqrt{ \\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2}} = (67 - 49) - 2.403 \\times \\sqrt{ \\frac{17^2}{135} + \\frac{18^2}{35}} = 9.887[\/latex].<\/p>\n

                                                  Thus, the corresponding 99% confidence interval for [latex]\\mu_1 - \\mu_2[\/latex] is [latex](9.887, \\infty)[\/latex].
                                                  \nInterpretation: we are 99% confident that the difference in average grades is at least 9.887 between attendees and non-attendees.<\/li>\n

                                                5. Does the interval in part (b) support the conclusion in part a)?
                                                  \nIn part a), we reject [latex]H_0[\/latex] at the 1% significance level and claim that [latex]\\mu_1 - \\mu_2 \\: \\gt \\: 0[\/latex].
                                                  \nIn part b), since the entire interval is above 0, we can claim that [latex]\\mu_1 - \\mu_2 \\: \\gt \\: 0[\/latex] with 99% confidence, which supports the results obtained in part a).<\/li>\n
                                                6. Based on the interval obtained in part b), can we claim that the class average of attendees is at least 5 marks higher than that of the non-attendees? How about 10 marks higher?
                                                  \nWe can claim that the class average of attendees is at least 5 marks higher than that of the non-attendees since the entire interval is above 5. However, we cannot claim that the class average of attendees is at least 10 marks higher than that of the non-attendees since the interval contains 10.<\/a><\/p>\n
                                                  \"A
                                                  Figure 9.3<\/strong>: Confidence Interval of difference in Class Average. [Image Description (See Appendix D Figure 9.3)<\/a>]<\/figcaption><\/figure>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n

                                                  9.2.2 Pooled Two-Sample t<\/em> Test and t<\/em> Interval<\/strong><\/h2>\n

                                                  If the two population standard deviations are equal, i.e., [latex]\\sigma_1 = \\sigma_2 = \\sigma[\/latex], we can pool the two samples together to get a better estimate of the common standard deviation [latex]\\sigma[\/latex]<\/p>\n

                                                  [latex]\\hat{\\sigma} = s_p = \\sqrt{ \\frac{(n_1 - 1)s_1^2 + (n_2 -1)s_2^2 } {(n_1 - 1) + (n_2 -1)}}[\/latex]<\/p>\n

                                                  where the term [latex](n_1 - 1)s_1^2 = \\sum_{\\text{sample 1}}(x - \\bar{x}_1)^2[\/latex] is the variation of the data within sample 1, and [latex](n_2 - 1)s_2^2 = \\sum_{\\text{sample 2}}(x - \\bar{x}_2)^2[\/latex] is the variation of the data within sample 2.<\/p>\n

                                                  Recall that the standard deviation of [latex]\\bar{X}_1 - \\bar{X}_2[\/latex] is [latex]\\sigma_{\\scriptsize \\bar{X}_1 - \\bar{X}_2} = \\sqrt{ \\frac{\\sigma_1^2}{n_1} + \\frac{\\sigma_2^2}{n_2} }[\/latex]. Thus, if [latex]\\sigma_1 = \\sigma_2 = \\sigma[\/latex], then [latex]\\sigma_{\\scriptsize \\bar{X}_1 - \\bar{X}_2}[\/latex] reduces to [latex]\\sqrt{ \\frac{\\sigma^2}{n_1} + \\frac{\\sigma^2}{n_2} } = \\sigma \\sqrt{ \\frac{1}{n_1} + \\frac{1}{n_2} }[\/latex]. Estimating [latex]\\sigma[\/latex] with [latex]s_p[\/latex] leads to the pooled test statistic:<\/p>\n

                                                  [latex]t = \\frac{(\\bar{X}_1 - \\bar{X}_2) - (\\mu_1 - \\mu_2)}{s_p \\sqrt{\\frac{1}{n_1} + \\frac{1}{n_2} } } \\sim \\text{$t$ distribution}[\/latex]<\/p>\n

                                                  with [latex]df = (n_1 - 1) + (n_2 -1) = n_1 + n_2 -2[\/latex].<\/p>\n

                                                  The assumption [latex]\\sigma_1 = \\sigma_2[\/latex] is very difficult to verify. Some textbooks suggest a rule of thumb:<\/p>\n

