Question

Determine the margin of error for a confidence interval to estimate the population mean with nequals24 and s​ = 13.5 for the confidence levels below. ​a) 80​% ​b) 90​% ​c) 99​%

Answer 1

Solution :

Given that,

sample standard deviation = s = 13.5

sample size = n = 24

Degrees of freedom = df = n - 1 = 24 - 1 =23

a) At 80% confidence level

= 1 - 80%

=1 - 0.80 =0.20

/2 = 0.10

t/2,df = t0.10,23 = 1.319

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

= 1.319 * ( 13.5 / 24)

Margin of error = E = 3.63

b) At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

t/2,df = t0.05,23 = 1.714

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

= 1.714 * ( 13.5 / 24)

Margin of error = E = 4.72

c) At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

t/2,df = t0.005,23 = 2.807

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

= 2.807 * ( 13.5 / 24)

Margin of error = E = 7.74