Question

The director of research and development is testing a new drug. She wants to know if there is evidence at the 0.02 level that the drug stays in the system for more than 319 minutes. For a sample of 68 patients, the mean time the drug stayed in the system was 324 minutes. Assume the population variance is known to be 625.

1 State the null and alternative hypotheses.

2  

Find the value of the test statistic. Round your answer to two decimal places.

3  

Specify if the test is one-tailed or two-tailed.

4  

Find the P-value of the test statistic. Round your answer to four decimal places.

5

Identify the level of significance for the hypothesis test.

6 Make the decision to reject or fail to reject the null hypothesis.

Answer 1

Given

1) Hypothesis:

H0 : = 319

H1 : > 319

2) n = 68

X_bar = 324

2 = 625

= 2 = 625 = 25

Test statistic Z value

z = (x_bar - ​​​​​​) /(/n)

= (324 - 319)/(25/68=

z = 1.65

3) the above claim is one tailed. So test is one tailed.

4) p value for Z test statistic is 0.0495

P value = 0.0495

5) the p value is greater than therefore p value is greater than 0.02 level of significance.

6) conclusion:

The p value (0.0495) is greater than 0.02 levels of significance hence do not reject null hypothesis. There is not enough evidence to claim that the drug stay in system for more than 319 minutes at 0.02 significance level.

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