Question

The report "New Study Shows Need for Americans to Focus on Securing Online Accounts and Backing Up Critical Data" (PRNewswire, October 29,2009 ) reported that only \(25 \%\) of Americans change computer passwords quarterly, in spite of a recommendation from the National Cyber Security Alliance that passwords be changed at least once every 90 days. For purposes of this exercise, assume that the \(25 \%\) figure is correct for the population of adult Americans. a. A random sample of size \(n=200\) will be selected from this population and \(\hat{p}\), the proportion who change passwords quarterly, will be calculated. What are the mean and standard deviation of the sampling distribution of \(\hat{p} ?\) b. Is the sampling distribution of \(\hat{p}\) approximately normal for random samples of size \(n=200 ?\) Explain. c. Suppose that the sample size is \(n=50\) rather than \(n=200 .\) Does the change in sample size affect the mean and standard deviation of the sampling distribution of \(\hat{p} ?\) If so, what are the new values of the mean and standard deviation? If not, explain why not. d. Is the sampling distribution of \(\hat{p}\) approximately normal for random samples of size \(n=50 ?\) Explain.

Short answer

a. For \( n=200 \), the mean of the sampling distribution will be \( 0.25 \), and the standard deviation will be \( 0.034 \). b. Yes, the sampling distribution will be approximately normal for \( n=200 \). c. With a change in sample size to \( n=50 \), the mean remains the same, \( 0.25 \) while the standard deviation increases to \( 0.069 \). d. Yes, the sampling distribution will also be approximately normal for \( n=50 \).

Step-by-step solution

Determine the population proportion

According to the exercise scenario, the population proportion (\( p \)) of Americans changing their passwords every quarter is reported to be \(0.25\) or \(25\%\).

Calculate the mean and standard deviation of the sampling distribution for n=200

As per the properties of the sampling distribution of sample proportions, the mean of this distribution \( µ_{\hat{p}} \) equals the population proportion \( p \), so \( µ_{\hat{p}} = p = 0.25 \). The standard deviation of the sampling distribution \( σ_{\hat{p}} \) is calculated as \( σ_{\hat{p}} = \sqrt{(\frac{p(1-p)}{n})} = \sqrt{(\frac{0.25 * 0.75}{200})} = 0.034 \).

Check for normality of the sample proportion with n=200

Following the Central Limit Theorem (CLT), the sampling distribution of the sample proportion is approximately normally distributed if both \( np ≥ 10 \) and \( n(1 - p) ≥ 10 \). Let's check: \( np = 200 * 0.25 = 50 ≥ 10 \), and \( n(1 - p) = 200 * 0.75 = 150 ≥10 \). Therefore, it can be stated that the sampling distribution of \( \hat{p} \) is approximately normal for samples of size \( n = 200 \).

Calculate the mean and standard deviation for n=50

When the sample size changes to \( n = 50 \), the mean of the sampling distribution (\( µ_{\hat{p}} \)) remains the same as the population proportion \( p = 0.25 \). The standard deviation changes as \( σ_{\hat{p}} = \sqrt{(\frac{p(1-p)}{n})} = \sqrt{(\frac{0.25 * 0.75}{50})} = 0.069 \).

Check for normality of the sample proportion with n=50

By repeating the previous checks, we find: \( np = 50 * 0.25 = 12.5 ≥ 10 \), and \( n(1 - p) = 50 * 0.75 = 37.5 ≥10 \). Therefore, it can also be stated that the sampling distribution of \( \hat{p} \) is approximately normal for samples of size \( n = 50 \).