Question

A random sample of 15 size AA batteries for toys yield a mean of 2.68 hours with standard deviation, 1.24 hours.

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

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

Answer 1

Given that,

= 2.68

s =1.24

n = 15

Degrees of freedom = df = n - 1 = 15- 1 = 14

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

_{t /2  df} = t_0.005, 14= 2.977 ( using student t table)

critical value =2.977

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

= 2.977* ( 1.24/ 15)

E=0.9531

Margin of error = E =0.9531