                                                  If the ratio of the larger to the smaller sample standard deviation is less than 2, then the assumption is considered to be reasonable, i.e., [latex]\\frac{\\max \\{ s_1, s_2\\}}{\\min \\{ s_1, s_2\\}} < 2[\/latex]<\/strong>.<\/p>\n

                                                  \n

                                                  Assumptions<\/strong>:<\/p>\n

                                                    \n
                                                  1. Simple random samples.<\/li>\n
                                                  2. Independent samples.<\/li>\n
                                                  3. Normal populations or large sample sizes [latex](n_1 \\geq 30, n_2 \\geq 30)[\/latex].<\/li>\n
                                                  4. Equal population standard deviations. This assumption is reasonable if [latex]\\frac{\\max \\{ s_1, s_2\\}}{\\min \\{ s_1, s_2\\}} < 2[\/latex].<\/li>\n<\/ol>\n

                                                    Steps<\/strong>:<\/p>\n

                                                      \n
                                                    1. Set up the hypotheses:\n
                                                      \n\n\n\n\n\n
                                                      Two-tailed test<\/strong><\/td>\nRight (upper)-tailed test<\/strong><\/td>\nLeft (lower)-tailed test<\/strong><\/td>\n<\/tr>\n
                                                      \n
                                                      [latex]H_0: \\mu_1 - \\mu_2 = \\Delta_0[\/latex]<\/div>\n<\/td>\n
                                                      		\n
                                                      [latex]H_0: \\mu_1 - \\mu_2 \\leq \\Delta_0[\/latex]<\/div>\n<\/td>\n
                                                      		\n
                                                      [latex]H_0: \\mu_1 - \\mu_2 \\geq \\Delta_0[\/latex]<\/div>\n<\/td>\n<\/tr>\n
                                                      \n
                                                      [latex]H_a: \\mu_1 - \\mu_2 \\neq \\Delta_0[\/latex]<\/div>\n<\/td>\n
                                                      		\n
                                                      [latex]H_a: \\mu_1 - \\mu_2 \\: \\gt \\: \\Delta_0[\/latex]<\/div>\n<\/td>\n
                                                      		\n
                                                      [latex]H_a: \\mu_1 - \\mu_2 \\: \\lt \\: \\Delta_0[\/latex]<\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

                                                      Note that [latex]\\Delta_0[\/latex] can be zero or any value you want to test.<\/li>\n

                                                    2. State the significance level [latex]\\alpha[\/latex].<\/li>\n
                                                    3. Compute the value of the test statistic: [latex]t_o = \\frac{(\\bar{x}_1 - \\bar{x}_2) - \\Delta_0}{s_p \\sqrt{ \\frac{1}{n_1} + \\frac{1}{n_2}}}[\/latex], with [latex]df = n_1 + n_2 \u2013 2[\/latex] and [latex]s_p = \\sqrt{ \\frac{(n_1 - 1)s_1^2 + (n_2 -1)s_2^2 } {(n_1 - 1) + (n_2 -1)}}[\/latex].<\/li>\n
                                                    4. Use the t-score table (Table IV) to find the P-value or rejection region.
                                                      \n\n\n\n\n\n\n\n\n
                                                      <\/td>\n\n
                                                      Two-tailed<\/div>\n<\/th>\n
                                                      		\n
                                                      Right-tailed<\/div>\n<\/th>\n
                                                      		\n
                                                      Left-tailed<\/div>\n<\/th>\n<\/tr>\n<\/thead>\n
                                                      Null<\/th>\n\n
                                                      [latex]H_0: \\mu_1 - \\mu_2 = \\Delta_0[\/latex]<\/div>\n<\/td>\n
                                                      		\n
                                                      [latex]H_0: \\mu_1 - \\mu_2 \\leq \\Delta_0[\/latex]<\/div>\n<\/td>\n
                                                      		\n
                                                      [latex]H_0: \\mu_1 - \\mu_2 \\geq \\Delta_0[\/latex]<\/div>\n<\/td>\n<\/tr>\n
                                                      Alternative<\/th>\n\n
                                                      [latex]H_a: \\mu_1 - \\mu_2 \\neq \\Delta_0[\/latex]<\/div>\n<\/td>\n
                                                      		\n
                                                      [latex]H_a: \\mu_1 - \\mu_2 \\: \\gt \\: \\Delta_0[\/latex]<\/div>\n<\/td>\n
                                                      		\n
                                                      [latex]H_a: \\mu_1 - \\mu_2 \\: \\lt \\: \\Delta_0[\/latex]<\/div>\n<\/td>\n<\/tr>\n
                                                      P-value<\/th>\n\n
                                                      [latex]2P(t \\geq |t_o|)[\/latex]<\/div>\n<\/td>\n
                                                      		\n
                                                      [latex]P(t \\geq t_o)[\/latex]<\/div>\n<\/td>\n
                                                      		\n
                                                      [latex]P(t \\leq t_o)[\/latex]<\/div>\n<\/td>\n<\/tr>\n
                                                      Rejection region<\/th>\n[latex]t \\geq t_{\\alpha \/ 2}[\/latex] or [latex]t \\leq - t_{\\alpha \/ 2}[\/latex]<\/td>\n\n
                                                      [latex]t \\geq t_{\\alpha}[\/latex]<\/div>\n<\/td>\n
                                                      		\n
                                                      [latex]t \\leq - t_{\\alpha}[\/latex]<\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n
                                                    5. Decision: Reject the null [latex]H_0[\/latex] if P-value [latex]\\leq \\alpha[\/latex] or [latex]t_o[\/latex] falls in the rejection region.<\/li>\n
                                                    6. Conclusion.<\/li>\n<\/ol>\n

