Question

1. If n=28, x¯ (x-bar)=50, and s=6, find the margin of error at a 95% confidence level (use at least two decimal places)

2. What is the margin of error for a poll with a sample size of 2400 people? Round your answer to the nearest hundredth of a percent.

%

3. If you want a poll to have a margin of error of 3.24%, how large will your sample have to be? Round your answer to the nearest whole number.

__people

Answer 1

Solution :

1)

sample size = n = 28

Degrees of freedom = df = n - 1 = 27

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,27 = 2.052

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

= 2.052 * ( 6/ 28)

= 2.33

2)

sample size = n = 2400

Degrees of freedom = df = n - 1 = 2399

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,2399 = 1.961

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

= 1.961 * ( 6/ 2400)

= 0.2402

= 24.02%

3)

Margin of error = E = 0.0324

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96

sample size = n = (Z/2* / E) 2

n = (1.96 * 6/ 0.0324)2

n = 362.96

n = 363

Sample size = 363