Question

The report "Audience Insights: Communicating to Teens (Aged \(12-17)^{\prime \prime}\) (www.cdc.gov, 2009 ) described teens' attitudes about traditional media, such as TV, movies, and newspapers. In a representative sample of American teenage girls, \(41 \%\) said newspapers were boring. In a representative sample of American teenage boys, \(44 \%\) said newspapers were boring. Sample sizes were not given in the report. a. Suppose that the percentages reported were based on samples of 58 girls and 41 boys. Is there convincing evidence that the proportion of those who think that newspapers are boring is different for teenage girls and boys? Carry out a hypothesis test using \(\alpha=.05\). b. Suppose that the percentages reported were based on samples of 2,000 girls and 2,500 boys. Is there convincing evidence that the proportion of those who think that newspapers are boring is different for teenage girls and boys? Carry out a hypothesis test using \(\alpha=0.05\). c. Explain why the hypothesis tests in Parts (a) and resulted in different conclusions.

Short answer

Based on the hypothesis tests, there is convincing evidence in both samples that the proportion of teenage girls and boys who found newspapers boring is different, with boys being more likely to find newspapers boring. The larger sample gave more assurance of this difference.

Step-by-step solution

State the Hypotheses

For both parts, the null and alternative hypotheses are as follows: Null hypothesis \(H_0\): The proportion of boys and girls who find newspapers boring is the same, i.e., \(P_{boys} = P_{girls}\). Alternative hypothesis \(H_a\): The proportion of boys and girls who find newspapers boring is different, i.e., \(P_{boys} \neq P_{girls}\). Let's denote \(p_1\) as the proportion in the girls' sample and \(p_2\) as the proportion in the boys' sample.

Compute the Test Statistics for small sample

For the first part, i.e., 58 girls and 41 boys, the proportions are \(p_1 = 41/58 = 0.71\) and \(p_2 = 44/41 = 1.07\). First calculate the pooled proportion \(p = (p_1 \times n_1 + p_2 \times n_2) / (n_1 + n_2) = (0.71 \times 58 + 1.07 \times 41) / (58 + 41) = 0.88\).Then, calculate the standard error SE = \(\sqrt {p(1 - p)(1/n_1 + 1/n_2)} = \sqrt {0.88(1 - 0.88)(1/58 + 1/41)} = 0.12\).Lastly compute the Z-score \(Z = (p_2 - p_1) / SE = (1.07 - 0.71) / 0.12 = 2.5\).

Compute the Test Statistics for large sample

For the second part, i.e., 2000 girls and 2500 boys, the proportions are \(p_1 = 0.41\) and \(p_2 = 0.44\). In these cases, we directly consider the given percentage values as these are large samples. Compute the pooled proportion \(p = (p_1 \times n_1 + p_2 \times n_2) / (n_1 + n_2) = (0.41 \times 2000 + 0.44 \times 2500) / (2000 + 2500) = 0.42\).Calculate the standard error SE = \(\sqrt {p(1 - p)(1/n_1 + 1/n_2)} = \sqrt {0.42(1 - 0.42)(1/2000 + 1/2500)} = 0.01\).Finally, compute the Z-score \(Z = (p_2 - p_1) / SE = (0.44 - 0.41) / 0.01 = 3.0\).

Conclusions based on p-value

For each case, calculate the p-value using the Z-score. If the p-value is less than the significance level (\(\alpha = 0.05\)), then reject the null hypothesis; otherwise, fail to reject it. In practice, p-values would be found via a Z-table or calculator, but cannot be calculated straight from the Z-score. However, Z-scores above 2 or below -2 typically correspond to p-values less than 0.05.In the small sample, \(Z = 2.5\) so p-value < \(\alpha = 0.05\) and the null hypothesis would be rejected - there appears to be convincing evidence for a difference in attitudes between girls and boys. In the large sample, the Z-score is even higher (\(Z = 3.0\)), so the p-value is even smaller and thus, again, the null hypothesis would be rejected. The larger sample size in the second test gives even more assurance that the difference in proportions is not due to sampling variability.

Explaining the differences

The reasons for different conclusions in the two tests are mainly due to the sample size. Larger sample size reduces the standard error and therefore gives a larger Z-score. This is why a larger sample gives more convincing evidence against the null hypothesis. However, both tests reject the null hypothesis, implying that there likely is a difference in the proportions.