Question

Exercise 3 A test for the variance can be done based on the following rejection rules: Case (a) Case (b) Case (c) H0 : σ 2 ≤ σ 2 0 (or σ 2 = σ 2 0 ) H0 : σ 2 ≥ σ 2 0 (or σ 2 = σ 2 0 ) H0 : σ 2 = σ 2 0 H1 : σ 2 > σ2 0 H1 : σ 2 < σ2 0 H1 : σ 2 6= σ 2 0 Reject if (n − 1)S 2 σ 2 0 > χ2 n−1,α (n − 1)S 2 σ 2 0 < χ2 n−1,1−α (n − 1)S 2 σ 2 0 < χ2 n−1,1−α/2 or (n − 1)S 2 σ 2 0 > χ2 n−1,α/2 Use this test to solve the following problem: A pharmaceutical company is concerned about the variability of the content of the active ingredient in a new allergy medication they plan to submit for approval to the FDA. Even though certain level of variability is not a problem, if there is evidence that the standard deviation of the content exceeds 0.1mg, the FDA would not approve the new medication. The pharmaceutical company’s managers want to make sure that the new medication will get the approval before submitting it because, otherwise, the reputation of the company would be seriously damaged and, therefore, conduct a test at the 1% significance level to try to prove that, at this level of significance, they can prove that the variability is below the standards required by the FDA. (a) Propose a null and an alternative hypothesis to test the safety of the medication (note that the test is written for the variance and not for the standard deviation!). (b) The company takes a sample of 100 pills and measures the content of the active ingredient in all of them, obtaining a standard deviation of 0.085mg. Is this enough evidence at the 1% significance level to say that the variability is below the FDA requirements? (c) Find and interpret the p-value of this test. Should the company submit the medication?

Answer 1

Give FDA would not approve the new medication if the standard deviation of the content exceeds 0.1mg,

so variance σ₀² = (σ₀)² = 0.1² = 0.01mg²

a) Hypothesis:

H₀ : σ ² ≥ σ₀²

H₁ : σ ² < σ₀²

b)

Test and CI for One Variance

Method

σ: standard deviation of Sample

The chi-square method is valid only for the normal distribution.

Descriptive Statistics

N	StDev	Variance	99% Upper Bound for σ using Chi-Square
100	0.0850	0.00723	0.1016

Test

Null hypothesis	H₀: σ² = 0.01
Alternative hypothesis	H₁: σ² < 0.01

Method	Test Statistic	DF	P-Value
Chi-Square	71.53	99	0.017

c) Since p-value is greater than 0.01 (level of significance) hence we fail to reject the null hypothesis and conclude that there is not enough evidence at the 1% significance level to say that the variability is below the FDA requirements.

company should not submit the medication.