Question

a study was done using a treatment group and a placebo group. the results are shown in the table. assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. complete parts (a) and (b) below. use a 0.05 significance level for both parts. a. test the claim that the two samples are from populations with the same mean. what are the null and alternative hypotheses? oa. ( h_{0}: mu_{1}=mu_{2} ) ( h_{1}: mu_{1}
eq mu_{2} ) ob. ( h_{0}: mu_{1}
eq mu_{2} ) ( h_{1}: mu_{1}<mu_{2} ) oc. ( h_{0}: mu_{1}=mu_{2} ) ( h_{1}: mu_{1}>mu_{2} ) od. ( h_{0}: mu_{1}<mu_{2} ) ( h_{1}: mu_{1} geq mu_{2} )

Explanation:

To test the claim that two populations have the same mean, the null hypothesis (\(H_0\)) typically states that the means are equal (\(\mu_1 = \mu_2\)). The alternative hypothesis (\(H_1\)) for a two - tailed test (since we are just testing if they are the same, not a directional difference like greater than or less than) is that the means are not equal (\(\mu_1
eq\mu_2\)).

Looking at the options:

• Option A: \(H_0:\mu_1 = \mu_2\) and \(H_1:\mu_1

eq\mu_2\) matches our reasoning for a test of equal means (two - tailed test).

• Option B: The null hypothesis here is \(\mu_1

eq\mu_2\) which is incorrect as the null should be the claim of equality.

• Option C: The alternative hypothesis is \(\mu_1>\mu_2\) which is a one - tailed (right - tailed) test, not what we need for testing if means are the same.
• Option D: The null and alternative hypotheses are set up incorrectly for a test of equal means.

Answer:

A. \(H_0:\mu_1=\mu_2\), \(H_1:\mu_1
eq\mu_2\)