A random sample of 4 homes for sale in NW Gainesville between 16000 and 20000 square feet had a mean of $156100 and a standard deviation of $13660. Construct a 95% confidence interval for the average home of this size.

Using the t-distribution, it is found that the 95% confidence interval for the average price of a home of this size is ($134,364, $177,836).

In this problem, we have the standard deviation for the sample, thus, the t-distribution is used.

The first step is finding the number of degrees of freedom, which is the sample size subtracted by 1, thus .

Then, we find the critical value for a 95% confidence interval with 3 df, which looking at a t-table or using a calculator is given by t = 3.1824.

The margin of error is:

Thus:

The confidence interval is:

Then

The 95% confidence interval is ($134,364, $177,836).

A similar problem is given at link

142713.2≤x≤169486.8

Step-by-step explanation:

Confidence interval is expressed according to the equation

CI = xbar± z•s/√n

xbar is the mean = $15610

z is the z-score at 95% CI = 1.96

s is the standard deviation = 13660

n is the sample size= 4

Substitute into the formula

CI = 156100±(1.96×13660/√4)

CI = 156100±(1.96×13660/2)

CI = 156100±(1.96×6830)

CI = 156100±(13386.8)

CI = (156100-13386.8, 156100-13386.8)

CI = (142,713.2, 169,486.8)

Hence the 95% confidence interval for the average home of this size is 142713.2≤x≤169486.8

i can you with this sweetie!

step-by-step explanation: