A quality control inspector keeps a tally sheet of the number of acceptable and unacceptable products that come off two different production lines. The completed sheet is shown bel... A quality control inspector keeps a tally sheet of the number of acceptable and unacceptable products that come off two different production lines. The completed sheet is shown below. a. Can the inspector infer at the 5% significance level that production line 1 is doing a better job than production line 2? b. What is the p-value of the test? c. Estimate with 95% confidence the difference in population proportions. Read more

Steps to Solve

1. Calculate the sample proportions of acceptable products for each production line. To do this, divide the number of acceptable products by the total number of products for each line. $p_1 = \frac{152}{152 + 48} = \frac{152}{200} = 0.76$ $p_2 = \frac{136}{136 + 54} = \frac{136}{190} \approx 0.7158$

2. State the null and alternative hypotheses. The null hypothesis ($H_0$) is that there is no difference in the proportion of acceptable products between the two production lines: $p_1 - p_2 = 0$. The alternative hypothesis ($H_1$) is that production line 1 is doing a better job than production line 2, meaning the proportion of acceptable products from line 1 is greater than that from line 2: $p_1 - p_2 > 0$. This is a one-tailed test.

3. Calculate the pooled proportion. The pooled proportion is used to estimate the standard error when testing the difference between two proportions. $p = \frac{152 + 136}{200 + 190} = \frac{288}{390} \approx 0.7385$

4. Calculate the test statistic (z-score). The z-score is calculated as follows: $z = \frac{(p_1 - p_2) - (0)}{\sqrt{p(1-p)(\frac{1}{n_1} + \frac{1}{n_2})}}$ $z = \frac{(0.76 - 0.7158)}{\sqrt{0.7385(1-0.7385)(\frac{1}{200} + \frac{1}{190})}}$ where $n_1 = 200$ and $n_2 = 190$ $z = \frac{0.0442}{\sqrt{0.7385 * 0.2615 * (0.005 + 0.00526)}}$ $z = \frac{0.0442}{\sqrt{0.19313 * 0.01026}} = \frac{0.0442}{\sqrt{0.001981}} \approx \frac{0.0442}{0.0445} \approx 0.993$

5. Determine the p-value. Since it's a right-tailed test ($p_1 - p_2 > 0$), the p-value is the probability of observing a z-score as high as or higher than the calculated z-score. Using a z-table or calculator, the p-value for $z \approx 0.993$ is approximately $1 - 0.8398 = 0.1602$.

6. Compare the p-value to the significance level. The significance level ($\alpha$) is 5% or 0.05. Since the p-value (0.1602) is greater than the significance level (0.05), we fail to reject the null hypothesis.

7. Calculate the confidence interval for the difference in population proportions. The formula for the confidence interval is: $(p_1 - p_2) \pm z_{\alpha/2} \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}$ For a 95% confidence interval, $z_{\alpha/2} = 1.96$. $(0.76 - 0.7158) \pm 1.96 \sqrt{\frac{0.76(1-0.76)}{200} + \frac{0.7158(1-0.7158)}{190}}$ $0.0442 \pm 1.96 \sqrt{\frac{0.76 * 0.24}{200} + \frac{0.7158 * 0.2842}{190}}$ $0.0442 \pm 1.96 \sqrt{\frac{0.1824}{200} + \frac{0.2034}{190}}$ $0.0442 \pm 1.96 \sqrt{0.000912 + 0.001071}$ $0.0442 \pm 1.96 \sqrt{0.001983}$ $0.0442 \pm 1.96 * 0.0445$ $0.0442 \pm 0.0872$ The 95% confidence interval is $(-0.043, 0.1314)$.