A Statistics professor has observed that for several years students score an average of 114 points out of 150 on the semester exam. A salesman suggests that he try a statistics software package that gets students more involved with computers, predicting that it will increase students' scores. The software is expensive, and the salesman offers to let the professor use it for a semester to see if the scores on the final exam increase significantly. The professor will have to pay for the software only if he chooses to continue using it. In the trial course that used this software, 218 students scored an average of 117 points on the final with a standard deviation of 8.7 points.a. What is the test statistic?b. What is the P-value?c. What is the appropriate conclusion?i. Fail to reject H_0. The change is not statistically significant. The software does not appear to improve exam scores.ii. Fail to reject H_0. The change is statistically significant. The software does appear to improve exam scores.iii. Reject H_0. The change is not statistically significant. The software does not appear to improve exam scores.iv. Reject H_0. The change is statistically significant. The software does appear to improve exam scores.

We are given that a Statistics professor has observed that for several years students score an average of 114 points out of 150 on the semester exam. The software is expensive, and the salesman offers to let the professor use it for a semester to see if the scores on the final exam increase significantly.

In the trial course that used this software, 218 students scored an average of 117 points on the final with a standard deviation of 8.7 points.

Let [tex]\mu[/tex] = mean scores on the final exam.

SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \leq[/tex] 114 points {means that the mean scores on the final exam does not increases after using software}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 114 points {means that the mean scores on the final exam increase significantly after using software}

The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;

T.S. = [tex]\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample average points = 117 points

s = sample standard deviation = 8.7 points

n = sample of students = 218

So, test statistics = [tex]\frac{117-114}{\frac{8.7}{\sqrt{218} } }[/tex] ~ [tex]t_2_1_7[/tex]

= 5.091

Now, P-value of the test statistics is given by the following formula;

P-value = P( [tex]t_2_1_7[/tex] > 5.091) = Less than 0.05%

Since, in the question we are not given with the level of significance at which hypothesis can be tested, so we assume it to be 5%. Now at 5% significance level, the t table gives critical value of 1.645 at 217 degree of freedom for right-tailed test. Since our test statistics is higher than the critical value of t as 5.091 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the change is statistically significant. The software does appear to improve exam scores.