Question

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

Sample​ size,

nequals

100

​;

sample​ mean,

x overbar

equals70.0​cm; sample standard​ deviation,

sequals

5.0cm

Round to one decimal place

Answer 1

Solution :

Given that,

= 70 cm

s =5 cm

n = Degrees of freedom = df = n - 1 = 100- 1 = 99

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,99 =1.984 ( using student t table)

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

= 1.984* ( 5/ 100)

= 0.992

margin of error E=1 (ROUND)

The 95% confidence interval is,

- E < < + E

70 - 1 < < 70+ 1

69 < < 71

( 69 , 71)