Proportions A study selected independent random samples of 50 dogs and 60 cats and found that 38 of the dogs and 15 of the cats could be trained to fetch a ball. Do these data provide convincing evidence that the proportion of all dogs that can be trained to fetch a ball is greater than the proportion of all cats that can be trained to fetch a ball? Use \( a=0.05 \). Calculate the test statistic and P-value for this test, and then complete this concluding statement. If needed, use the z-table to answer the question. Because the P-value of \( \square \) is \( \square \) \( a=0.05 \) we \( \square \) \( \mathrm{H}_{0} \) We \( \square \) convincing evidence to conclude that the proportion of all dogs that can be trained to fetch a ball is \( \square \) the proportion of all cats that can be trained to fetch a balf.

To determine whether the proportion of dogs that can be trained to fetch a ball is greater than that of cats, we'll perform a hypothesis test for the difference between two proportions. ### **Step 1: Define the Hypotheses** - **Null Hypothesis (\( H_0 \))**: \( p_{\text{dogs}} = p_{\text{cats}} \) - **Alternative Hypothesis (\( H_a \))**: \( p_{\text{dogs}} > p_{\text{cats}} \) ### **Step 2: Calculate Sample Proportions** - **Dogs**: - Sample size (\( n_1 \)) = 50 - Number trained (\( x_1 \)) = 38 - Sample proportion (\( \hat{p}_1 \)) = \( \frac{38}{50} = 0.76 \) - **Cats**: - Sample size (\( n_2 \)) = 60 - Number trained (\( x_2 \)) = 15 - Sample proportion (\( \hat{p}_2 \)) = \( \frac{15}{60} = 0.25 \) ### **Step 3: Calculate the Pooled Proportion (\( \hat{p} \))** \[ \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{38 + 15}{50 + 60} = \frac{53}{110} \approx 0.4818 \] ### **Step 4: Calculate the Standard Error (SE)** \[ SE = \sqrt{\hat{p}(1 - \hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{0.4818 \times 0.5182 \times \left(\frac{1}{50} + \frac{1}{60}\right)} \approx 0.0956 \] ### **Step 5: Calculate the Test Statistic (z)** \[ z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.76 - 0.25}{0.0956} \approx 5.34 \] ### **Step 6: Determine the P-value** A \( z \)-score of 5.34 is extremely high. Consulting the standard normal distribution table or using statistical software, the P-value is: \[ P\text{-value} < 0.0001 \] ### **Step 7: Make a Decision** Compare the P-value with the significance level (\( \alpha = 0.05 \)): \[ \text{Since } P\text{-value} < \alpha, \text{ we reject } H_0. \] ### **Conclusion** Because the P-value of **\( < 0.0001 \)** is **\( < 0.05 \)**, we **reject** \( H_0 \). We have convincing evidence to conclude that the proportion of all dogs that can be trained to fetch a ball is **greater than** the proportion of all cats that can be trained to fetch a ball. --- **Filled-in Statement:** Because the P-value of **\( < 0.0001 \)** is **\( < 0.05 \)**, we **reject** \( \mathrm{H}_{0} \). We have convincing evidence to conclude that the proportion of all dogs that can be trained to fetch a ball is **greater than** the proportion of all cats that can be trained to fetch a ball.