Question

A manufacturer of a Pharmaceutical claims that their pain killer pills have 5 mg of Codeine. Their research quality control manager takes a random sample of 20 pills. She calculates the sample average to be 4.227 mg. with a standard deviation of .217 lb. At the 0.1 level of significance, is the manufacturer correct?

Answer 1

Solution:

Here, we have to use one sample t test for the population mean.

The null and alternative hypotheses are given as below:

Null hypothesis: H0: their pain killer pills have 5 mg of Codeine.

Alternative hypothesis: Ha: their pain killer pills do not have 5 mg of Codeine.

H0: µ = 5 versus Ha: µ ≠ 5

This is a two tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 5

Xbar = 4.227

S = 0.217

n = 20

df = n – 1 = 19

α = 0.10

Critical value = - 1.7291 and 1.7291

(by using t-table or excel)

t = (Xbar - µ)/[S/sqrt(n)]

t = (4.227 - 5)/[ 0.217/sqrt(20)]

t = -15.9307

P-value = 0.0000

(by using t-table)

P-value < α = 0.1

So, we reject the null hypothesis

There is not sufficient evidence to conclude that their pain killer pills have 5 mg of Codeine.

At the 0.1 level of significance, the manufacturer is not correct.