Question

Rhino viruses typically cause common colds. In a test of the effectivness of echinacea, 37 of the 42 subjects treated with echinacea developed rhinovirus infections. In a placebo group, 94 of the 11 developed rhinovirus infections. Use 0.01 significance level to test the claim that echinacea on rhinovirus infections.

1. Test the cliam using a hypothosis test- Consider the first sample of subjects to be treated with echinacea and the second sample to be the sample of subjects treated with placebo. What are the null and alternative hypotheses for the hypothesis test?

2. z=

3. p value=

Answer 1

Solution:-

n₁ = 42

n₂ = 110

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P₁ = P₂
Alternative hypothesis: P₁ ≠ P₂

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the proportion from population 1 is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p₁ * n₁ + p₂ * n₂) / (n₁ + n₂)

p = 0.8618

SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.0626

z = (p₁ - p₂) / SE

z = 0.42

where p₁ is the sample proportion in sample 1, where p₂ is the sample proportion in sample 2, n₁ is the size of sample 1, and n₂ is the size of sample 2.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than - 0.42 or greater than 0.42

Thus, the P-value = 0.675

Interpret results. Since the P-value (0.675) is greater than the significance level (0.01), we have to accept the null hypothesis.