Question

# PLEASE DOUBLE CHECK ANSWER, AND NO HANDWRITTEN ANSWERS. THANK YOU!! The director of research and development...

The director of research and development is testing a new medicine. She wants to know if there is evidence at the 0.02 level that the medicine relieves pain in more than 361 seconds. For a sample of 70 patients, the mean time in which the medicine relieved pain was 364 seconds. Assume the population variance is known to be 361. Make the decision to reject or fail to reject the null hypothesis.

Sample size = n = 70

Sample mean = = 364

Population variance = 361

Population standard deviation = = 19

Claim: The medicine relieves pain in more than 361 seconds.

The null and alternative hypothesis is

$H0:\mu\leq 361$

$H1:\mu>361$

Level of significance = 0.02

Here population standard deviation is known so we have to use z-test statistic.

Test statistic is

$z =\frac{364-361 }{19 /\sqrt{70}}=\frac{3}{2.270934}=1.32$

Critical value = 2.05   ( Using z table)

The critical value > Test statistic | z | we fail to reject the null hypothesis.

Conclusion:

The medicine relieves pain in NOT more than 361 seconds.

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