Question

Tylenol needs to make sure that the standard deviation for the amount of acetaminophen in its regular strength tablet must be less than 5 mg. To test this, the quality control manager randomly selects 20 pills and found that the standard deviation amount of acetaminophen was 3.89 mg. Test the claim at the α=0.10 level of significance. Note: The data is normal with no outliers.

Answer 1

Here, we have to use Chi square test for the population variance or standard deviation.

The null and alternative hypotheses for this test are given as below:

Null hypothesis: H0: The standard deviation for the amount of acetaminophen in its regular strength tablet is 5 mg.

Alternative hypothesis: Ha: The standard deviation for the amount of acetaminophen in its regular strength tablet is less than 5 mg.

H0: σ = 5 versus Ha: σ < 5

This is a lower tailed test.

The test statistic formula is given as below:

Chi-square = (n – 1)*S^2/ σ2

From given data, we have

n = 20

S = 3.89

σ = 5

Chi-square = (20 - 1)*3.89^2/5^2

Chi-square = 11.5004

We are given

Level of significance = α = 0.05

df = n – 1

df = 19

Critical value = 11.6509

(by using Chi square table or excel)

Test statistic Chi-square = 11.5004 < Lower Critical value = 11.6509

So, we reject the null hypothesis

There is sufficient evidence to conclude that the standard deviation for the amount of acetaminophen in its regular strength tablet is less than 5 mg.