Question

The director of research and development is testing a new drug. She wants to know if there is evidence at the 0.050.05 level that the drug stays in the system for more than 307307 minutes. For a sample of 6767 patients, the mean time the drug stayed in the system was 315315 minutes. Assume the population standard deviation is 2323.

Step 5 of 6:

Identify the level of significance for the hypothesis test.

Answer 1

Solution :

The null and alternative hypothesis is ,

= 307

= 315

= 23

n = 67

This will be a right tailed test because the alternative hypothesis is showing a specific direction

This is the right tailed test .

The null and alternative hypothesis is ,

H0 :   = 307

Ha :   > 307

Test statistic = z

= ( - ) / / n

= (315 - 307) / 23 / 67

= 2.85

p(Z > 2.85) = 1 - P (Z < 2.85) = 1 - 0.9978 = 0.0022

P-value = 0.0022

= 0.05  

0.0022 < 0.05

P-value <

Reject the null hypothesis .