Final answer:
The 95% confidence interval for the mean number of hours a union member is absent per month, given a sample mean of 9.3 hours, standard deviation of 2.3 hours, and sample size of 301, is approximately 9.04 to 9.56 hours. None of the provided options A to D match this interval.
Step-by-step explanation:
The student's question is about constructing a 95% confidence interval for the actual mean number of hours a labor union member is absent per month. The sample mean is 9.3 hours, the standard deviation is 2.3 hours, and the sample size is 301 members. To calculate the 95% confidence interval, we use the formula mean ± z*(standard deviation/sqrt(sample size)), where z is the z-score corresponding to a 95% confidence level. Since the sample size is large (>30), we can use the z-distribution to approximate the t-distribution. For a 95% confidence interval, the z-score is approximately 1.96. Now we can compute the margin of error:
Margin of Error = 1.96 * (2.3/sqrt(301)) = 1.96 * (2.3/17.35) ≈ 1.96 * 0.132 = 0.2592
The confidence interval is then:
Lower Limit = 9.3 - 0.2592 = 9.0408
Upper Limit = 9.3 + 0.2592 = 9.5592
Thus, the 95% confidence interval for the actual mean number of hours a union member is absent per month is approximately 9.04 to 9.56 hours, which is not exactly reflected in the provided options A to D. Therefore, based on the calculated values, none of the provided options A, B, C, or D is correct.