Question

The results for daily intakes of milk (in ounces) for ten randomly selected people were xˉ=18.23 and s=6.11. Find a 99% confidence interval for the population standard deviation σ. Assume that the distribution of ounces of milk intake daily is normally distributed.

What is the confidence interval?

lower limit

=Incorrect answer:

3.774

and the upper limit =

Incorrect answer:

13.916

Answer 1

solution

Given that,

= 18.23

s =6.11

n = 10

Degrees of freedom = df = n - 1 =10 - 1 = 9

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2  df = t0.005,9 = 3.250 ( using student t table)

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

=3.250 * (6.11 / 10) = 6.28 rounded 2 decimal

The 99% confidence interval is,

- E < < + E

18.23 - 6.28 < <18.23 + 6.28

11.95< < 24.51

lower limit 11.95

upper limit 24.51