Answer :
To determine the appropriate conclusions for this hypothesis test, we need to analyze the given information and compare it to the significance level.
Step-by-Step Explanation:
Hypotheses Setup:
- The null hypothesis [tex]H_0[/tex]: [tex]\mu \geq 250,000[/tex].
- The alternative hypothesis [tex]H_a[/tex]: [tex]\mu < 250,000[/tex].
- This is a left-tailed test since we want to test if the mean is less than $250,000.
Significance Level:
- The test is conducted at a 5% significance level, or [tex]\alpha = 0.05[/tex].
Test Statistic and p-value:
- The test statistic is given as [tex]t = -2.082[/tex].
- The p-value is provided as 0.026.
Decision Rule:
- Compare the p-value to the significance level [tex]\alpha[/tex]:
- If [tex]\text{p-value} < \alpha[/tex], reject [tex]H_0[/tex].
- If [tex]\text{p-value} \geq \alpha[/tex], fail to reject [tex]H_0[/tex].
- Compare the p-value to the significance level [tex]\alpha[/tex]:
Comparison and Conclusion:
- In this case, [tex]0.026 < 0.05[/tex], so we reject [tex]H_0[/tex].
- At the 5% significance level, the data provide sufficient evidence to support the claim that the average price of a home in this zip code is less than $250,000.
Chosen Multiple Choice Options:
- Reject H0.
- At the 5% significance level, the data provide sufficient evidence to support the claim that the average price of a home in a certain zip code is less than $250,000.
These selections are appropriate because the p-value is less than the significance level, leading us to reject the null hypothesis.