Question

Suppose that \(20 \%\) of the customers of a cable television company watch the Shopping Channel at least once a week. The cable company does not know the actual proportion of all customers who watch the Shopping Channel at least once a week and is trying to decide whether to replace this channel with a new local station. The company plans to take a random sample of 100 customers and to use \(p\) as an estimate of the population proportion. a. Show that \(\sigma_{p}\), the standard deviation of \(\hat{p},\) is equal to 0.0400 b. If for a different sample size, \(\sigma_{p}=0.0231\), would you expect more or less sample-to-sample variability in the sample proportions than when \(n=100 ?\) c. Is the sample size that resulted in \(\sigma_{p}=0.0231\) larger than 100 or smaller than \(100 ?\) Explain your reasoning.

Short answer

The calculated standard deviation \( \sigma_{p} \) is 0.0400, confirming the value given in the question. The smaller standard deviation of 0.0231 suggests less variability in the sample proportions and indicates a larger sample size than originally used (i.e., more than 100).

Step-by-step solution

Compute the standard deviation of the proportion

The standard deviation \( \sigma_{p} \) of a sample proportion \( p \) is given by the square root of \( (p(1-p))/n \), where \( n \) is the number of samples. In this case, \( p = 0.2 \) (20% of the customers watch the Shopping Channel at least once a week) and \( n = 100 \). Therefore, \( \sigma_{p} = \sqrt{(0.2*(1-0.2))/100}=0.0400 \)

Evaluate the effect of sample size on variability

With a smaller standard deviation \( \sigma_{p}=0.0231 \) than the original calculation, the sample proportions would exhibit less variability than when \( n=100 \). This is because the standard deviation \( \sigma_{p} \) decreases as the sample size \( n \) increases, assuming the proportion \( p \) remains constant. So lowering the standard deviation means increasing sample size resulting in less variability.

Compare the sample size

The sample size that resulted in \( \sigma_{p}=0.0231 \) is larger than 100. As mentioned above, for a fixed proportion, the standard deviation of the proportion is inversely related to the square root of the sample size. Therefore, increasing the sample size \( n \) reduces the standard deviation \( \sigma_{p} \). Hence, if \( \sigma_{p} \) is smaller, then the sample size must be greater than the original size of 100.