Question

A random sample of 6 homes in Gainesville, Florida between 1800 and 2200 square feet had a mean of 212990 and a standard deviation of 14500. There are no outliers in the sample data set. Construct a 95% confidence interval for the average price of a home in Gainesville of this size.

Answer 1

Answer:

The 95% confidence interval for the average price of a home in Gainesville of this size is between 183,772.5 square feet and 242,207.5 square feet.

Step-by-step explanation:

We have the standard deviation of the sample, so we use the students t-distribution.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 6 - 1 = 5

Now, we have to find a value of T, which is found looking at the t table, with 5 degrees of freedom(y-axis) and a confidence level of 0.95($t_{95}$). So we have T = 2.015

The margin of error is:

M = T*s = 2.015*14500 = 29217.5.

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 212990 - 29217.5 = 183,772.5 square feet

The upper end of the interval is the sample mean added to M. So it is 212990 + 29217.5 = 242,207.5 square feet

The 95% confidence interval for the average price of a home in Gainesville of this size is between 183,772.5 square feet and 242,207.5 square feet.