                                                      A [latex](1 - \\alpha) \\times 100%[\/latex] two-sample t<\/em> confidence interval for [latex]\\mu_1 - \\mu_2[\/latex] is<\/p>\n

                                                      \n\n\n\n\n\n\n\n
                                                      \n
                                                      Two-tailed<\/div>\n<\/th>\n
                                                      		\n
                                                      Right-tailed<\/div>\n<\/th>\n
                                                      		\n
                                                      Left-tailed<\/div>\n<\/th>\n<\/tr>\n<\/thead>\n
                                                      \n
                                                      [latex]H_0: \\mu_1 - \\mu_2 = \\Delta_0[\/latex]<\/div>\n<\/td>\n
                                                      		\n
                                                      [latex]H_0: \\mu_1 - \\mu_2 \\leq \\Delta_0[\/latex]<\/div>\n<\/td>\n
                                                      		\n
                                                      [latex]H_0: \\mu_1 - \\mu_2 \\geq \\Delta_0[\/latex]<\/div>\n<\/td>\n<\/tr>\n
                                                      \n
                                                      [latex]H_a: \\mu_1 - \\mu_2 \\neq \\Delta_0[\/latex]<\/div>\n<\/td>\n
                                                      		\n
                                                      [latex]H_a: \\mu_1 - \\mu_2 \\: \\gt \\: \\Delta_0[\/latex]<\/div>\n<\/td>\n
                                                      		\n
                                                      [latex]H_a: \\mu_1 - \\mu_2 \\: \\lt \\: \\Delta_0[\/latex]<\/div>\n<\/td>\n<\/tr>\n
                                                      [latex]\\small{(\\bar{x}_1 - \\bar{x}_2) \\pm t_{\\alpha \/ 2} s_p \\sqrt{ \\frac{1}{n_1} + \\frac{1}{n_2}}}[\/latex]<\/td>\n[latex]\\small{\\left((\\bar{x}_1 - \\bar{x}_2) - t_{\\alpha} s_p \\sqrt{ \\frac{1}{n_1} + \\frac{1}{n_2}}, \\infty \\right)}[\/latex]<\/td>\n[latex]\\small{\\left(- \\infty , (\\bar{x}_1 - \\bar{x}_2) + t_{\\alpha} \u00a0s_p \\sqrt{ \\frac{1}{n_1} + \\frac{1}{n_2}}\\right)}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n

                                                       <\/p>\n

                                                      \n
                                                      \n

                                                      Example: Pooled Two-Sample t Test and Interval<\/p>\n<\/header>\n

                                                      \n

                                                      Some students attend class regularly, but some do not. An instructor wants to compare the class averages for those who attend lectures regularly ([latex]\\mu_1[\/latex]) with those who do not ([latex]\\mu_2[\/latex]). A simple random sample of size [latex]n_1=135[\/latex] is selected from the attendees, and a simple random sample of size [latex]n_2=35[\/latex] is taken from the non-attendees. The sample mean and sample standard deviation for attendees are [latex]\\bar{x}_1 = 67, s_1 = 17[\/latex]; and for non-attendees are [latex]\\bar{x}_2 = 49, s_2 = 18[\/latex].<\/p>\n

