Answer to Question #169484 in Statistics and Probability for Sunera

Question

A statistics practitioner took a random sample of 50 observations from a population whose

standard deviation is 25 and computed the sample mean to be 100.

a) Estimate the population mean with 90% confidence.

b) Repeat part (a) using a 95% confidence level.

c) Repeat part (a) using a 99% confidence level.

d) Describe the effect on the confidence interval estimate of increasing the confidence level.

Answer 1

a)

sample mean 'x̄=100.00

sample size   n=50.00

std deviation σ=25.000

std errror ='σx=$σ/\sqrt{n}$ =3.5355

for 90 % CI value of z=1.645

margin of error E=z*std error =5.815

lower bound=sample mean-E=94.18

Upper bound=sample mean+E=105.82

from above 90% confidence interval for population mean =(94.18,105.82)

b)

for 95 % CI value of z=1.960

margin of error E=z*std error =6.930

lower bound=sample mean-E=93.07

Upper bound=sample mean+E=106.93

from above 95% confidence interval for population mean =(93.07,106.93)

c)

for 99 % CI value of z=2.576

margin of error E=z*std error =9.107

lower bound=sample mean-E=90.89

Upper bound=sample mean+E=109.11

from above 99% confidence interval for population mean =(90.89,109.11)

d)increasing the confidence level, increases the width of confidence interval