Every year, all incoming high school freshmen in a large school district take a math placement test. For this year's test, the district has prepared two possible versions: Version 1 that covers more material than last year's test and Version 2 test that is similar to last year's test. The district suspects that the mean score for Version 1 will be less than the mean score for Version 2. To examine this, over the summer the district randomly selects 95 incoming freshmen to come to its offices to take Version 1, and it randomly selects 85 incoming freshmen to come take Version 2 . The 95 incoming freshmen taking Version 1 score a mean of 113.2 points with a standard deviation of 15.2 . The 85 incoming freshmen taking Version 2 score a mean of 116.8 points with a standard deviation of 19.2 . Assume that the population standard deviations of the test scores from the two versions can be estimated to be the sample standard deviations, since the samples that are used to compute them are quite large. At the 0.05 level of significance, is there enough evidence to support the claim that the mean test score, $\mu_{1}$, for Version 1 is less than the mean test score, $\mu_{2}$, for Version 2 ? Perform a one-tailed test. Then complete the parts below.
Carry your intermediate computations to at least three decimal places. (If necessary, consult a list of formulas.)
(a) State the null hypothesis $H_{0}$ and the alternative hypothesis $H_{1}$.
\[
\begin{array}{l}
H_{0}: \square \\
H_{1}: \square
\end{array}
\]
(b) Determine the type of test statistic to use.
(Choose one) $\mathbf{V}$
(c) Find the value of the test statistic. (Round to three or more decimal places.)
Q
(d) Find the critical value at the 0.05 level of significance. (Round to three or more decimal places.)
\begin{tabular}{ccc}
$\mu$ & $\sigma$ & $\rho$ \\
$\bar{x}$ & $s$ & $\hat{p}$ \\
$\square^{0}$ & $\square \square$ & $\frac{\square}{\square}$ \\
$\square=\square$ & $\square \leq \square$ & $\square \geqslant \square$ \\
$\square \neq \square$ & $\square< \square$ & $\square> \square$ \\
$x$ & & $\sigma$
\end{tabular}
(e) Can we support the claim that the mean test score for Version 1 is less than the mean test score of Version 2?
Yes ONo
Answer
Conclude the test. There is not enough evidence to support the claim that the mean test score for Version 1 is less than the mean test score for Version 2.
Step 1 ：State the null hypothesis $H_{0}$ and the alternative hypothesis $H_{1}$. The null hypothesis is that the mean score for Version 1 is equal to the mean score for Version 2, and the alternative hypothesis is that the mean score for Version 1 is less than the mean score for Version 2. So, we have: \n $H_{0}: \mu_{1} = \mu_{2}$ \n $H_{1}: \mu_{1} < \mu_{2}$
Step 2 ：Determine the type of test statistic to use. We are comparing the means of two independent samples and we know the standard deviations of the samples. Therefore, we will use a t-test.
Step 3 ：Calculate the t statistic. The formula for the t statistic in a two-sample t-test is: \n $t = \frac{\bar{x}_{1} - \bar{x}_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}}$ \n Substituting the given values, we get: \n $t = \frac{113.2 - 116.8}{\sqrt{\frac{15.2^{2}}{95} + \frac{19.2^{2}}{85}}} = -1.384$
Step 4 ：Find the critical value at the 0.05 level of significance. The degrees of freedom for a two-sample t-test is the smaller of $n_{1} - 1$ and $n_{2} - 1$. So, $df = min(95 - 1, 85 - 1) = 84$. Looking up the t value for 84 degrees of freedom and a one-tailed test at the 0.05 level of significance, we find that the critical value is -1.653.
Step 5 ：Compare the t statistic to the critical value. Since the t statistic of -1.384 is greater than the critical value of -1.653, we do not reject the null hypothesis.
Step 6 ：Conclude the test. There is not enough evidence to support the claim that the mean test score for Version 1 is less than the mean test score for Version 2.
