Question

The director of research and development is testing a new drug. She wants to know if there is evidence at the 0.05 level that the drug stays in the system for more than 375 minutes. For a sample of 38 patients, the mean time the drug stayed in the system was 382 minutes. Assume the population standard deviation is 24.

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

We have to test, if the drug stays in the system for more than 375 minutes i.e.,

.....................(step 1)

Given, the sample size, n = 38

sample mean, = 382(minutes)

population standard deviation, = 24.

test statistic:

............................(step 2)

The test is a one tailed test. ....................................(step 3)

for this testing, p-value of the test statistic z is P(Z > 1.80) = 0.0359. .........................(step 4)

Using the significance level of 0.05, ..............................(step 5)

decision rule: ...............................(step 6)

we reject the null hypothesis if value of z is greater than Z0.05 = 1.645 (value from standard normal table) or if p-value < 0.05.

Since, z = 1.80 > 1.645 and also p-value=0.0359 < 0.05, we reject the null hypothesis at 0.05 level of significance.