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A real-estate agent compares the selling prices of homes in a municipality in southwestern PA compared to the average selling price of a home in PA(125,000)to see if there is a difference. The results of her study are shown. Is there enough evidence to reject the claim that the average cost of a home in Scott Township is different than the average population price in PA?

Scott: Xbar = 93,430 o = 5,602 n = 55

a) what interval method would you use to create your interval and why?

User Amarnath
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Final answer:

The appropriate interval method for comparing the average cost of a home in Scott Township to the average in Pennsylvania is a z-interval, given a known population standard deviation and a sample size greater than 30. However, a hypothesis test would be more relevant in this context to check if there is a significant difference from the state average.

Step-by-step explanation:

To determine if there is enough evidence to reject the claim that the average cost of a home in Scott Township is different from the average population price in Pennsylvania, we can use a confidence interval method.

The most appropriate interval method for this scenario would be a z-interval, as we are dealing with a known population standard deviation and sample size (n) is large (n > 30).

A z-interval can be calculated using the formula:

Z-interval = Xbar ± (z* × (o / sqrt(n)))

Where Xbar is the sample mean, o is the population standard deviation, n is the sample size, and z* is the z-score corresponding to the desired confidence level.

Generally, for a 95% confidence interval, the z-score used is approximately 1.96.

However, in this particular case, since we are interested in testing if there is a difference compared to a specified value (the state average), a hypothesis test would be more suitable than a confidence interval.

A two-tailed hypothesis test using a t-test might be considered if the population standard deviation were unknown, but with a known standard deviation, the z-test is applicable.

User Yevt
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