Questions: Test the claim about the difference between two population means μ1 and μ2 at the level of significance α. Assume the samples are random and independent, and the populations are normally distributed.
Claim: μ1=μ2 ; α=0.05. Assume σ1^2=σ2^2 Sample statistics: x̄1=32.3, s1=3.6, n1=14 and x̄2=34.6, s2=2.3, n2=18
Identify the null and alternative hypotheses. Choose the correct answer below. A. H0: μ1 ≥ μ2 B. H0: μ1=μ2 Ha: μ1<μ2 Ha: μ1 ≠ μ2 C. H0: μ1>μ2 D. H0: μ1 ≠ μ2 Ha: μ1 ≤ μ2 Ha: μ1=μ2 E. H0: μ1 ≤ μ2 F. H0: μ1<μ2 Ha: μ1>μ2
Transcript text: Test the claim about the difference between two population means $\mu_{1}$ and $\mu_{2}$ at the level of significance $\alpha$. Assume the samples are random and independent, and the populations are normally distributed.
Claim: $\mu_{1}=\mu_{2} ; \alpha=0.05$. Assume $\sigma_{1}^{2}=\sigma_{2}^{2}$
Sample statistics: $\bar{x}_{1}=32.3, s_{1}=3.6, n_{1}=14$ and
\[
\bar{x}_{2}=34.6, s_{2}=2.3, n_{2}=18
\]
Identify the null and alternative hypotheses. Choose the correct answer below.
A. $\mathrm{H}_{0}: \mu_{1} \geq \mu_{2}$ B. $H_{0}: \mu_{1}=\mu_{2}$ $\mathrm{H}_{\mathrm{a}}: \mu_{1}<\mu_{2}$ $\mathrm{H}_{\mathrm{a}}: \mu_{1} \neq \mu_{2}$
C. $\mathrm{H}_{0}: \mu_{1}>\mu_{2}$ D. $H_{0}: \mu_{1} \neq \mu_{2}$ $H_{a}: \mu_{1} \leq \mu_{2}$ $H_{a}: \mu_{1}=\mu_{2}$
E. $H_{0}: \mu_{1} \leq \mu_{2}$ F. $H_{0}: \mu_{1}<\mu_{2}$ $\mathrm{H}_{\mathrm{a}}: \mu_{1}>\mu_{2}$
Solution
Solution Steps
Step 1: Define the Hypotheses
We are testing the claim about the difference between two population means \( \mu_1 \) and \( \mu_2 \). The null and alternative hypotheses are defined as follows:
The critical value for a two-tailed test at the significance level \( \alpha = 0.05 \) with \( df = 30 \) is:
\[
\text{Critical Value} = 2.0423
\]
Step 7: Conclusion
Since the P-value \( 0.0 \) is less than the significance level \( \alpha = 0.05 \), we reject the null hypothesis \( H_0 \). This indicates that there is significant evidence to suggest that the means of the two populations are different.