{"id":1095,"date":"2021-06-12T18:35:15","date_gmt":"2021-06-12T22:35:15","guid":{"rendered":"https:\/\/openbooks.macewan.ca\/rcommander\/?post_type=chapter&p=1095"},"modified":"2024-02-08T14:30:25","modified_gmt":"2024-02-08T19:30:25","slug":"10-3-one-proportion-z-interval","status":"publish","type":"chapter","link":"https:\/\/openbooks.macewan.ca\/introstats\/chapter\/10-3-one-proportion-z-interval\/","title":{"raw":"10.3 One-Proportion z Interval","rendered":"10.3 One-Proportion z Interval"},"content":{"raw":"
\r\n\r\nAssumptions<\/strong>:\r\n
    \r\n \t
  1. A simple random sample.<\/li>\r\n \t
  2. Large sample size: both the number of successes [latex]x[\/latex] and the number of failures [latex]n-x[\/latex] are at least 5.<\/li>\r\n<\/ol>\r\nNote: Recall that one proportion inferences require [latex]np \\geq 5[\/latex] and [latex]n(1-p) \\geq 5[\/latex]. However, [latex]p[\/latex] is generally unknown, and estimated with [latex]\\hat{p} = \\frac{x}{n}[\/latex]. Thus, since [latex]n \\hat{p} = n \\frac{x}{n} = x[\/latex] and [latex]n(1 - \\hat{p}) = n \\left( 1 - \\frac{x}{n} \\right) = n \\left( \\frac{n-x}{n} \\right) = n-x [\/latex], the sample is deemed sufficiently large if [latex]n \\hat{p} = x \\geq 5[\/latex] and [latex]n(1 - \\hat{p}) = n -x \\geq 5[\/latex]. We require at least 5 successes and at least 5 failures in the sample.\r\n\r\nA point estimate for the population proportion [latex]p[\/latex] is the sample proportion [latex]\\hat{p} = \\frac{x}{n}[\/latex]. Therefore, a [latex](1 \u2013 \\alpha) \\times 100\\%[\/latex] confidence interval for the population proportion p<\/em> is\r\n
    \r\n\r\n\r\n\r\n\r\n
    \r\n
    Two-Tailed<\/strong><\/div><\/td>\r\n
    		\r\n
    Upper-Tailed<\/strong><\/div><\/td>\r\n
    		\r\n
    Lower-Tailed<\/strong><\/div><\/td>\r\n<\/tr>\r\n
    [latex]\\left(\\hat{p} - z_{\\alpha \/ 2} \\sqrt{ \\frac{\\hat{p} (1 - \\hat{p})}{n}}, \\hat{p} + z_{\\alpha \/ 2} \\sqrt{ \\frac{\\hat{p} (1 - \\hat{p})}{n}} \\right)[\/latex]<\/td>\r\n[latex]\\left(\\hat{p} - z_{\\alpha} \\sqrt{ \\frac{\\hat{p} (1 - \\hat{p})}{n}}, 1 \\right)[\/latex]<\/td>\r\n[latex]\\left(0 , \\hat{p} + z_{\\alpha} \\sqrt{ \\frac{\\hat{p} (1 - \\hat{p})}{n}} \\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNote: Since the range of proportion is between 0 and 1, the right-end point of the upper-tailed interval is bounded by 1 and the left-end point of the lower-tailed interval is bounded by 0.\r\n\r\n<\/div>\r\n<\/div>\r\n \r\n
    \r\n

    Example: One-Proportion Z Interval<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nA credit card company sent out [latex]n=400[\/latex] advertisements, and [latex]x=30[\/latex] customers responded. Obtain a 95% confidence interval for the proportion of respondents.\r\n\r\nCheck the assumptions:\r\n
      \r\n \t
    1. We have a simple random sample (SRS).<\/li>\r\n \t
    2. Both the number of successes [latex]x=30[\/latex] and number of failures [latex]n-x = 400 - 30 = 370[\/latex] are greater than 5.<\/li>\r\n<\/ol>\r\nThe sample proportion is\r\n

      [latex]\\hat{p} = \\frac{x}{n} = \\frac{30}{400} = 0.075.[\/latex]<\/p>\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 respondents is\r\n

      [latex] \\hat{p} \\pm z_{\\alpha \/ 2} \\sqrt{\\frac{\\hat{p} (1 - \\hat{p})}{n}} = 0.075 \\pm 1.96 \\times \\sqrt{\\frac{0.075 (1 - 0.075)}{400}} = (0.049, 0.101).[\/latex]<\/p>\r\nInterpretation<\/strong>: We are 95% confident that the proportion of respondents is somewhere between 0.049 and 0.101, i.e., we are 95% confident that the percentage of respondents is somewhere between 4.9% and 10.1%.\r\n\r\n<\/div>\r\n<\/div>","rendered":"

