Final answer:
The hypothesis test involves comparing the average income of households in the area to $25,000. The null hypothesis is that the average income is $25,000 and the alternative hypothesis is that the average income is greater than $25,000. Based on a random sample of 9 households, a z-test is used to compare the sample mean with the population mean. The decision is to reject the null hypothesis, suggesting that there is enough evidence to support the claim that the average income of households in the area is greater than $25,000.
Step-by-step explanation:
The answer to this question is option C. H₀: μ = 25000, H₁: μ > 25000, z critical, z test, Reject H₀. The null hypothesis (H₀) states that the average income of households in the area is equal to $25,000. The alternative hypothesis (H₁) states that the average income of households in the area is greater than $25,000.
To test this hypothesis, a random sample of 9 households is taken. The z-test is used to compare the sample mean with the population mean. The critical value for a one-tailed test at a significance level of 0.05 is found from the z-table. If the calculated test statistic (z-test) is greater than the critical value, then the null hypothesis is rejected.
In this case, the decision is to reject the null hypothesis (H₀) because the calculated test statistic (z-test) is greater than the critical value. This suggests that there is enough evidence to support the claim that the average income of households in the area is greater than $25,000.