Question

Mrs. Warrack, a retired archeologist, enjoys hiking in Badger Creed every other day. Usually she travels 3 miles per hike. During the past year Mrs. Warrack has recorded the times of her hikes. For a sample of 90 times, the mean was x ̅=23.18 minutes and the standard deviation was s = 1.87 minutes. Let μ mean the mean hiking time for the entire distribution of Mrs. Warrack’s 3-mile hike times (taken over the past year )    

How many degrees of freedom are there?    

Find a 0.95 confidence interval μ. Use t*= 1.987 .

Answer 1

Solution :

Given that,

= 23.18

s = 1.87

n = 90

Degrees of freedom = df = n - 1 = 90 - 1 = 89

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,89 =1.987

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

= 1.987 * (1.87 / 90) = 0.392 (rounded)

The 95% confidence interval is,

- E < < + E

23.18 - 0.392 < < 23.18 + 0.392

22.788 < < 23.572

(22.788, 23.572 )