An SAT prep course claims to improve the test score of students. The table below shows the scores for seven students the first two times they took the verbal SAT. Before taking the SAT for the second time, each student took a course to try to improve his or her verbal SAT scores. Do these results support the claim that the SAT prep course improves the students' verbal SAT scores? Let d=(verbal SAT scores prior to taking the prep course)−(verbal SAT scores after taking the prep course)d=(verbal SAT scores prior to taking the prep course)−(verbal SAT scores after taking the prep course). Use a significance level of α=0.01 for the test. Assume that the verbal SAT scores are normally distributed for the population of students both before and after taking the SAT prep course. Student 1 2 3 4 5 6 7 Score on first SAT 530 410 380 600 480 440 380 Score on second SAT 560 460 400 620 500 520 430
Step 1 of 5: State the null and alternative hypotheses for the test.
Step 2 of 5: Find the value of the standard deviation of the paired differences. Round your answer to one decimal place
Step 3 of 5: Compute the value of the test statistic. Round your answer to three decimal places.
Step 4 of 5: Determine the decision rule for rejecting the null hypothesis H0. Round the numerical portion of your answer to three decimal places.
Reject H0 if (t, ItI) (<,>) ____________
Step 5 of 5: Make the decision for the hypothesis test. (Reject or Fail to Reject Null Hypothesis)
step 1 of 5:
Ho : mean = 0
Ha: mean < 0
step 2 of 5: standard deviation = 22.678
Step 3 of 5 : test statistic t = -4.5
step 4 of 5: Critical value for alpha = 0.01, df = 6 is -3.143
reject Ho if test statistic t < -3.143
step 5 of 5: reject null hypothesis
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