{"id":1954,"date":"2021-09-16T17:43:32","date_gmt":"2021-09-16T17:43:32","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/?post_type=chapter&p=1954"},"modified":"2023-12-05T09:29:47","modified_gmt":"2023-12-05T09:29:47","slug":"a-population-proportion-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/a-population-proportion-2\/","title":{"raw":"Plus-Four Confidence Interval for p","rendered":"Plus-Four Confidence Interval for p"},"content":{"raw":"
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Learning Outcomes<\/h3>\r\n
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  • Using the formula for creating a confidence interval or technology, construct a confidence interval for a population proportion using the \"plus-four\" method<\/li>\r\n<\/ul>\r\n<\/section><\/div>\r\n

    \"Plus Four\" Confidence Interval for p<\/em><\/h2>\r\nThere is a certain amount of error introduced into the process of calculating a confidence interval for a proportion. Because we do not know the true proportion for the population, we are forced to use point estimates to calculate the appropriate standard deviation of the sampling distribution. Studies have shown that the resulting estimation of the standard deviation can be flawed.\r\n\r\nFortunately, there is a simple adjustment that allows us to produce more accurate confidence intervals. We simply pretend that we have four additional observations. Two of these observations are successes and two are failures. The new sample size, then, is n<\/em> + 4, and the new count of successes is x<\/em> + 2.\r\n\r\nComputer studies have demonstrated the effectiveness of this method. It should be used when the confidence level desired is at least 90% and the sample size is at least ten.\r\n
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    Example 3<\/h3>\r\nA random sample of 25 statistics students was asked: \"Have you smoked a cigarette in the past week?\" Six students reported smoking within the past week. Use the \"plus-four\" method to find a 95% confidence interval for the true proportion of statistics students who smoke.\r\n\r\n \r\n\r\nThe first solution is step-by-step.\r\n[reveal-answer q=\"950372\"]Show Solution A[\/reveal-answer]\r\n[hidden-answer a=\"950372\"]\r\n\r\nSix students out of 25 reported smoking within the past week, so x<\/em> = 6 and n<\/em> = 25. Because we are using the \"plus-four\" method, we will use x<\/em> = 6 + 2 = 8 and n<\/em> = 25 + 4 = 29.\r\n\r\np'<\/em> =\u00a0[latex]\\dfrac{{x}}{{n}} =\\dfrac{{8}}{{29}} \\approx[\/latex] 0.276\r\n\r\nq'<\/em> = 1 - p'<\/em> - 1 - 0.276 = 0.724\r\n\r\nSince CL = 0.95, we know [latex]\\sigma[\/latex] = 1 - 0.95 = 0.05 and [latex]\\dfrac{a}{2}[\/latex] = 0.025.\r\n\r\n[latex]\\displaystyle{z}_{0.025}={1.96}[\/latex]\r\n\r\nEPB<\/em> = [latex]{(z_{\\frac{a}{2}})}{\\sqrt {\\frac{p'q'}{n}}}[\/latex] = (1.96)[latex]{\\sqrt{\\frac{0.276(0.724)}{29}}} \\approx[\/latex] 0.163\r\n

    p'<\/i> -\u00a0EPB\u00a0<\/em><\/span>=\u00a0<\/span>0.60 -\u00a0<\/span>0.036\u00a0<\/span>=\u00a0<\/span>0.564<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\r\n

    p'\u00a0<\/em>+\u00a0EBP\u00a0<\/em><\/span>=\u00a0<\/span>0.60\u00a0<\/span>+\u00a0<\/span>0.036\u00a0<\/span>=\u00a0<\/span>0.636<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\r\n

    We are 95% confident that the true proportion of all statistics students who smoke cigarettes is between 0.113 and 0.439.<\/p>\r\n[\/hidden-answer]\r\n\r\n \r\n\r\nThe second solution uses a function of the TI-83, 83+, or 84 calculators.\r\n[reveal-answer q=\"454704\"]Show Solution B[\/reveal-answer]\r\n[hidden-answer a=\"454704\"]\r\n\r\nPress STAT and arrow over to TESTS.\r\n\r\nArrow down to A:1-PropZint. Press ENTER.\r\n\r\nRemember that the plus-four method assume an additional four trials: two successes and two failures. You do not need to change the process for calculating the confidence interval; simply update the values of x<\/em> and n<\/em> to reflect these additional trials.\r\n\r\nArrow down to x<\/em> and enter eight.\r\n\r\nArrow down to n<\/em> and enter 29.\r\n\r\nArrow down to C-Level and enter 0.95.\r\n\r\nArrow down to Calculate and press ENTER.\r\n\r\nThe confidence interval is (0.113, 0.439).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

