Question

A sample of size 9 was taken from some population, and the sample standard deviation was 5. The confidence interval for the true mean came out to be ( 23.4 , 29.6 ). What was the confidence level?

Answer 1

Solution:

n = 9

s = 5

d.f. = n - 1 = 9 - 1 = 8

S.E. = s/n = 5/9 = 1.66666666667

confidence interval for the true mean came out to be ( 23.4 , 29.6 )

Margin of error = (Upper Limit - Lower Limit)/2

= (29.6 + 23.4)/2

= 3.1

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

3.1 = t_/2,df * 1.66666666667

t_/2,df = 1.860

But d.f = 8

Use t table, See 8 d.f.and see where is the value 1.860. It is at 0.05

So , /2 = 0.05

= 0.10

Now , confidence level = 1 - = 1 - 0.10 = 0.90 = 90%

Answer: The confidence level is 0.90 i.e. 90%