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 319minutes. 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.)Find the value of the test statistic. Round your answer to two decimal places.

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

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

4.)Identify the level of significance for the hypothesis test.

5.)Reject or fail to reject the null hypothesis?

Answer 1

Solution :

The null and alternative hypothesis are

H₀ : = 319 ........... Null hypothesis

H_a : > 319 ........... Alternative hypothesis

Here, n = 68, = 324, ²= 625, = 25

The test statistic t is

z = ( - )/[/n]

= [324 - 319]/[25 /68]

= 1.65

The value of the test​ statistic z = 1.65

Now ,

> sign in Ha indicates that the test is ONE TAILED.

z = 1.65

So , using calculator ,

p-value = P(z > 1.65)

= 0.0495

p value = 0.0495

Since,

p value is greater than the significance level 0.02

Decision:

Fail to Reject the null hypothesis H0