                                                        \n
                                                      1. Is it reasonable to conduct a pooled two-sample t-test<\/i>\u00a0to test whether those who attend lectures have a higher average<\/strong>? If yes, run the test at the 1% significance level.<\/li>\n<\/ol>\n

                                                        Check the assumptions<\/strong>:<\/p>\n

                                                          \n
                                                        1. We have simple random samples.<\/li>\n
                                                        2. The two samples are independent.<\/li>\n
                                                        3. We have large sample sizes [latex](n_1 = 135 > 30, n_2 =35 > 30)[\/latex].<\/li>\n
                                                        4. Equal standard deviation [latex]\\frac{\\max \\{ s_1, s_2 \\} }{\\min \\{ s_1, s_2 \\}} = \\frac{\\max \\{ 17, 18 \\}}{\\min \\{ 17, 18 \\}} = \\frac{18}{17} < 2[\/latex].<\/li>\n<\/ol>\n

                                                          It is reasonable to conduct a pooled two-sample t-test since all the assumptions for pooled two-sample t-test\u00a0are met.
                                                          \n
                                                          \nSteps<\/strong>:<\/p>\n

                                                            \n
                                                          1. \n
                                                              \n
                                                            1. Set up the hypotheses: [latex]H_0: \\mu_1 - \\mu_2 \\leq 0[\/latex] versus [latex]H_a: \\mu_1 - \\mu_2 \\: \\gt \\: 0[\/latex]. This is a right-tailed test.<\/li>\n
                                                            2. The significance level is [latex]\\alpha=0.01[\/latex].<\/li>\n
                                                            3. Compute the value of the test statistic:
                                                              \n[latex]t_o = \\frac{(\\bar{x}_1 - \\bar{x}_2) - \\Delta_0}{s_p \\sqrt{ \\frac{1}{n_1} + \\frac{1}{n_2}}} = \\frac{(67-49) - 0}{17.207 \\sqrt{ \\frac{1}{135} + \\frac{1}{35}}} = 5.515[\/latex] with [latex]df = n_1 + n_2 \u2013 2 = 135+35\u20132=168[\/latex] (not given in Table IV, use df=100), and with
                                                              \n[latex]s_p = \\sqrt{ \\frac{(n_1 - 1)s_1^2 + (n_2 -1)s_2^2 } {n_1 + n_2 -2}} = \u00a0\\sqrt{ \\frac{(135 - 1)17^2 + (35 -1)18^2 } {135 + 35 - 2}} = 17.207.[\/latex]<\/li>\n
                                                            4. Find the P-value. For a right-tailed test with the observed test statistics [latex]t_o=5.515[\/latex], the P-value is the area to the right of [latex]t_o[\/latex] i.e., p-value [latex]=P(t \\geq t_o) = P(t \\geq 5.515) < 0.0005[\/latex], since [latex]t_o=5.515 \\: \\gt \\: 3.390 (t_{0.0005})[\/latex] with [latex]df=100[\/latex].<\/li>\n
                                                            5. Decision: Since the P-value [latex]<0.0005< 0.01(\\alpha)[\/latex] reject the null hypothesis [latex]H_0[\/latex]<\/li>\n
                                                            6. Conclusion: At the 1% significance level, the data provide sufficient evidence that those who attend lectures have a higher average<\/strong>.<\/li>\n<\/ol>\n<\/li>\n
                                                            7. Obtain a confidence interval for the difference between the class average for attendees and non-attendees, [latex]\\mu_1 - \\mu_2[\/latex], corresponding to the test in part a).
                                                              \nPart a) contains a right-tailed test at the 1% significance level. Therefore, we should obtain a 99% upper-tailed interval [latex]((\\bar{x}_1 - \\bar{x}_2) - t_{\\alpha} s_p \\sqrt{\\frac{1}{n_1} + \\frac{1}{n_2} }, \\infty)[\/latex], with [latex]\\alpha=0.01, df=100[\/latex], and [latex]t_{0.01}=2.364[\/latex]. The lower bound for the upper-tailed interval is [latex]\\begin{align*} (\\bar{x}_1 - \\bar{x}_2) - t_{\\alpha} s_p \\sqrt{\\frac{1}{n_1} + \\frac{1}{n_2} } &= (67 - 49) - 2.364 \\times 17.207 \\times \\sqrt{\\frac{1}{135} + \\frac{1}{35}} \\\\ &= 10.284. \\end{align*}[\/latex] Thus, the corresponding 99% confidence interval for [latex]\\mu_1 - \\mu_2[\/latex] is \u00a0[latex](10.284, \\infty)[\/latex].
                                                              \nInterpretation<\/strong>: we are 99% confident that the difference in average grades is at least 10.284 between attendees and non-attendees.<\/li>\n
                                                            8. Based on the confidence interval in part b), can we claim that the class average of attendees is at least 10 marks higher than that of the non-attendees?
                                                              \nYes, since the entire interval is above 10, we can claim that [latex]\\mu_1 - \\mu_2 \\: \\gt \\: 10[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n

