Question

The director of research and development is testing a new medicine. She wants to know if there is evidence at the 0.05 level that the medicine relieves pain in more than 330 seconds. For a sample of 53 patients, the mean time in which the medicine relieved pain was 334 seconds. Assume the standard deviation is known to be 20 . Step 1 of 5 : Enter the hypotheses:

Answer 1

Solution :

Given that,

Population mean = = 330

Sample mean = = 334

Sample standard deviation = s = 20

Sample size = n = 53

Level of significance = = 0.05

This is a right tailed test.

The null and alternative hypothesis is,

Ho: 330

Ha: 330

The test statistics,

t = ( - )/ (s/)

= ( 334 - 330 ) / ( 20 /53)

= 1.456

Critical value of  the significance level is α = 0.05, and the critical value for a right-tailed test is

= 1.675

Since it is observed that |t| =1.456 = 1.675, it is then concluded that the null hypothesis is fail to reject.

P- Value = 0.0757

The p-value is p = 0.0757 > 0.05, it is concluded that the null hypothesis is fail to reject.

Conclusion:

It is concluded that the null hypothesis Ho is fail to reject. Therefore, there is not enough evidence to claim that the medicine

relieves pain in more than 330 seconds. at 0.05 significance level.