Question

How much money does the average professional football player fan spend on food at a single football​ game? That question was posed to 58 randomly selected football fans. The sample results provided a sample mean and standard deviation of​ $15.00 and​ $3.00, respectively. Find the​ 95% confidence interval for u.

Answer 1

Solution :

Given that,

= $15

s = $3

n = 58

Degrees of freedom = df = n - 1 = 58 - 1 = 57

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,57 = 2.002

Margin of error = E = t/2,df * (s /n)

= 2.002 * (3 / 58)

= 0.8

The 95% confidence interval estimate of the population mean is,

- E < < + E

15 - 0.8 < < 15 + 0.8

14.2 < < 15.8

(14.2,15.8)  

The 95% confidence interval 14.2 to 15.8