Question

A machine shop that manufactures toggle levers has both a day and a night shift. A toggle lever is defective if a standard nut cannot be screwed onto the threads. Let \(p_{1}\) and \(p_{2}\) be the proportion of defective levers among those manufactured by the day and night shifts, respectively. We shall test the null hypothesis, \(H_{0}: p_{1}=p_{2}\), against a two-sided alternative hypothesis based on two random samples, each of 1000 levers taken from the production of the respective shifts. Use the test statistic \(Z^{*}\) given in Example \(6.5 .3 .\) (a) Sketch a standard normal pdf illustrating the critical region having \(\alpha=0.05\). (b) If \(y_{1}=37\) and \(y_{2}=53\) defectives were observed for the day and night shifts, respectively, calculate the value of the test statistic and the approximate \(p-\) value (note that this is a two-sided test). Locate the calculated test statistic on your figure in part (a) and state your conclusion. Obtain the approximate \(p\) -value of the test.

Short answer

The calculated z-score is approximately -2.1 which falls in the critical region of the standard normal distribution curve. The calculated p-value is approximately 0.036, which is less than the level of significance 0.05. Hence, the null hypothesis \( H_0: p_1 = p_2 \) is rejected. There is evidence to suggest that the proportions of defective levers are different between the day and night shifts.

Step-by-step solution

Draw Normal PDF

Draw a normal distribution curve or probability density function. Mark the critical points using the z-scores for standard deviations. For \(\alpha = 0.05\) in a two-tailed test, half the significance level becomes \(\alpha/2 = 0.025\). The z-scores for \(\alpha/2\) are approximately -1.96 and 1.96

Calculate the Observed Proportions

Calculate the observed proportions \( p_1 \) and \( p_2 \) for the day and night shifts respectively. \( p_1 = y_1/n_1 = 37/1000 = 0.037 \) and \( p_2 = y_2/n_2 = 53/1000 = 0.053 \)

Calculate the Pooled Proportion

Calculate the pooled proportion \( p \) using the formula \( \frac{y_1 + y_2}{n_1 + n_2} = \frac{37+53}{1000+1000} = 0.045 \)

Calculate the Test Statistic

Calculate the test statistic \( z^* \) using the formula \( z^* = \frac{p_1 - p_2}{\sqrt{\frac{p(1-p)}{n_1} + \frac{p(1-p)}{n_2}}} = \frac{0.037-0.053}{\sqrt{\frac{0.045(1-0.045)}{1000} + \frac{0.045(1-0.045)}{1000}}} \approx -2.1 \)

Calculate p-value and Make Conclusion

The p-value is the probability that the z-score is less than or equal to the absolute value of the test statistic. For \( z = -2.1 \), we look up the value from the Z-table, and we find that \( P(Z \leq -2.1) \approx 0.018 \). However, since this is a two-tailed test, we need to double this value to get the final p-value, which is 0.036. Since the p-value is less than 0.05, we reject the null hypothesis in favour of the alternative that the proportions between the two shifts are not equal.