                                                               <\/p>\n

                                                              \"\"<\/div>\n
                                                              <\/div>\n
                                                              \n
                                                              \n
                                                              \n

                                                              Exercise: Two-Sample Test<\/p>\n<\/header>\n

                                                              \n

                                                              The following table summarizes the operative times of neurosurgeries conducted by a dynamic system (Z-plate) and a static system (ALPS plate).<\/p>\n

                                                              Table 9.1<\/strong>: Operating Time of Dynamic and Static System<\/p>\n

                                                              \n\n\n\n\n\n\n
                                                              \n
                                                              Dynamic<\/strong><\/div>\n<\/td>\n
                                                              		\n
                                                              Static<\/strong><\/div>\n<\/td>\n<\/tr>\n
                                                              \n
                                                              [latex]\\bar{x}_1 = 400[\/latex]<\/div>\n<\/td>\n
                                                              		\n
                                                              [latex]\\bar{x}_2 = 480[\/latex]<\/div>\n<\/td>\n<\/tr>\n
                                                              \n
                                                              [latex]s_1 = 85[\/latex]<\/div>\n<\/td>\n
                                                              		\n
                                                              [latex]s_2 = 40[\/latex]<\/div>\n<\/td>\n<\/tr>\n
                                                              \n
                                                              [latex]n_1 = 60[\/latex]<\/div>\n<\/td>\n
                                                              		\n
                                                              [latex]n_2 = 30[\/latex]<\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n
                                                                \n
                                                              1. Test at the 5% significance level whether the dynamic system (Z-plate) has a lower mean operative time than the static system (ALPS plate).<\/li>\n
                                                              2. Obtain a confidence interval for the difference in mean operative time between the dynamic and the static systems, [latex]\\mu_1 - \\mu_2[\/latex], corresponding to the test in part a).<\/li>\n<\/ol>\n
                                                                \nShow\/Hide Answer<\/summary>\n
                                                                  \n
                                                                1. Check the assumptions<\/strong>:\n
                                                                    \n
                                                                  1. We have simple random samples.<\/li>\n
                                                                  2. The two samples are independent.<\/li>\n
                                                                  3. We have large sample sizes [latex]n_1 = 60 > 30, n_2 = 30 \\geq 30[\/latex].<\/li>\n
                                                                  4. Equal standard deviations [latex]\\frac{\\max \\{ s_1, s_2 \\} }{\\min \\{ s_1, s_2 \\}} = \\frac{\\max \\{ 85, 40 \\}}{\\min \\{ 85, 40 \\}} = \\frac{85}{40} \\: \\gt \\: 2[\/latex].<\/li>\n<\/ol>\n

                                                                    Since the equal standard deviation assumption is violated, we should use the non-pooled two-sample t-test<\/i>.<\/p>\n

