Question

The process of producing muscle relaxant medication yields tablets with varying amounts of the active ingredient. It is claimed that the average amount of active ingredient per tablet is at least 120 milligrams. The Consumer Watchdog Bureau tests a random sample of 50 tablets. The mean content of the active ingredient for this sample is 116.2 milligrams and the standard deviation is 17 milligrams. What is the approximate p-value for the appropriate test?

	0.0602
	0.1204
	0.4397
	0.5603
	0.05

Answer 1

Solution:

The test statistics t is given by ..

t =  

= (116.2 - 120)/(17/50)

= -1.581

Now ,

n = 50

df = n - 1 = 50 - 1 = 49

Read the claim "the average amount of active ingredient per tablet is at least 120 milligrams "

One tailed left sided test

So ,

p value = 0.0602