Question

The mean time it takes aspirin to relieve headache pain is known to be 30 minutes. A drug company has developed a new drug that they claim provides relief in less time. Government scientists tested the new drug on a sample of 25 individuals with headaches. For this sample, the mean time to relief was 26 minutes. The population standard deviation for the new drug is known to be 7 minutes. A level of significance of .01 is to be used to test if the new drug relieves pain faster than aspirin.

6. Based on your test statistic, should the null hypothesis be rejected? Explain

7. What is your conclusion with respect to the effectiveness of the new drug?

8. Determine the p-value of the sample mean. SHOW ALL WORK.

9. Based on the P-value, should the null hypothesis be rejected? On what comparison did you base your decision? Explain.

10. If the standard deviation of 7 minutes used in this test had been the sample’s standard deviation rather than the population’s standard deviation, what would the critical value for the test have been?

Answer 1

6) The test statistic here is computed as:

For 0.01 level of significance, as this is a one tailed test, we have from the standard normal tables here:
P(Z < -2.326) = 0.01

As the test statistic value is less than the critical value of the test, therefore the test is significant here and we can reject the null hypothesis here. Therefore null hypothesis is rejected here is the correct answer here.

7) As the test is significant here, we can conclude here that we have sufficient evidence that the new drug relieves pain faster than aspirin.

8) The p-value is computed from the standard normal tables here as:

p = P( Z < -2.8571) = 0.0021

Therefore 0.0021 is the required p-value here.

9) As the p-value here is 0.0021 < 0.01 which is the level of significance, therefore the test is significant and we can reject the null hypothesis here and conclude that we have sufficient evidence here that the new drug relieves pain faster than aspirin.

10) As we dont have the population standard deviation, we will use the t distribution tables here to get the critical value as:

For n - 1 = 24 degrees of freedom, we get from the t distribution tables here:

P( t24 < -2.492) = 0.01

Therefore -2.492 is the required critical value here.