{"id":1104,"date":"2021-06-12T20:11:39","date_gmt":"2021-06-13T00:11:39","guid":{"rendered":"https:\/\/openbooks.macewan.ca\/rcommander\/?post_type=chapter&p=1104"},"modified":"2024-02-08T14:31:51","modified_gmt":"2024-02-08T19:31:51","slug":"10-5-one-proportion-z-test-for-p","status":"publish","type":"chapter","link":"https:\/\/openbooks.macewan.ca\/introstats\/chapter\/10-5-one-proportion-z-test-for-p\/","title":{"raw":"10.5 One-Proportion z Test for p","rendered":"10.5 One-Proportion z Test for p"},"content":{"raw":"The assumptions and steps of a one-proportion [latex]z[\/latex] test are as follows.\r\n
\r\n\r\nAssumptions<\/strong>:\r\n
    \r\n \t
  1. A simple random sample.<\/li>\r\n \t
  2. Both [latex]np_0[\/latex] and [latex]n(1-p_0)[\/latex] are at least 5, where [latex]p_0[\/latex] is the hypothesized value of [latex]p[\/latex] under the null [latex]H_0[\/latex].<\/li>\r\n<\/ol>\r\nSteps to perform a one-proportion <\/strong>z test<\/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<\/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: p = p_0[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]H_0: p \\leq p_0[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]H_0: p \\geq p_0[\/latex]<\/div><\/td>\r\n<\/tr>\r\n
      \r\n
      [latex]H_a: p \\neq p_0[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]H_a: p \\: \\gt \\:p_0[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]H_a: p \\: \\lt \\:p_0[\/latex]<\/div><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div><\/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]z_o = \\frac{\\hat{p} - p_0}{\\sqrt{\\frac{p_0 (1 - p_0)}{n} }}[\/latex], with [latex]\\hat{p} = \\frac{x}{n}[\/latex].<\/li>\r\n \t
    4. Find the P-value or<\/strong> 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: p = p_0[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]H_0: p \\leq p_0[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]H_0: p \\geq p_0[\/latex]<\/div><\/td>\r\n<\/tr>\r\n
      Alternative<\/th>\r\n\r\n
      [latex]H_a: p \\neq p_0[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]H_a: p \\: \\gt \\:p_0[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]H_a: p \\: \\lt \\:p_0[\/latex]<\/div><\/td>\r\n<\/tr>\r\n
      P-value<\/th>\r\n\r\n
      [latex]2P(Z \\geq |z_o|)[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]P(Z \\geq z_o)[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]P(Z \\leq z_o)[\/latex]<\/div><\/td>\r\n<\/tr>\r\n
      Rejection region<\/th>\r\n[latex]Z \\geq z_{\\alpha \/ 2}[\/latex] or [latex]Z \\leq - z_{\\alpha \/ 2}[\/latex]<\/td>\r\n\r\n
      [latex]Z \\geq z_{\\alpha }[\/latex]<\/div><\/td>\r\n
      		\r\n
      [latex]Z \\leq - z_{\\alpha }[\/latex]<\/div><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n \t
    5. Reject the null [latex]H_0[\/latex] if the P-value [latex] \\leq \\alpha[\/latex] or [latex]z_o[\/latex] falls in the rejection region.<\/li>\r\n \t
    6. Conclusion.<\/li>\r\n<\/ol>\r\n<\/div>\r\n \r\n
      \r\n

