The principal of a large high school wants to improve student test scores, so he asks one of his science teachers to try a new method of teaching. Thirty-one students take a pretest on the first day of science class. After 6 weeks using the new instruction, the same students take a posttest. The mean difference (posttest – pretest) in scores is 4.8 points with a standard deviation of 9.7 points. Assuming the conditions for inference are met, what is the P-value for testing the hypotheses H0 :μdiff = 0; Ha :μdiff > 0?Find the t-table here.
Because the same 31 students took both tests, use a one-sample t-test on the differences. The test statistic is t = (4.8 − 0) / (9.7/√31) ≈ 2.76, with 30 degrees of freedom. For the right-tailed alternative, the P-value is P(t₃₀ ≥ 2.76) ≈ 0.0049, so the correct choice is 0.0049.
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