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.

Step 1 of 6: State the null and alternative hypotheses.

Step 2 of 6:

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

Step 3 of 6:

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

Step 4 of 6:

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

Step 5 of 6:

Identify the level of significance for the hypothesis test.

Step 6 of 6:

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

Answer 1

1)

H0: = 319

Ha: > 319

2)

Test statistics

z = ( - ) / sqrt ( / n)

= ( 324 - 319 ) / sqrt ( 625 / 68)

= 1.65

3)

This is one tailed test.

4)

p-value = P(Z > z)

= 1 - P(Z < z)

= 1 - P(Z < 1.65)

= 1 - 0.9505

= 0.0495

5)

Level of significance = 0.02

6)

Since p-value > 0.02 level, fail to reject the null hypothesis