Question

Determine the margin of error for a 99​% confidence interval to estimate the population mean when s​ = 43 for the sample sizes below. ​a) n=12 ​b) n=25 ​c) n=46 ​a) The margin of error for a 99​% confidence interval when n=12 is _.

Answer 1

Solution :

Given that,

sample standard deviation = s = 43

a) sample size = n = 12

Degrees of freedom = df = n - 1 = 12 - 1 = 11

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

t/2,df = t_0.005,11 = 3.106

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

= 3.106 * (43 / 12)

Margin of error = E = 38.55

b) sample size = n = 25

Degrees of freedom = df = n - 1 = 25 - 1 = 24

t/2,df = t_0.005,24 = 2.797

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

= 2.797 * (43 / 25)

Margin of error = E = 24.05

c) sample size = n = 46

Degrees of freedom = df = n - 1 = 46 - 1 = 45

t/2,df = t_0.005,45 = 2.690

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

= 2.690 * (43 / 46)

Margin of error = E = 17.05