                                                                    Steps<\/strong>:<\/p>\n

                                                                      \n
                                                                    1. Set up the hypotheses: [latex]H_0: \\mu_1 - \\mu_2 \\geq 0[\/latex] versus [latex]H_a: \\mu_1 - \\mu_2 < 0[\/latex].\nThis is a left-tailed test.<\/li>\n
                                                                    2. The significance level is [latex]\\alpha = 0.05[\/latex].<\/li>\n
                                                                    3. Compute the value of the test statistic:[latex]t_o = \\frac{(\\bar{x}_1 - \\bar{x}_2) - \\Delta_0}{\\sqrt{ \\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2}}} = \\frac{(400 - 480) - 0}{\\sqrt{ \\frac{85^2}{60} + \\frac{40^2}{30}}} = -6.069[\/latex] with[latex]df = \\frac{ \\left( \\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2} \\right)^2}{\\frac{1}{n_1 - 1} \\left( \\frac{s_1^2}{n_1} \\right)^2 + \\frac{1}{n_2 - 1} \\left( \\frac{s_2^2}{n_2} \\right)^2 } = \\frac{ \\left( \\frac{85^2}{60} + \\frac{40^2}{30} \\right)^2}{\\frac{1}{60 - 1} \\left( \\frac{85^2}{60} \\right)^2 + \\frac{1}{30 - 1} \\left( \\frac{40^2}{30} \\right)^2 } = 87.797,[\/latex] rounded down to [latex]df = 87[\/latex].<\/li>\n
                                                                    4. Find the P-value. For a left-tailed test with the observed test statistics [latex]t_o = \u2013 6.069[\/latex], the P-value is the area to the left of [latex]t_o[\/latex], i.e., [latex]\\mbox{P-value} = P(t \\leq t_o) = P(t \\leq -6.069) = P(t \\geq 6.069) < 0.0005,[\/latex] since [latex]6.069 \\: \\gt \\: 3.406(t_{0.0005})[\/latex].<\/li>\n
                                                                    5. Decision: Since the P-value [latex]< 0.0005 < 0.05 (\\alpha)[\/latex], reject the null hypothesis [latex]H_0[\/latex].<\/li>\n
                                                                    6. Conclusion: At the 5% significance level, the data provide sufficient evidence that the dynamic system (Z-plate) has a lower mean operative time than the static system (ALPS plate).<\/li>\n<\/ol>\n<\/li>\n
                                                                    7. For a left-tailed test at the 5% significance level, the corresponding confidence interval is a 95% lower-tailed interval [latex]( - \\infty, (\\bar{x}_1 - \\bar{x}_2) + t_{\\alpha} \\sqrt{\\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2} })[\/latex] with the upper confidence bound [latex](\\bar{x}_1 - \\bar{x}_2) + t_{\\alpha} \\sqrt{\\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2} } = (400 - 480) + 1.663 \\times \\sqrt{ \\frac{85^2}{60} + \\frac{40^2}{30}} = -58.079.[\/latex] Note that for [latex]df=87[\/latex], [latex]t_{0.05}=1.663[\/latex]. Therefore, the 95% lower-tailed interval is [latex]( - \\infty, (\\bar{x}_1 - \\bar{x}_2) + t_{\\alpha} \\sqrt{\\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2} }) = (- \\infty , -58.079)[\/latex].
                                                                      \nInterpretation<\/strong>: we are 95% confident that the difference in mean operative time between the dynamic and the static systems is below -58.097. Since the entire interval is below 0, we can claim that [latex]\\mu_1 - \\mu_2 < 0[\/latex], which supports the conclusion of the hypothesis test in part a).<\/li>\n<\/ol>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n
                                                                      \"\"<\/div>\n

                                                                      It is safer to use the non-pooled two-sample [latex]t[\/latex] test if we are not sure whether the two population standard deviations are equal. Use the pooled two-sample [latex]t[\/latex] test only if we have evidence that the population standard deviations are equal. For example, we can use the pooled two-sample [latex]t[\/latex] test when we compare two independent groups in a one-way ANOVA (analysis of variance) analysis since equal standard deviation is one of the assumptions of the one-way ANOVA F test which will be covered in Chapter 13.<\/p>\n","protected":false},"author":19,"menu_order":2,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1017","chapter","type-chapter","status-publish","hentry"],"part":1007,"_links":{"self":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/1017","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/users\/19"}],"version-history":[{"count":108,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/1017\/revisions"}],"predecessor-version":[{"id":5595,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/1017\/revisions\/5595"}],"part":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/parts\/1007"}],"metadata":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/1017\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/media?parent=1017"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapter-type?post=1017"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/contributor?post=1017"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/license?post=1017"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}