Question

The analysis of a random sample of 550 households in a suburb of a large city indicates that a 92% confidence interval for the mean family income is ($45,700, $59,150). Could this information be used to conduct a test of the null hypothesis H₀: µ = 44,000 against the alternative hypothesis H_a: µ ≠ 44,000 at the α = 0.08 level of significance?

Yes, because $44,000 is not contained in the 92% confidence interval, the null hypothesis would be rejected in favor of the alternative, and it could be concluded that the mean family income is significantly different from $44,000 at the α = 0.08 level

Yes, because $44,000 is not contained in the 92% confidence interval, the null hypothesis would not be rejected, and it could be concluded that the mean family income is not significantly different from $44,000 at the α = 0.08 level

No, because the value of σ is not known

No, because it is not known whether the data are Normally distributed

No, because the entire data set is needed to do this test

Answer 1

Given that, the null and alternative hypotheses are,

H0 : μ = 44000

Ha : μ ≠ 44000

The 92% confidence interval for the mean family income is ($45,700, $59,150).

Since, population mean = 44000 is not lies in above inteval we should reject the null hypothesis.

Conclusion : Yes, because $44,000 is not contained in the 92% confidence interval, the null hypothesis would be rejected in favor of the alternative, and it could be concluded that the mean family income is significantly different from $44,000 at the α = 0.08 level