Question

A random sample of 85 group leaders, supervisors, and similar personnel at General Motors revealed that, on average, they spent 6.5 years in a particular job before being promoted. The standard deviation of the sample was 1.7 years. Construct a 95% confidence interval.

Answer 1

Explanation:

From the information given,

Number of personnel sampled, n = 85

Mean or average = 6.5

Standard deviation of the sample = 1.7

We want to determine the confidence interval for the mean number of years that personnel spent in a particular job before being promoted.

For a 95% confidence interval, the confidence level is 1.96. This is the z value and it is determined from the normal distribution table. We will apply the following formula to determine the confidence interval.

z×standard deviation/√n

= 1.96 × 6.5/√85

= 1.38

The confidence interval for the mean number of years spent before promotion is

The lower end of the interval is 6.5 - 1.38 = 5.12 years

The upper end is 6.5 + 1.38 = 7.88 years

Therefore, with 95% confidence interval, the mean number of years spent before being promoted is between 5.12 years and 7.88 years