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 that those without a high school degree43 percent versus 25 percent." For purposes of this exercise, suppose that these percentages were based on random samples of size 200 from each of the two groups of interest (college graduates and those without a high school degree). Is there convincing evidence that the proportion experiencing a sunburn is higher for college graduates than it is for those without a high school degree? Answer based on a test with a \(.05\) significance level.

Short answer

Determine the test statistic \(Z\) using the given proportions and sample sizes, find the corresponding \(P\) -value. If the \(P\) -value is less than the significance level \(0.05\), we have enough evidence to say that the proportion of college graduates who experience sunburn is significantly higher than that of those without a high school degree.

Step-by-step solution

Identify Sample Sizes and Observed Proportions

The sample size for both groups is 200. The observed proportion of sunburn for the group with college degrees is 0.43 (43%) and for the group without a high school degree is 0.25 (25%).

State the Null and Alternative Hypotheses

The null hypothesis (\(H_0\)) would state that there is no difference between the proportions i.e. the proportion of sunburn for college graduates equals the proportion for those without a high school degree. The alternative hypothesis (\(H_a\)), which we are trying to provide evidence for, would state that the proportion of sunburn is higher for the college graduates than for those without the high school degree.

Compute the test statistic

The test statistic for a difference in proportions is the Z-statistic given by the formula: Z = \(\frac{{(\hat{p_1} - \hat{p_2}) - 0}}{{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{{n_1}}+\frac{1}{{n_2}})}}}\) where \(\hat{p_1}\) and \(\hat{p_2}\) are the sample proportions, n1 and n2 are the sample sizes, and \(\hat{p}\) is the pooled sample proportion given by \(\hat{p} = \frac{{\hat{p_1}n_1 + \hat{p_2}n_2}}{{n_1 + n_2}}\). For this exercise, we compute the test statistic using the given data.

Determine the P-value

The P-value is the probability of observing a sample statistic as extreme, or more so, than the one calculated given that the null hypothesis is true. It can be found using a standard normal (Z) distribution table or using software. If the computed P-value is less than the level of significance, we reject the null hypothesis - else we fail to reject.

Draw Conclusion

Based on the P-value, we decide if there's convincing evidence that the proportion experiencing a sunburn is higher for college graduates than it is for those without a high school degree. If the P-value < signifiacnce level, we reject the null hypothesis and support the alternative. Otherwise, we lack enough evidence to say for sure.