Question

Assume that population means are to be estimated from the samples described. Use the sample results to approximate the margin of error and​ 95% confidence interval.

Sample size=1,048​,

sample mean=​$46,209​,

sample standard deviation=​$21,000

The margin of error is $ __

Answer 1

Solution :

Given that,

= 46,209​

s =21,000

n = 1,048

Degrees of freedom = df = n - 1 = 1,048 - 1 = 1,047

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,_1,047 =1.962

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

= 1.962 * (21,000 / 1,048)

= 648.69

Margin of error = 648.69