Question

Isaac is a professional swimmer who trains, in part, by running. She would like to     estimate the average number of miles she runs in each week. For a random sample     of 20 weeks, the mean is x = 17.5 miles with standard deviation s = 3.8 miles. Find      a 99% confidence interval for the population mean number of weekly miles Isaac runs.
(a) 15.01 to 19.99 miles (b) 15.07 to 19.93 miles      (c) 15.34 to 19.66 miles (d) 15.31 to 19.69 miles   (e) 15.08 to 19.92 miles

Answer 1

solution

Given that,

= 17.5

s =3.8

n = 20

Degrees of freedom = df = n - 1 =20 - 1 =19

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2,df= t0.005, 19= 2.861    ( using student t table)

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

= 2.861* (3.8 / 20) = 2.43

The 99% confidence interval estimate of the population mean is,

- E < < + E

17.5-2.43 < < 17.5+ 2.43

15.07 < < 19.93

( 15.07 , 19.93)