Question

"Smartest People Often Dumbest About Sunburns" is the headline of an article that appeared in the San Luis Obispo Tribune (July 19,2006\()\). The article states that "those with a college degree reported a higher incidence of sunburn than those without a high school degree-43\% versus \(25 \% . "\) Suppose that these percentages were based on independent random samples of size 200 from each of the two groups of interest (college graduates and those without a high school degree). a. Are the sample sizes large enough to use the largesample confidence interval for a difference in population proportions? b. Estimate the difference in the proportion of people with a college degree who reported sunburn and the corresponding proportion for those without a high school degree using a \(90 \%\) confidence interval. c. Is zero included in the confidence interval? What does this suggest about the difference in the two population proportions? d. Interpret the confidence interval in the context of this problem.

Short answer

a. Yes, the sample sizes are large enough to use the large sample confidence interval for a difference in population proportions based on the calculation of \(np > 5 \) and \(n(1-p) > 5\). b. The estimate of the difference in proportion is 0.119 to 0.241 with a 90% confidence interval. c. Zero is not included in the confidence interval, implying a significant difference in the reporting of sunburn between the two population groups. d. The confidence interval indicates that a college graduate is 11.9% to 24.1% more likely to report sunburn than a person without high school.

Step-by-step solution

Checking Sample Size

To determine if the sample sizes are large enough, the samples need to meet the criterion of \(np > 5 \) and \(n(1−p) > 5\). For college graduates, \(np= 200(0.43) = 86\), and \(n(1 - p) = 200(1 - 0.43) = 114\). For people without a high school diploma, \(np = 200(0.25) = 50\), and \(n(1 - p) = 200(1 - 0.25) =150\). The results for both groups meet the condition, so the sample sizes are large enough.

Calculating the Confidence Interval

The formula to estimate the difference is given as: \(\hat{P1} - \hat{P2} ± Z*\sqrt{\( \hat{P1}(1 - \hat{P1})/n1 + \hat{P2}(1-\hat{P2})/n2\)}\). Here, \(\hat{P1}\) and \(\hat{P2}\) are the sample proportions, and \( n1 \) and \( n2 \) are the sample sizes. The values for \( Z \) is \(1.645\) for a 90% confidence interval. The values are plugged into the formula, giving the result: \(0.43 - 0.25 ± 1.645*\sqrt{\(0.43*(1 - 0.43)/200 + 0.25*(1-0.25)/200}\) = \(0.18 ± 0.061\), which brings us to the interval of \(0.119\) to \(0.241\).

Checking For Zero

The calculated confidence interval here does not include zero. This indicates that the difference in the population proportions is statistically significant- implying that there is a significant difference between college graduates and those without a high school degree when it comes to reporting sunburn.

Interpretation of the Results

The confidence interval indicates that with 90% confidence, the difference in proportion of people who reported sunburn between college graduates and those without high school degrees ranges from 11.9% to 24.1%. In other words, a college graduate is 11.9% to 24.1% more likely to report sunburn than a person without a high school degree.