Question

1. The faculty of a university mathematics department is concerned about the performance of students in the introductory calculus offered by the department and required of all science and engineering majors. Historically, class averages on the test have been about 75, a passing grade but indicative that students may not be learning the material as well as they need to in order to go on to the next course. The chair would like to raise the average to at least 80. The department decides to implement a tutoring program in which each section of the course has assigned to it an advanced student who has done very well in the course previously. That student attends the section assigned and is available for student consultation 10 hours each week outside of class. A random sample of 50 students from the first semester of the program is selected and the test score for each student in the sample is recorded. The sample of scores is provided below.

78      75      74      73      70      75      70      71      77      78

74      77      81      76      78      72      73      73      75      77

72      82      80      74      72      77      77      73      79      75

74      80      78      76      79      81      75      71      74      78

75      71      75      77      72      74      77      75      77      78

What do these 50 scores suggest about the chair’s concern? Be specific about assumptions, hypotheses being tested, test statistic used, and the critical region for your test. Interpret the results of your test in regard to the concerns of the chair. (You may do this test by hand or get SPSS to do it for you. If you use SPSS, please provide the output it produces along with your interpretation of the results.) As in problem 2, you need not consider any confidence intervals.

Answer 1

First, let's calculate the sample mean of test scores of 50 students.

The sample mean = 75.5 (calculating by summing test scores and dividing by 50) and Sample Standard Deviation = 2.978 (Using Sample Standard Deviation formula =

The chair wants to certain that the new approach helped students get better and improved the overall average score.

Hence, we want to choose our null hypothesis such that rejecting it, we can conclude the above consideration.

So, Null hypothesis, H₀: Population mean of test score <=75

Alternate Hypothesis, H₁: Population mean of test score > 75

The current sample mean = 75.5

Samples, n = 50 and sample standard deviation = 2.978

Calculating t-value using the above information,

t = 1.187

We are using t-test for the process.

Calculating probability using t-table, we get 0.1205. This probability is greater than the standard 95% significant level (0.05)

Hence, we can not reject the hypothesis that the population mean <75

Thus, nothing conclusive can be said about the new approach improving the overall score.