Question

ch 22 tv news program conducts a call - in poll about a proposed city ban on smoking in public places. of the 2467 callers, 1900 were opposed to the ban. which of the following statements are true with respect to using this sample to estimate p, the proportion of all tv news viewers that favor such a ban on smoking in public places?
there appear to be no violations to any of the assumptions made in using the methods of this chapter. we should now be able to use this sample to estimate p.
n is so large that both the count of successes (np^), and the count of failures(n(1 - p^)), are 15 or more, so its okay to use this sample to estimate p.
the population is much larger than the sample, so its okay to use this sample to estimate p.
there is no way this sample can be viewed as an srs of all tv news viewers, so we cant use this sample to estimate p.

Explanation:

• A call - in poll is a form of voluntary response sampling. In voluntary response sampling, people choose to participate (in this case, by calling in).
• Simple Random Sampling (SRS) requires that every individual in the population has an equal and known probability of being selected. In a call - in poll, the sample is not selected randomly in the sense of SRS. People who have a strong opinion (either for or against, but in this case, those against may be more likely to call in) are more likely to participate.
• For the first option: There are violations. The sampling method (call - in poll) is a non - SRS method.
• For the second option: Just because \(n = 2467\) (the sample size) is large, the fact that it is a non - SRS (voluntary response) makes it not okay to use for estimating \(p\) in the context of standard sampling - based proportion estimation (which assumes some form of random sampling).
• For the third option: While the population (all TV news viewers) is likely larger than the sample, the non - random (voluntary response) nature of the sample makes it unsuitable for estimating \(p\) using standard methods that assume random sampling.

Answer:

There is no way this sample can be viewed as an SRS of all TV news viewers, so we can't use this sample to estimate \(p\).