Question

In order for a new drug to be sold on the market, the variance of the active ingredient in each dose should be 0.03mg. A random sample of 25tablets with a dosage strength of 56.57mg is taken. The variance of the active ingredient from this sample is found to be 0.01920.0192. Does the data suggests at α=0.005 that the variance of the drug in the tablets is less than the desired amount? Assume the population is normally distributed.

Step 3 of 5:

Determine the value of the test statistic. Round your answer to three decimal places.

Answer 1

We have to test the hypothesis that

Variance of active ingredient is less than 0.03

i.e. Null Hypothesis -

against

Alternative Hypothesis - ( Left-tailed test)

We use chi-square test to test significance of population variance.

The value of chi-square test statistic is

Given :

Value of chi-square test statistic = 15.36

Degrees of freedom = n-1 =24

Alpha: level of significance = 0.05

Since test is left-tailed and value of test statistic is 15.36, p-value is obtained by

p-value = 0.0902

Decision: Since p-value > level of significance, we failed to reject the null hypothesis.

Conclusion: There is insufficient evidence to claim that the Variance of active ingredient is less than 0.03.