Question

If n=18, ¯xx¯(x-bar)=40, and s=10, find the margin of error at a 99% confidence level

Give your answer to two decimal places.

Answer 1

solution

s =10

n = 18

Degrees of freedom = df = n - 1 = 18- 1 =17

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2  df = t0.005,17 = 2.899    ( using student t table)

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

= 2.899* (10 / 18)

E = 6.83

Margin of error = E = 6.83