Question

test the claim about the difference between two population means $\mu_1$ and $\mu_2$ at the level of sig samples are random and independent, and the populations are normally distributed. claim: $\mu_1 = \mu_2$; $\alpha = 0.01$. assume $\sigma_1^2 = \sigma_2^2$ sample statistics: $\bar{x}_1 = 34.3$, $s_1 = 3.7$, $n_1 = 10$ and $\bar{x}_2 = 36.2$, $s_2 = 2.2$, $n_2 = 15$ identify the null and alternative hypotheses. choose the correct answer below. \\(\bigcirc\\) a. $h_0: \mu_1 \
eq \mu_2$ $h_a: \mu_1 = \mu_2$ \LXI0 b. $h_0: \mu_1 \leq \mu_2$ $h_a: \mu_1 > \mu_2$ \LXI1 c. $h_0: \mu_1 > \mu_2$ $h_a: \mu_1 \leq \mu_2$ \LXI2 d. $h_0: \mu_1 < \mu_2$ $h_a: \mu_1 \geq \mu_2$ \LXI3 e. $h_0: \mu_1 \geq \mu_2$ $h_a: \mu_1 < \mu_2$ \LXI4 f. $h_0: \mu_1 = \mu_2$ $h_a: \mu_1 \
eq \mu_2$

Explanation:

Step1: Recall Hypothesis Testing Basics

The null hypothesis (\(H_0\)) is a statement of equality or no difference, and the alternative hypothesis (\(H_a\)) is what we are trying to find evidence for. The claim here is \(\mu_1 = \mu_2\), so the null hypothesis should reflect this claim (or a statement that includes equality). The alternative hypothesis will be the opposite of the null, considering the type of test. Since the claim is \(\mu_1 = \mu_2\) and we are testing for a difference (because the alternative should be two - tailed if there's no direction specified other than the claim of equality), we check the options.

Step2: Analyze Each Option

• Option A: \(H_0:\mu_1

eq\mu_2\) is incorrect because the null should be the claim (equality), not inequality.

• Option B: \(H_0:\mu_1\leq\mu_2\) and \(H_a:\mu_1 > \mu_2\) is a one - tailed test (right - tailed), but our claim is equality, so this is wrong.
• Option C: \(H_0:\mu_1>\mu_2\) and \(H_a:\mu_1\leq\mu_2\) is a left - tailed test, not matching the claim of equality.
• Option D: \(H_0:\mu_1<\mu_2\) and \(H_a:\mu_1\geq\mu_2\) is a right - tailed test, incorrect.
• Option E (first E): \(H_0:\mu_1\geq\mu_2\) and \(H_a:\mu_1 < \mu_2\) is a left - tailed test, incorrect.
• Option F (the last E, let's correct the label, it should be F? Wait, the original has an E and then F? Wait, the last option is \(H_0:\mu_1=\mu_2\) and \(H_a:\mu_1

eq\mu_2\), which is a two - tailed test. Since the claim is \(\mu_1=\mu_2\), the null hypothesis is \(H_0:\mu_1 = \mu_2\) and the alternative is the opposite (two - tailed, since we are testing if there is a difference from equality), so this option is correct.

Answer:

F. \(H_0:\mu_1=\mu_2\) \(H_a:\mu_1
eq\mu_2\) (Note: There was a label error in the original, the correct option with \(H_0:\mu_1 = \mu_2\) and \(H_a:\mu_1
eq\mu_2\) is the one that should be chosen. If we consider the original labels, the last option (let's assume it's F, but in the given, it's labeled as E again, but the content is \(H_0:\mu_1=\mu_2\), \(H_a:\mu_1
eq\mu_2\))