Suppose Diana, an educational researcher at a local university, wants to test the impact of a new Spanish course that integrates cultural-immersion teaching techniques with standard teaching practices. She selects a simple random sample of 72 freshmen and divides them into 36 pairs, matched on IQ and high school GPA. She randomly selects one member of each pair to take the new course, while the other member in the pair takes the traditional course. Next, Diana records the course grade, tallied on a scale from 0 to 4, for all sample members at the end of the semester, and she computes the difference in grades between the members in each matched pair by subtracting the traditional course grade from the new course grade. She wants to determine if the new Spanish course improves or weakens student performance. She runs a matched-pairs t-test to test the null hypothesis, H0:μ=0, against the alternative hypothesis, H1:μ≠0, where μ is the mean course grade difference for the student population.
The sample statistics for Diana's test are summarized in the table.
Variable description Sample mean Sample standard deviation Standard error estimate
traditional course grade x⎯⎯trad= 3.33496 strad=2.02198 SEtrad=0.33700
new course grade x⎯⎯new=3.45287 snew=2.11043 SEnew=0.35174
difference (new − traditional) x⎯⎯=0.11791 s=0.31452 SE=0.05242
Although Diana does not know the standard deviation of the underlying population of course grade differences, she assumes that the population is normally distributed because the sample data are symmetric, single-peaked, and contain no outliers.
Compute the t-statistic for Diana's matched-pairs t-test. Provide your answer with precision to three decimal places.
Compute the p-value of the t-statistic using software. You may find some software manuals helpful. Provide your answer precise to three decimal places.
t=
p =
Select the accurate statement regarding Diana's test decision if she tests at a significance level of α=0.05.
Diana should fail to reject the null hypothesis because the mean grade difference is too small.
Diana should reject the null hypothesis because the p-value is greater than the value α=0.05.
Diana should reject the null hypothesis because the p-value is less than the value α=0.05.
Diana should fail to reject the null hypothesis because the p-value is greater than 0.025.
Diana should fail to reject the null hypothesis because the p-value is greater than the value α=0.05.