      Example: One-Proportion z Test<\/p>\r\n\r\n<\/header>\r\n

      \r\n\r\nLet [latex]p[\/latex] be the proportion of athletes wearing blue suits who win a judo match. Randomly select n<\/em> = 100 Olympic judo matches and suppose 55 winners wore a blue suit. The other 45 wore a white suit.\r\n
        \r\n \t
      1. Test at the 5% significance level whether a color bias exists.<\/strong>\r\nIf there is no colour bias, the proportions of blue and white winners should be 0.5 and 0.5. Therefore, letting p<\/em> be the proportion of winners in blue, the hypotheses are [latex]H_0: p = 0.5[\/latex] versus [latex]H_a: p \\neq 0.5[\/latex].\r\nCheck the assumptions<\/strong>:\r\n
          \r\n \t
        1. We have a simple random sample (SRS).<\/li>\r\n \t
        2. Both [latex]np_0 = 100 \\times 0.5 = 50[\/latex] and [latex]n (1 - p_0) = 100 \\times (1-0.5) = 50[\/latex] are at least 5.<\/li>\r\n<\/ol>\r\nSteps<\/strong>:\r\n
            \r\n \t
          1. Set up the hypotheses. [latex]H_0: p = 0.5[\/latex] versus [latex]H_a: p \\neq 0.5[\/latex]<\/li>\r\n \t
          2. State the significance level is [latex]\\alpha = 0.05[\/latex].<\/li>\r\n \t
          3. Compute the test statistic:\r\n

            [latex]\\hat{p} = \\frac{x}{n} = \\frac{55}{100} = 0.55, z_o = \\frac{\\hat{p} - p_0}{\\sqrt{\\frac{p_0 (1 - p_0)}{n}}} = \\frac{0.55 - 0.5}{ \\sqrt{ \\frac{0.5(1-0.5)}{100} }} = 1.[\/latex]<\/p>\r\n<\/li>\r\n \t

          4. Find the P-value. For a two-tailed test, the P-value is twice the area to the right of the absolute value of the observed test statistic [latex]z_o[\/latex]. That is:\r\nP-value = [latex]2P(Z \\geq |z_o|)[\/latex] [latex]= 2P(Z \\geq 1)[\/latex] [latex]= 2P(Z \\leq -1)[\/latex] [latex]= 2 \\times 0.1587[\/latex] [latex]= 0.3174[\/latex].<\/li>\r\n \t
          5. Decision: Since the P-value [latex] = 0.3174 \\: \\gt \\: 0.05(\\alpha)[\/latex], we cannot reject the null [latex]H_0[\/latex].<\/li>\r\n \t
          6. Conclusion: At the 5% significance level, we do not have sufficient evidence that color bias exists in judging Olympic Judo matches.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
          7. Obtain a confidence interval <\/strong>corresponding to the test in part (a).\r\n<\/strong>For a two-tailed test at the 5% significance level, we obtain a 95% two-tailed interval.\r\nThe sample proportion is [latex]\\hat{p} = \\frac{x}{n} = \\frac{55}{100} = 0.55[\/latex].\r\n

            [latex]1 - \\alpha = 0.95 \\Longrightarrow \\alpha = 0.05 \\Longrightarrow z_{\\alpha \/ 2} = z_{0.025} = 1.96[\/latex].<\/p>\r\nA 95% confidence interval for the proportion of winners in blue is\r\n

            [latex]\\hat{p} \\pm z_{\\alpha \/ 2} \\sqrt{\\frac{\\hat{p} (1 - \\hat{p})}{n}} = 0.55 \\pm 1.96 \\times \\sqrt{\\frac{0.55(1-0.55)}{100}} = (0.4525, 0.6475)[\/latex].<\/p>\r\nInterpretation<\/strong>: We are 95% confident that the proportion of winners in blue is somewhere between 0.4525 and 0.6475, i.e., we are 95% confident that the percentage of winners in blue is somewhere between 45.25% and 64.75%.<\/li>\r\n \t

          8. Does this interval support the conclusion of the one-proportion z<\/strong> test?\r\n<\/strong>Yes. In part a), we failed to reject [latex]H_0: p=0.5[\/latex] at the 5% significance level. Similarly, in part b), the 95% confidence interval [latex](0.4525, 0.6475)[\/latex] contains the hypothesized value [latex]p_0 = 0.5[\/latex], meaning we don't have sufficient evidence to claim that [latex]p[\/latex] is significantly different from 0.5.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>","rendered":"

            The assumptions and steps of a one-proportion [latex]z[\/latex] test are as follows.<\/p>\n

            \n

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

              \n
            1. A simple random sample.<\/li>\n
            2. Both [latex]np_0[\/latex] and [latex]n(1-p_0)[\/latex] are at least 5, where [latex]p_0[\/latex] is the hypothesized value of [latex]p[\/latex] under the null [latex]H_0[\/latex].<\/li>\n<\/ol>\n

              Steps to perform a one-proportion <\/strong>z test<\/strong>:<\/p>\n

                \n
              1. Set up the hypotheses:\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: p = p_0[\/latex]<\/div>\n<\/td>\n
                		\n
                [latex]H_0: p \\leq p_0[\/latex]<\/div>\n<\/td>\n
                		\n
                [latex]H_0: p \\geq p_0[\/latex]<\/div>\n<\/td>\n<\/tr>\n
                \n
                [latex]H_a: p \\neq p_0[\/latex]<\/div>\n<\/td>\n
                		\n
                [latex]H_a: p \\: \\gt \\:p_0[\/latex]<\/div>\n<\/td>\n
                		\n
                [latex]H_a: p \\: \\lt \\:p_0[\/latex]<\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/li>\n
              2. State the significance level [latex]\\alpha[\/latex].<\/li>\n
              3. Compute the value of the test statistic: [latex]z_o = \\frac{\\hat{p} - p_0}{\\sqrt{\\frac{p_0 (1 - p_0)}{n} }}[\/latex], with [latex]\\hat{p} = \\frac{x}{n}[\/latex].<\/li>\n
              4. Find the P-value or<\/strong> 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: p = p_0[\/latex]<\/div>\n<\/td>\n
                		\n
                [latex]H_0: p \\leq p_0[\/latex]<\/div>\n<\/td>\n
                		\n
                [latex]H_0: p \\geq p_0[\/latex]<\/div>\n<\/td>\n<\/tr>\n
                Alternative<\/th>\n\n
                [latex]H_a: p \\neq p_0[\/latex]<\/div>\n<\/td>\n
                		\n
                [latex]H_a: p \\: \\gt \\:p_0[\/latex]<\/div>\n<\/td>\n
                		\n
                [latex]H_a: p \\: \\lt \\:p_0[\/latex]<\/div>\n<\/td>\n<\/tr>\n
                P-value<\/th>\n\n
                [latex]2P(Z \\geq |z_o|)[\/latex]<\/div>\n<\/td>\n
                		\n
                [latex]P(Z \\geq z_o)[\/latex]<\/div>\n<\/td>\n
                		\n
                [latex]P(Z \\leq z_o)[\/latex]<\/div>\n<\/td>\n<\/tr>\n
                Rejection region<\/th>\n[latex]Z \\geq z_{\\alpha \/ 2}[\/latex] or [latex]Z \\leq - z_{\\alpha \/ 2}[\/latex]<\/td>\n\n
                [latex]Z \\geq z_{\\alpha }[\/latex]<\/div>\n<\/td>\n
                		\n
                [latex]Z \\leq - z_{\\alpha }[\/latex]<\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n
              5. Reject the null [latex]H_0[\/latex] if the P-value [latex]\\leq \\alpha[\/latex] or [latex]z_o[\/latex] falls in the rejection region.<\/li>\n
              6. Conclusion.<\/li>\n<\/ol>\n<\/div>\n

                 <\/p>\n

                \n
                \n

                Example: One-Proportion z Test<\/p>\n<\/header>\n

                \n

                Let [latex]p[\/latex] be the proportion of athletes wearing blue suits who win a judo match. Randomly select n<\/em> = 100 Olympic judo matches and suppose 55 winners wore a blue suit. The other 45 wore a white suit.<\/p>\n

                  \n
                1. Test at the 5% significance level whether a color bias exists.<\/strong>
                  \nIf there is no colour bias, the proportions of blue and white winners should be 0.5 and 0.5. Therefore, letting p<\/em> be the proportion of winners in blue, the hypotheses are [latex]H_0: p = 0.5[\/latex] versus [latex]H_a: p \\neq 0.5[\/latex].
                  \nCheck the assumptions<\/strong>:<\/p>\n
                    \n
                  1. We have a simple random sample (SRS).<\/li>\n
                  2. Both [latex]np_0 = 100 \\times 0.5 = 50[\/latex] and [latex]n (1 - p_0) = 100 \\times (1-0.5) = 50[\/latex] are at least 5.<\/li>\n<\/ol>\n

