Question

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, 95 new hires took the test and scored a mean of 112.5 points with a standard deviation of 17.1. last year, 55 new hires took the test and scored a mean of 117.5 points with a standard deviation of 18.9. assume that the population 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:

Explanation:

Step1: Identify the given values

Let $n_1 = 95$, $\bar{x}_1=112.5$, $s_1 = 17.1$, $n_2=55$, $\bar{x}_2 = 117.5$, $s_2=18.9$. For a 95% confidence - interval, the critical value $z_{\alpha/2}=1.96$.

Step2: Calculate the standard error

The formula for the standard error $SE$ of the difference in means is $SE=\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}$.
\[

$$\begin{align*} SE&=\sqrt{\frac{17.1^{2}}{95}+\frac{18.9^{2}}{55}}\\ &=\sqrt{\frac{292.41}{95}+\frac{357.21}{55}}\\ &=\sqrt{3.078 + 6.495}\\ &=\sqrt{9.573}\\ &\approx3.094 \end{align*}$$

\]

Step3: Calculate the margin of error

The margin of error $E = z_{\alpha/2}\times SE$.
$E=1.96\times3.094\approx6.064$.

Step4: Calculate the confidence - interval

The difference in sample means is $\bar{x}_1-\bar{x}_2=112.5 - 117.5=- 5$.
The lower limit of the confidence interval is $(\bar{x}_1-\bar{x}_2)-E=-5 - 6.064=-11.064\approx - 11.06$.
The upper limit of the confidence interval is $(\bar{x}_1-\bar{x}_2)+E=-5 + 6.064 = 1.064\approx1.06$.

Answer:

Lower limit: $-11.06$
Upper limit: $1.06$