Solutions

Please note that the steps show rounded numbers, but that the final answers to the problems are calculated without rounding.

Problem Part Solution
1 - \(H_0:p_1 = p_2\)
2 - \(z = \frac{(\hat{p}_1-\hat{p}_2)-(p_1-p_2)}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1}+\frac{1}{n_2})}}\)
\(\hat{p}_1 = \text{Sample proportion for group 1}\)
\(\hat{p}_2 = \text{Sample proportion for group 2}\)
\(\hat{p} = \text{Overall sample proportion}\)
\(n_1 = \text{Sample size for group 1}\)
\(n_2 = \text{Sample size for group 2}\)
3 - \((\hat{p}_1 - \hat{p}_2) \pm z^* \sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}\)
\(\hat{p}_1 = \text{Sample proportion for group 1}\)
\(\hat{p}_2 = \text{Sample proportion for group 2}\)
\(n_1 = \text{Sample size for group 1}\)
\(n_2 = \text{Sample size for group 2}\)
\(z^* = \text{z value for a confidence interval}\)
4 A \(296*0.213=63 >10\)
\(251*0.108=27>10\)
\(296(1-0.213)=233>10\)
\(251(1-0.108)=224>10\)
4 B \(H_0:p_1 = p_2\)
\(H_0:p_1 \neq p_2\)
4 C \(z = 3.309\)
4 D \(\text{P-value} = 0.001\)
4 E NormalDistApplet
Students should include a sketch of normal distribution curve with both to the left and right of the two z-scores shaded.
4 F reject the null hypothesis
4 G We have sufficient evidence to say that there is a difference in the proportions of men who use labels and women who use labels.
5 - (0.045, 0.165)
6 - We are 95% confident that the true difference in the proportions of men being label users and women being label users is between 0.045 and 0.165.