Question

random sample of 19 size AA batteries for toys yield a mean of 3.45 hours with standard deviation, 1.23 hours.

(a) Find the critical value, ??, for a 99% CI. ?? =

(b) Find the margin of error for a 99% CI.

Answer 1

c )solution

Given that,

= 3.45

s =1.23

n = 19

Degrees of freedom = df = n - 1 = 19 - 1 = 18

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2  df = t0.005,18= 2.878    ( using student t table)

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

= 2.878* (1.23 / 19) = 0.812