Question

In a random sample of 8 people with advanced degrees in biology, the mean monthly income was $4744 and the standard deviation was $580. Assume the monthly incomes are normally distributed. Construct a 95% confidence interval for the population mean monthly income for people with advanced degrees in biology.

Answer 1

Answer: u= ( 4342.08, 5145.92).

Step-by-step explanation: the population mean is estimated using the sample by the formulae assuming a 95% confidence level

u = x' + Zα/2 * (√σ/n) or x' - Zα/2 * (√σ/n)

u = estimated population mean

x' = sample mean = 4744

n = sample size =8

σ = sample standard deviation. = 580

α = level of significance = 1- confidence level = 1-0.95= 0.05

Zα/2 = z score from the normal distribution table for a 2 tailed test = 1.96

First boundary value for interval

u = 4744 + 1.96 ( 580/√8)

u = 4744 + 1.96 * (205.0609)

u = 4744 + 401.92

u = 5145.92

Second boundary value for interval

u = 4744 - 1.96 ( 580/√8)

u = 4744 - 1.96 * (205.0609)

u = 4744 - 401.92

u = 4342.08

Thus the confidence interval for population mean is

u= ( 4342.08, 5145.92).