Question

Rhino viruses typically cause common colds. In a test of the effectiveness of? echinacea, 46 of the 51subjects treated with echinacea developed rhinovirus infections. In a placebo? group,

84 of the 98 subjects developed rhinovirus infections. Use a 0.05 significance level to test the claim that echinacea has an effect on rhinovirus infections.

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

Identify the test statistic

Identify the? P-value.

What is the conclusion based on the hypothesis? test?

The? P-value is ? the significance level of ?=0.05, so ? the null hypothesis. There ? sufficient evidence to support the claim that echinacea treatment has an effect.

Test the claim by constructing an appropriate confidence interval.

What is the conclusion based on the confidence? interval?

What is the conclusion based on the confidence? interval?

Based on the? results, does echinacea appear to have any effect on the infection? rate?

Echinacea does appear to have a significant effect on the infection rate. There is evidence that it lowers the infection rate.

Echinacea does appear to have a significant effect on the infection rate. There is evidence that it increases the infection rate.

Echinacea does not appear to have a significant effect on the infection rate.

The results are inconclusive.

Answer 1

Solution:-

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 one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. 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 = 0.87248

SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.05759
z = (p₁ - p₂) / SE

z = 0.78

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 one-tailed test, the P-value is the probability that the z-score is greater than 0.78.

Thus, the P-value = 0.218

Interpret results. Since the P-value (0.218) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Echinacea does not appear to have a significant effect on the infection rate.