Question

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.

Answer 1

Answer:

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

Answer 2

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 sample mean is $\overline{x} = 156100$.
• The sample standard deviation is $s = 13660$.
• Sample of 4 homes, thus $n = 4$.

The first step is finding the number of degrees of freedom, which is the sample size subtracted by 1, thus $df = 4 - 1 = 3$.

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:

$M = t\frac{s}{\sqrt{n}}$

Thus:

$M = 3.1824\frac{13660}{\sqrt{4}} = 21736$

The confidence interval is:

$\overline{x} \pm M$

Then

$\overline{x} - M = 156100 - 21736 = 134364$

$\overline{x} + M = 156100 + 21736 = 177836$

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

A similar problem is given at https://brainly.com/question/15180581