      \n

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

        \n
      1. A simple random sample.<\/li>\n
      2. Large sample size: both the number of successes [latex]x[\/latex] and the number of failures [latex]n-x[\/latex] are at least 5.<\/li>\n<\/ol>\n

        Note: Recall that one proportion inferences require [latex]np \\geq 5[\/latex] and [latex]n(1-p) \\geq 5[\/latex]. However, [latex]p[\/latex] is generally unknown, and estimated with [latex]\\hat{p} = \\frac{x}{n}[\/latex]. Thus, since [latex]n \\hat{p} = n \\frac{x}{n} = x[\/latex] and [latex]n(1 - \\hat{p}) = n \\left( 1 - \\frac{x}{n} \\right) = n \\left( \\frac{n-x}{n} \\right) = n-x[\/latex], the sample is deemed sufficiently large if [latex]n \\hat{p} = x \\geq 5[\/latex] and [latex]n(1 - \\hat{p}) = n -x \\geq 5[\/latex]. We require at least 5 successes and at least 5 failures in the sample.<\/p>\n

        A point estimate for the population proportion [latex]p[\/latex] is the sample proportion [latex]\\hat{p} = \\frac{x}{n}[\/latex]. Therefore, a [latex](1 \u2013 \\alpha) \\times 100\\%[\/latex] confidence interval for the population proportion p<\/em> is<\/p>\n

        \n\n\n\n\n
        \n
        Two-Tailed<\/strong><\/div>\n<\/td>\n
        		\n
        Upper-Tailed<\/strong><\/div>\n<\/td>\n
        		\n
        Lower-Tailed<\/strong><\/div>\n<\/td>\n<\/tr>\n
        [latex]\\left(\\hat{p} - z_{\\alpha \/ 2} \\sqrt{ \\frac{\\hat{p} (1 - \\hat{p})}{n}}, \\hat{p} + z_{\\alpha \/ 2} \\sqrt{ \\frac{\\hat{p} (1 - \\hat{p})}{n}} \\right)[\/latex]<\/td>\n[latex]\\left(\\hat{p} - z_{\\alpha} \\sqrt{ \\frac{\\hat{p} (1 - \\hat{p})}{n}}, 1 \\right)[\/latex]<\/td>\n[latex]\\left(0 , \\hat{p} + z_{\\alpha} \\sqrt{ \\frac{\\hat{p} (1 - \\hat{p})}{n}} \\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

        Note: Since the range of proportion is between 0 and 1, the right-end point of the upper-tailed interval is bounded by 1 and the left-end point of the lower-tailed interval is bounded by 0.<\/p>\n<\/div>\n<\/div>\n

         <\/p>\n

        \n
        \n

        Example: One-Proportion Z Interval<\/p>\n<\/header>\n

        \n

        A credit card company sent out [latex]n=400[\/latex] advertisements, and [latex]x=30[\/latex] customers responded. Obtain a 95% confidence interval for the proportion of respondents.<\/p>\n

        Check the assumptions:<\/p>\n

          \n
        1. We have a simple random sample (SRS).<\/li>\n
        2. Both the number of successes [latex]x=30[\/latex] and number of failures [latex]n-x = 400 - 30 = 370[\/latex] are greater than 5.<\/li>\n<\/ol>\n

          The sample proportion is<\/p>\n

          [latex]\\hat{p} = \\frac{x}{n} = \\frac{30}{400} = 0.075.[\/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 respondents is<\/p>\n

          [latex]\\hat{p} \\pm z_{\\alpha \/ 2} \\sqrt{\\frac{\\hat{p} (1 - \\hat{p})}{n}} = 0.075 \\pm 1.96 \\times \\sqrt{\\frac{0.075 (1 - 0.075)}{400}} = (0.049, 0.101).[\/latex]<\/p>\n

          Interpretation<\/strong>: We are 95% confident that the proportion of respondents is somewhere between 0.049 and 0.101, i.e., we are 95% confident that the percentage of respondents is somewhere between 4.9% and 10.1%.<\/p>\n<\/div>\n<\/div>\n","protected":false},"author":19,"menu_order":3,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1095","chapter","type-chapter","status-publish","hentry"],"part":1075,"_links":{"self":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/1095","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":26,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/1095\/revisions"}],"predecessor-version":[{"id":5310,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/1095\/revisions\/5310"}],"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\/1095\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/media?parent=1095"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapter-type?post=1095"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/contributor?post=1095"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/license?post=1095"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}