A random sample of 5 homes in Newberry, Florida had a mean of $169,900 and a standard deviation of $21,756. Construct a 95% confidence interval for the average home of this size in Newberry. A. (144885.25, 194914.75).
B. (150440.842, 189359.16).
C. (150830, 188969.98).
D. (142886, 196914).

Using the t-distribution, it is found that the 95% confidence interval for the average price of a home of this size in Newberry is of (142886, 196914), given by option D.

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

First, we find the number of degrees of freedom, which is the sample size subtracted by 1, so df = 5 - 1 = 4.

Then, we find the critical value for the 95% confidence interval with 4 df, which looking at the t-table or using a calculator is given by t = 2.7765.

The margin of error is:

In this case:

The confidence interval is:

Then

.

.

Thus, the correct option is D. (142886, 196914).

A similar problem is given at link

D. (142886, 196914).

Step-by-step explanation:

To solve for this question, we would use t score instead of z score. The reason is because, our sample size is small.

The formula for Confidence Interval =

C.I = Mean ± t score × standard deviation/√n

n = 5 samples

Degrees of freedom = n - 1 = 5 - 1 = 4

The t score for a 95% confidence interval with degrees of freedom 5 is : 2.776

Hence,

Confidence Interval =

169,900 ± 2.776 × 21,756/√5

169,900 ± 2.776 × 9.7295789837

169, 900 ± 27009.311259

Confidence Interval

169,900 - 27009

= 142886

169, 900 + 27009.311259

= 196914

Hence, the confidence interval = (142886, 196914)

Option D is the correct answer