Question

A researcher is interested in studying the incomes of poeple in a country. The population standard deviation is known to be $954. A random sample of 58 individuals resulted in a mean income of $19,824.

the upper limit of  90.1 % confidence interval?

Answer 1

Solution:

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± Z*σ/sqrt(n)

From given data, we have

Xbar = 19824

σ = 954

n = 58

Confidence level = 90%

Critical Z value = 1.6497

(by using z-table)

Confidence interval = Xbar ± Z*σ/sqrt(n)

Confidence interval = 19824 ± 1.6497*954/sqrt(58)

Confidence interval = 19824 ± 206.6545

Lower limit = 19824 - 206.6545 = 19617.35

Upper limit = 19824 + 206.6545 = 20030.65

Confidence interval = (19617.35, 20030.65)

Required upper limit = 20030.65