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

                      \n
                    1. Set up the hypotheses. [latex]H_0: p = 0.5[\/latex] versus [latex]H_a: p \\neq 0.5[\/latex]<\/li>\n
                    2. State the significance level is [latex]\\alpha = 0.05[\/latex].<\/li>\n
                    3. Compute the test statistic:\n

                      [latex]\\hat{p} = \\frac{x}{n} = \\frac{55}{100} = 0.55, z_o = \\frac{\\hat{p} - p_0}{\\sqrt{\\frac{p_0 (1 - p_0)}{n}}} = \\frac{0.55 - 0.5}{ \\sqrt{ \\frac{0.5(1-0.5)}{100} }} = 1.[\/latex]<\/p>\n<\/li>\n

                    4. Find the P-value. For a two-tailed test, the P-value is twice the area to the right of the absolute value of the observed test statistic [latex]z_o[\/latex]. That is:
                      \nP-value = [latex]2P(Z \\geq |z_o|)[\/latex] [latex]= 2P(Z \\geq 1)[\/latex] [latex]= 2P(Z \\leq -1)[\/latex] [latex]= 2 \\times 0.1587[\/latex] [latex]= 0.3174[\/latex].<\/li>\n
                    5. Decision: Since the P-value [latex]= 0.3174 \\: \\gt \\: 0.05(\\alpha)[\/latex], we cannot reject the null [latex]H_0[\/latex].<\/li>\n
                    6. Conclusion: At the 5% significance level, we do not have sufficient evidence that color bias exists in judging Olympic Judo matches.<\/li>\n<\/ol>\n<\/li>\n
                    7. Obtain a confidence interval <\/strong>corresponding to the test in part (a).
                      \n<\/strong>For a two-tailed test at the 5% significance level, we obtain a 95% two-tailed interval.
                      \nThe sample proportion is [latex]\\hat{p} = \\frac{x}{n} = \\frac{55}{100} = 0.55[\/latex].<\/p>\n

                      [latex]1 - \\alpha = 0.95 \\Longrightarrow \\alpha = 0.05 \\Longrightarrow z_{\\alpha \/ 2} = z_{0.025} = 1.96[\/latex].<\/p>\n

                      A 95% confidence interval for the proportion of winners in blue is<\/p>\n

                      [latex]\\hat{p} \\pm z_{\\alpha \/ 2} \\sqrt{\\frac{\\hat{p} (1 - \\hat{p})}{n}} = 0.55 \\pm 1.96 \\times \\sqrt{\\frac{0.55(1-0.55)}{100}} = (0.4525, 0.6475)[\/latex].<\/p>\n

                      Interpretation<\/strong>: We are 95% confident that the proportion of winners in blue is somewhere between 0.4525 and 0.6475, i.e., we are 95% confident that the percentage of winners in blue is somewhere between 45.25% and 64.75%.<\/li>\n

                    8. Does this interval support the conclusion of the one-proportion z<\/strong> test?
                      \n<\/strong>Yes. In part a), we failed to reject [latex]H_0: p=0.5[\/latex] at the 5% significance level. Similarly, in part b), the 95% confidence interval [latex](0.4525, 0.6475)[\/latex] contains the hypothesized value [latex]p_0 = 0.5[\/latex], meaning we don’t have sufficient evidence to claim that [latex]p[\/latex] is significantly different from 0.5.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n","protected":false},"author":19,"menu_order":5,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1104","chapter","type-chapter","status-publish","hentry"],"part":1075,"_links":{"self":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/1104","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":24,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/1104\/revisions"}],"predecessor-version":[{"id":5313,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/1104\/revisions\/5313"}],"part":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/parts\/1075"}],"metadata":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/1104\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/media?parent=1104"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapter-type?post=1104"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/contributor?post=1104"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/license?post=1104"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}