Confidence interval for the difference of population means: Use of the. The human resources department of a consulting firm gives a standard creativity test to a randomly selected group of new hires every year. This year, 60 new hires took the test and scored a mean of 112.9 points with a standard deviation of 13.1. Last year, 70 new hires took the test and scored a mean of 117.1 points with a standard deviation of 16.3. Assume that the pepulation standard deviations of the test scores of all new hires in the current year and the test scores of all new hires last year can be estimated by the sample standard deviations, as the samples used were quite large. Construct a $95 \%$ confidence interval for $\mu_{1}$ - $\mu_{2}$, the difference between the mean test score $\mu_{1}$ of new hires from the current year and the mean test score $\mu_{2}$ of new hires from last year. Then find the lower limit and upper limit of the $95 \%$ confidence interval. Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places. (If necessary, consult a list of formulas.) Lower limit: Upper limit: 5 Explanation Check Tcrms of
Solution
Step 1 ：Given that the sample mean, standard deviation, and size for this year are \(\bar{x}_1 = 112.9\), \(s_1 = 13.1\), and \(n_1 = 60\) respectively, and for last year are \(\bar{x}_2 = 117.1\), \(s_2 = 16.3\), and \(n_2 = 70\) respectively.
Step 2 ：We are asked to find a 95% confidence interval for the difference between the mean test score of new hires from the current year and the mean test score of new hires from last year, which is \(\mu_{1} - \mu_{2}\).
Step 3 ：The formula for a confidence interval for the difference of means is \((\bar{x}_1 - \bar{x}_2) \pm z \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\), where \(z\) is the z-score corresponding to the desired level of confidence.
Step 4 ：For a 95% confidence interval, the z-score is 1.96.
Step 5 ：Substituting the given values into the formula, we get the difference as -4.2 and the margin of error as 5.06.
Step 6 ：Subtracting the margin of error from the difference gives the lower limit of the confidence interval as -9.26.
Step 7 ：Adding the margin of error to the difference gives the upper limit of the confidence interval as 0.86.
Step 8 ：Thus, the 95% confidence interval for the difference between the mean test score of new hires from the current year and the mean test score of new hires from last year is \(\boxed{-9.26}\) to \(\boxed{0.86}\).
