To construct a 99% confidence interval for the average weekly hours employed of full-time undergraduate students, we can use the formula: Confidence Interval = Sample Mean ± (Z * Standard Deviation / √n). Given the sample mean of 10.4 hours per week for a sample of 18 students and a known standard deviation of 7.8 hours per week, the 99% confidence interval is approximately (3.43, 17.37).
To construct a 99% confidence interval for the average weekly hours employed of full-time undergraduate students, we can use the formula:
Confidence Interval = Sample Mean ± (Z * Standard Deviation / √n)
Given that we have a sample of 18 students with an average employment of 10.4 hours per week, and a known standard deviation of 7.8 hours per week, we can substitute these values into the formula to obtain:
Confidence Interval = 10.4 ± (Z * 7.8 / √18)
Since the distribution of hours employed is assumed to be normal, we can use the standard normal distribution to find the appropriate Z value. For a 99% confidence interval, the Z value is approximately 2.33. Substituting this value into the formula, we can calculate the confidence interval:
Confidence Interval = 10.4 ± (2.33 * 7.8 / √18)
Calculating the expression inside the parentheses, we get: 2.33 * 7.8 / √18 ≈ 6.97
Therefore, the 99% confidence interval for the average weekly hours employed of full-time undergraduate students is approximately 10.4 ± 6.97, or (3.43, 17.37).