Final answer:
To test whether a higher rate of sump pumps exists in Chicago compared to the national average, a one-tailed hypothesis test for proportions is conducted. The p-value obtained from the z-score will indicate whether the null hypothesis can be rejected, establishing if significant evidence is present.
Step-by-step explanation:
To determine whether there is significant evidence that homes with basements in Chicago have sump pumps at a higher rate than the average in the US, we can use a hypothesis test for proportions. We want to compare the sample proportion (p') to the population proportion (p).
Here are the steps for the hypothesis test:
- State the null hypothesis (H0): The proportion of homes with sump pumps in Chicago equals the national average (p = 0.5).
- State the alternative hypothesis (Ha): The proportion of homes with sump pumps in Chicago is greater than the national average (p > 0.5).
- Calculate the sample proportion (p') as the number of homes with sump pumps divided by the total number of homes surveyed: p' = 392/673.
- Use the standard error of the proportion (SE) to calculate the z-score for the sample proportion. SE = sqrt(p(1-p)/n), where n is the sample size.
- Find the p-value associated with the observed z-score.
- Compare the p-value to the significance level (α). If the p-value is less than α, reject the null hypothesis.
Since we're applying a one-tailed test (because the alternative hypothesis specifies a direction), the significance level α will typically be set at 0.05 for a 95% confidence level.
After calculating the z-score and finding the p-value, if the p-value is less than 0.05, we would conclude that there is significant evidence to suggest that a greater proportion of homes in Chicago with basements have sump pumps compared to the national average.