Reference no: EM131417130

In given Case Study, we learned that about 56% of American adults actually voted in the presidential election of 1992, whereas about 61% of a random sample claimed that they had voted. The size of the sample was not specified, but suppose it was based on 1600 American adults, a common size for such studies.

a. Into what interval of values should the sample proportion fall 68%, 95%, and almost all of the time?

b. Do you think that the observed value of 61% is reasonable, based on your answer to part (a)?

c. Now suppose that the sample had been of only 400 people. Compute a standardized score to correspond to the reported percentage of 61%. Comment on whether or not you believe that people in the sample could all have been telling the truth, based on your result.

Case Study
Do Americans Really Vote When They Say They Do?
On November 8, 1994, a historic election took place in which the Republican Party won control of both houses of Congress for the first time since 1952. But how many people actually voted? On November 28, 1994, Time magazine (p. 20) reported that in a telephone poll of 800 adults taken during the 2 days following the election, 56% claimed that they had voted.

Considering that only about 68% of adults are registered to vote, that isn't a bad turnout. But Time reported another disturbing fact along with the 56% turnout claimed in the survey. Time reported that, in fact, only 39% of American adults had voted, based on information from the Committee for the Study of the American Electorate. Could it be that the results of the poll simply reflected a sample that, by chance, voted with greater frequency than the general population? The sampling distribution for the sample proportion in this setting can answer that question.

Let's suppose that the truth about the population is, as reported by Time, that only 39% of American adults voted, so p = .39. Then the sampling distribution for a sample proportion tells us what kind of sample proportions we can expect in samples of 800 adults, the size used by the Time poll. The mean of the possibilities is p = .39, or 39%. The standard deviation of the possibilities is = .017, or 1.7%. Therefore, we are almost certain that the sample percentage based on a sample of 800 adults should fall within 3 × 1.7% = 5.1% of the truth of 39%.

In other words, if respondents were telling the truth, the sample percentage should be no higher than 44.1%-nowhere near the reported percentage of 56%! In fact, if we combine this sampling distribution with what we learned about normal curves in Chapter 8, we can say even more about how unlikely this sample result would be. If in truth only 39% (.39) of the population voted, the standardized score for the reported value of 56% (.56) is z = (.56 - .39)/.017 = 10.0. We know from Chapter 8 that it is virtually impossible to obtain a standardized score of 10.

Another example of the fact that reported voting tends to exceed actual voting occurred in the 1992 U.S. presidential election. According to the World Almanac (1995, p. 631), 61.3% of American adults reported voting in the 1992 election. In a footnote, the Almanac explains:

Total reporting voting compares with 55.9 percent of population actually voting for president, as reported by Voter News Service. Differences between data may be the result of a variety of factors, including sample size, differences in the respondents' interpretation of the questions, and the respondents' inability or unwillingness to provide correct information or recall correct information.

Unfortunately, because figures are not provided for the size of the sample, we cannot assess whether or not the difference between the actual percentage of 55.9 and the reported percentage of 61.3 can be explained by the natural variability among possible sample proportions.