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 less than 314 minutes. For a sample of 7 patients, the average time the drug stayed in the system was 309 minutes with a standard deviation of 21. Assume the population distribution is approximately normal.

Step 1 of 5:

State the null and alternative hypotheses.

Step 2 of 5:

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

Step 3 of 5:

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

Step 4 of 5:

Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.

Step 5 of 5:

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

Answer 1

Solution :

= 314

=309

S =21

n = 7

This is the left tailed test .

The null and alternative hypothesis is ,

H0 :    = 314

Ha : < 314

Test statistic = t

= ( - ) / S / n

= (309-314) / 21 / 7

= −0.630

Test statistic = t =−0.630

The critical value =−1.943

P-value =0.276

= 0.05  

P-value ≥

0.276 ≥ 0.05

Fail to reject the null hypothesis .

There is insufficient evidence to suggest that