Question

The manager of the Human Resources department of a large corporation wants to study absenteeism among clerical workers during the past year. A random sample of 25 clerical workers reveals an average absenteeism of 13.6 days and a standard deviation of 4.0 days. The sample also revealed that 9 of the 25 workers were absent at least 15 days during the year. Assume the number of absences per year are normally distributed. (a) Construct a 95% confidence interval estimate for the average number of days absent for clerical workers.

Answer 1

To construct a 95% confidence interval estimate for the average number of days absent for clerical workers, use the formula: CI = Xbar ± Z * (σ / sqrt(n)). Given the values of the sample mean, standard deviation, and sample size, we can calculate the Z-score and plug it into the formula to find the upper and lower bounds of the interval.

Step-by-step explanation:

To construct a 95% confidence interval estimate for the average number of days absent for clerical workers, we can use the formula:

CI = Xbar ± Z * (σ / sqrt(n))

Where:

• CI is the confidence interval
• Xbar is the sample mean
• Z is the Z-score corresponding to the desired confidence level
• σ is the population standard deviation
• n is the sample size

Given that the sample mean is 13.6, the standard deviation is 4.0, and the sample size is 25, we can calculate the Z-score using a Z-table or a calculator. The Z-score for a 95% confidence level is approximately 1.96.

Plugging in the values:

CI = 13.6 ± 1.96 * (4.0 / sqrt(25))

Simplifying the equation:

CI = 13.6 ± 1.96 * 0.8

Calculating the upper and lower bounds:

Upper bound = 13.6 + 1.96 * 0.8 = 15.35

Lower bound = 13.6 - 1.96 * 0.8 = 11.85

Therefore, the 95% confidence interval estimate for the average number of days absent for clerical workers is approximately 11.85 to 15.35.