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Step 1: State the hypotheses Null hypothesis (H0): There is no difference in the proportion of those who think newspapers are boring between girls and boys. (\(p_g = p_b\)) Alternative hypothesis (Ha): There is a difference in the proportion of those who think newspapers are boring between girls and boys. (\(p_g \neq p_b\)) Step 2: Calculate the test statistic We will use a two-proportion z-test to find the test statistic. \(z = \frac{(p_g - p_b) - 0}{\sqrt{\frac{p(1-p)}{n_g} + \frac{p(1-p)}{n_b}}}\) where \(p\) is the overall proportion of teens who think newspapers are boring. \(p = \frac{x_g + x_b}{n_g + n_b} = \frac{0.41(58) + 0.44(41)}{58 + 41} = \frac{23.8 + 18.04}{99} = 0.421\) Now, we'll find the z-score. \(z = \frac{(0.41-0.44)- 0}{\sqrt{\frac{0.421(1-0.421)}{58} + \frac{0.421(1-0.421)}{41}}} = -0.796\) Step 3: Determine the p-value and test the hypothesis Using a standard normal distribution table, we find that the p-value associated with a z-score of -0.796 is approximately 0.426 (two-tailed test). Since the p-value is greater than the set significance level \(\alpha = 0.05\), we fail to reject the null hypothesis. There is no convincing evidence that there is a difference in the proportion of those who think newspapers are boring between teenage girls and boys, given the sample sizes of 58 girls and 41 boys.

We will follow the same steps as before but for the larger sample sizes. Step 1: State the hypotheses No change from Part a. Step 2: Calculate the test statistic Find the overall proportion for the larger sample sizes: \(p = \frac{x_g + x_b}{n_g + n_b} = \frac{0.41(2000) + 0.44(2500)}{2000 + 2500} = \frac{820 + 1100}{4500} = 0.428\) Now, we will find the z-score for the larger sample sizes: \(z = \frac{(0.41-0.44)- 0}{\sqrt{\frac{0.428(1-0.428)}{2000} + \frac{0.428(1-0.428)}{2500}}} = -4.933\) Step 3: Determine the p-value and test the hypothesis Using a standard normal distribution table, we find that the p-value associated with a z-score of -4.933 is less than 0.0001 (two-tailed test). Since the p-value is less than the set significance level of \(\alpha = 0.05\), we reject the null hypothesis. There is convincing evidence that there is a difference in the proportion of those who think newspapers are boring between teenage girls and boys, given the larger sample sizes of 2000 girls and 2500 boys.

The reason Parts (a) and (b) resulted in different conclusions is the difference in sample sizes. In Part (a), we had smaller sample sizes of 58 girls and 41 boys, and the hypothesis test did not provide convincing evidence to conclude a difference between the proportions. However, in Part (b), when the sample sizes were increased to 2000 girls and 2500 boys, the hypothesis test provided convincing evidence of a difference in proportions. Larger sample sizes generally lead to a more precise estimate of the true population proportions and allow us to detect smaller differences that might be statistically significant. Thus, the increased sample size in Part (b) gave us enough power to reject the null hypothesis, whereas the smaller sample sizes in Part (a) were unable to do so.