7-1 Estimating a Population Proportion 317 Procedure for Constructing a Confidence Interval for p 1. Verify that the requirements in the preceding Key Elements box are satisfied. 2. Use technology or Table A-2 to find the critical value za>2 that corresponds to the desired confidence level. 3. Evaluate the margin of error E = za>2 2pnqn>n. 4. Using the value of the calculated margin of error E and the value of the sample proportion pn, find the values of the confidence interval limits pn - E and pn + E. Substitute those values in the general format for the confidence interval. 5. Round the resulting confidence interval limits to three significant digits. CP EXAMPLE 2 Constructing a Confidence Interval: Online Courses In the Chapter Problem, we noted that in a Sallie Mae survey of 950 undergraduate students, 53% take online courses. The sample results are n = 950 and pn = 0.53. a. Find the margin of error E that corresponds to a 95% confidence level. b. Find the 95% confidence interval estimate of the population proportion p. c. Based on the results, can we safely conclude that more than 50% of undergraduates take online courses? d. Assuming that you are an online reporter, write a brief statement that accurately describes the results, and include all of the relevant information. SOLUTION REQUIREMENT CHECK (1) The polling methods used by the Sallie Mae organization result in samples that can be considered to be simple random samples. (2) The conditions for a binomial experiment are satisfied because there is a fixed number of trials (950), the trials are independent (because the response from one undergraduate doesn’t affect the probability of the response from another undergraduate), there are two categories of outcome (an undergraduate either takes online courses or does not), and the probability remains constant, because P(undergraduate takes online courses) is fixed for a given point in time. (3) With 53% of the 950 undergraduates taking online courses, the number taking online courses is 53% of 950, which is 504 (from 0.53 * 950). If 504 of the 950 undergraduates take online courses, the other 446 do not, so the number of “successes” (504) and the number of “failures” (446) are both at least 5, as required. The check of requirements has been successfully completed. Technology The confidence interval and margin of error can be easily found using technology. From the Statdisk display we can see the required entries on the left and the results displayed on the right. Like most technologies, Statdisk requires a value for the number of successes, so we find 53% of 950 and round the result of 503.5 to the whole number 504. The results show that the margin of error is E = 0.03174, and the confidence interval is 0.499 6 p 6 0.562 (rounded). (The Wilson Score confidence interval included in the display will be discussed later in Part 2 of this section.) Statdisk Bias in Internet Surveys? Capitalizing on the widespread use of technology and social media, there is a growing trend to conduct surveys using only the Internet instead of using in-person interviews or phone calls to randomly selected subjects. Internet surveys are faster and much less expensive, and they provide important advantages in survey design and administration. But are Internet surveys biased because they use only subjects randomly selected from the 90% of the U.S. population that uses the Internet? The Pew Research Center studied this issue by comparing results from online polls to polls that included the offline population. It was found that the differences were generally quite small, but topics related to the Internet and technology resulted in much larger differences. We should be careful to consider consequences of bias with Internet surveys. continued
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