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.