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)