Question

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 1-test to test the null hypothesis, Ho:u= 0, against the alternative hypothesis, H :: #0, where u is the mean course grade difference for the student population. The sample statistics for Diana's test are summarized in the table. Variable Sample Sample standard Standard error description mean deviation estimate traditional course grade Xtrad = 3.28304 Strad = 2.24022 SEtrad = 0.37337 new course grade Xnew = 3.38589 Snew = 2.50336 SEnew = 0.41723 difference (new - traditional) x=0.10285 s= 0.34488 SE = 0.05748 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 4-statistic for Diana's matched-pairs 7-test. Provide your answer with precision to three decimal places.

Answer 1

(a)

t = 0.10285/ 0.05748

= 1.789

So,

Answer is:

t =

1.789

(b)

t score = 1.789

ndf = 36 - 1 = 35

Two Tail Test

By Technology, p value = 0.082

So,

Answer is:

p =

0.082

(c)
Correct option:

Diana should reject the null hypothesis because the P - value is less than the value $\alpha$ = 0.10.