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=100100​;

sample​ mean,

x overbarxequals=77.077.0

​cm; sample standard​ deviation,

sequals=5.05.0

cm

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 77.0

sample standard deviation = s = 5.0

sample size = n = 100

Degrees of freedom = df = n - 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 = t_0.025,99 = 1.984

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

= 1.984* ( 5 / 100)

= 0.992

The 95% confidence interval estimate of the population mean is,

- E < < + E

77 - 0.992 < < 77 + 0.992

76.008 < < 77.992

(76.008 , 77.992)