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. Construct a 95% confidence interval for the average price of a home in Gainesville of this size. Group of answer choices (201387, 224592) (197773, 228207) (196318, 229662) (196557, 229422)

Answer 1

Answer:

Option B)  (197773, 228207)    

Step-by-step explanation:

We are given the following in the question:

Sample mean, $\bar{x}$ = 212990

Sample size, n = 6

Alpha, α = 0.05

Sample standard deviation = 14500

95% Confidence interval:

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$  

Putting the values, we get,  

$t_{critical}\text{ at degree of freedom 5 and}~\alpha_{0.05} = \pm 2.5705$  

$212990 \pm 2.5705(\dfrac{14500}{\sqrt{6}} ) \\\\= 212990 \pm 15216.33 \\= (197773.67 ,228206.33)\\\approx (197773, 228207)$  

Option B)  (197773, 228207)

Answer 2

Using the t-distribution, it is found that the 95% confidence interval for the average price of a home in Gainesville of this size is (197773, 228207).

The first step is finding the number of degrees of freedom, which is the sample size subtracted by 1, so df = 6 - 1 = 5.

Now, we look at the t-table for the critical value for a 95% confidence interval with 5 df, which is t = 2.5706.

The margin of error is:

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

For this problem, $s = 14500, n = 6$, thus:

$M = 2.5706\frac{14500}{\sqrt{6}}$

$M = 15217$

The confidence interval is:

$\overline{x} \pm M$

For this problem, $\overline{x} = 212990$, then:

$\overline{x} - M = 212990 - 15217 = 197773$

$\overline{x} + M = 212990 + 15217 = 228207$

The correct option is (197773, 228207).

A similar problem is given at https://brainly.com/question/15872108