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    try it 3<\/h3>\r\nOut of a random sample of 65 freshmen at State University, 31 students have declared a major. Use the \"plus-four\" method to find a 96% confidence interval for the true proportion of freshmen at State University who have declared a major.\r\n\r\n \r\n\r\nThe first solution is step-by-step.\r\n\r\n[reveal-answer q=\"628897\"]Show Solution A[\/reveal-answer]\r\n[hidden-answer a=\"628897\"]\r\n\r\nUsing \"plus-four,\" we have x<\/em> = 31 + 2 = 33 and n<\/em> = 65 + 4 = 69.\r\n\r\nSince CL = 0.96, we know .\r\n\r\nz0.02 = 2.054\r\n\r\nWe are 96% confident that between 35.4% and 60.2% of all freshmen at State U have declared a major.\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\nThe second solution uses a function of the TI-83, 83+, or 84 calculators.\r\n\r\n[reveal-answer q=\"530491\"]Show Solution B[\/reveal-answer]\r\n[hidden-answer a=\"530491\"]\r\n\r\nPress STAT and arrow over to TESTS.\r\n\r\nArrow down to A:1-PropZint.\r\n\r\nPress ENTER.\r\n\r\nArrow down to x<\/em> and enter 33.\r\n\r\nArrow down to n<\/em> and enter 69.\r\n\r\nArrow down to C-Level and enter 0.96.\r\n\r\nArrow down to Calculate and press ENTER.\r\n\r\nThe confidence interval is (0.355, 0.602).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
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    Example 4<\/h3>\r\nThe Berkman Center for Internet & Society at Harvard recently conducted a study analyzing the privacy management habits of teen internet users. In a group of 50 teens, 13 reported having more than 500 friends on Facebook. Use the \"plus-four\" method to find a 90% confidence interval for the true proportion of teens who would report having more than 500 Facebook friends.\r\n\r\n \r\n\r\nThe first solution is step-by-step.\r\n[reveal-answer q=\"328722\"]Show Solution A[\/reveal-answer]\r\n[hidden-answer a=\"328722\"]\r\n\r\nUsing \"plus-four,\" we have x<\/em> = 13 + 2 = 15 and n<\/em> = 50 + 4 = 54.\r\n\r\np'<\/em> = [latex]{\\dfrac{15}{54}} \\approx[\/latex] 0.278\r\n\r\nq'<\/em> = 1 -\u00a0p'<\/em> = 1 - 0.241 = 0.722\r\n\r\nSince CL = 0.90, we know [latex]\\alpha[\/latex] = 1 - 0.90 = 0.10 and [latex]\\dfrac{a}{2}[\/latex] = 0.05.\r\n\r\nz0.05<\/sub> = 1.645\r\n\r\nEBP<\/em> = [latex]{\\left ( z_{\\frac{a}{2}} \\right )}{\\left ( \\sqrt{\\dfrac{p'q'}{n}} \\right )}[\/latex] = (1.645)[latex]{\\left ( \\sqrt{\\dfrac{(0.278)(0.722)}{54}} \\right )} {\\approx}[\/latex] 0.100\r\n

    p\u2032<\/em>\u00a0-\u00a0EPB<\/em>\u00a0= 0.278 - 0.100 = 0.178<\/p>\r\n

    p\u2032<\/em>\u00a0+\u00a0EPB<\/em>\u00a0= 0.278 + 0.100 = 0.378<\/p>\r\nWe are 90% confident that between 17.8% and 37.8% of all teens would report having more than 500 friends on Facebook.\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\nThe second solution uses a function of the TI-83, 83+, or 84 calculators.\r\n[reveal-answer q=\"391647\"]Show Solution B[\/reveal-answer]\r\n[hidden-answer a=\"391647\"]\r\n\r\nPress STAT and arrow over to TESTS.\r\n\r\nArrow down to A:1-PropZint. Press ENTER.\r\n\r\nArrow down to x<\/em> and enter 15.\r\n\r\nArrow down to n<\/em> and enter 54.\r\n\r\nArrow down to C-Level and enter 0.90.\r\n\r\nArrow down to Calculate and press ENTER.\r\n\r\nThe confidence interval is (0.178, 0.378).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

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    try it 4<\/h3>\r\nThe Berkman Center Study referenced in Example 4 talked to teens in smaller focus groups, but also interviewed additional teens over the phone. When the study was complete, 588 teens had answered the question about their Facebook friends with 159 saying that they have more than 500 friends. Use the \"plus-four\" method to find a 90% confidence interval for the true proportion of teens that would report having more than 500 Facebook friends based on this larger sample. Compare the results to those in Example 4.\r\n\r\n \r\n\r\nThe first solution is step-by-step.\r\n[reveal-answer q=\"472878\"]Show Solution A[\/reveal-answer]\r\n[hidden-answer a=\"472878\"]\r\n\r\nUsing \"plus-four,\" we have x = 159 + 2 = 161 and n = 588 + 4 = 592.\r\n\r\nSince CL = 0.90, we know .\r\n\r\nWe are 90% confident that between 24.2% and 30.2% of all teens would report having more than 500 friends on Facebook.\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\nThe second solution uses a function of the TI-83, 83+, or 84 calculators.\r\n[reveal-answer q=\"797162\"]Show Solution B[\/reveal-answer]\r\n[hidden-answer a=\"797162\"]\r\n\r\nPress STAT and arrow over to TESTS.\r\n\r\nArrow down to A:1-PropZint. Press ENTER.\r\n\r\nArrow down to x<\/em> and enter 161.\r\n\r\nArrow down to n<\/em> and enter 592.\r\n\r\nArrow down to C-Level and enter 0.90.\r\n\r\nArrow down to Calculate and press ENTER.\r\n\r\nThe confidence interval is (0.242, 0.302).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

    Conclusion<\/h3>\r\nThe confidence interval for the larger sample is narrower than the interval from Example 4. Larger samples will always yield more precise confidence intervals than smaller samples. The \"plus-four\" method has a greater impact on the smaller sample. It shifts the point estimate from 0.26 (13\/50) to 0.278 (15\/54). It has a smaller impact on the EPB, changing it from 0.102 to 0.100. In the larger sample, the point estimate undergoes a smaller shift: from 0.270 (159\/588) to 0.272 (161\/592). It is easy to see that the plus-four method has the greatest impact on smaller samples.","rendered":"
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    Learning Outcomes<\/h3>\n
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