computing the test statistic and p-value in minitab express

Do these data indicate that the design specifications have not been met? The hypotheses to be tested are H0: μ = 20 versus Ha: μ > 20, where μ = the true average activation time of the sprinkler system. A series of 12 fire alarm/sprinkler system tests results in an average activation time of 21.5 seconds.

The average activation time is supposed to be at most 20 seconds. these results are not sufficient evidence that the new curriculum has improved Math SAT scores.Ī sprinkler system is being installed in a newly renovated building on campus. although the results are statistically significant, they are not practically significant, since an increase of 20 points is fairly small.Ĭ. there is strong statistical evidence that the new curriculum has improved Math SAT scores in his state.ī. Since over 10,000 students in his state took the Math SAT, he can show that the P-value for testing whether the mean score in his state is more than the national average of 480 is less than 0.0001. His state recently adopted a new mathematics curriculum and he wonders if the improved scores are evidence that the new curriculum has been successful. The senator of a particular state notices that the mean score for students in his state who took the Math SAT is 500. Assume the distribution of scores is normal. Suppose the average Math SAT score for all students taking the exam this year is 480 with standard deviation 100. There is reason to believe that the standard drug is better than the new drug and hence the new drug should not be prescribed. The confidence interval contains the hypothesized value for μ and hence it is significant at the 0.05 level.Į. There is insufficient evidence to reject the claim that there is no difference between the new drug and the standard drug with respect to mean recovery time.ĭ. The experimenter should reject the claim that the new drug is the same as the standard drug with respect to mean recovery time.Ĭ. There is evidence at the 0.05 level of significance that the new drug is better than the standard drug.ī. What conclusion can be reached in this case concerning the new drug relative to the standard drug?Ī. The 95% confidence interval was found to be (24.6, 27.8). Following an extensive random experiment involving 65 patients, the data gathered were used to construct a 95% confidence interval estimate for the mean recovery time (in days) for patients on the new drug. It is known that the mean time to recovery for the standard drug is 26 days. A random experiment was conducted to see if a newly formulated drug produced a different effect on the mean time to recovery than that achieved using the standard drug.

Computing the test statistic and p-value in minitab express