Question

A business magazine samples 95 individuals responsible for economic forecasting for regional banks. The population is large enough that the with/without replacement distinction doesn’t matter. Suppose that the sample of 95 forecasts yields an average prediction of a 3.7% growth in real disposable income. Assume that the population standard deviation is 5%.

a) Calculate the lower bound for the 95% confidence interval for the population mean forecast.

b) Calculate the upper bound for the 95% confidence interval for the population mean forecast.

Answer 1

Solution:

Confidence interval for Population mean is given as below:

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

From given data, we have

x̄ = 3.7

σ = 5

n = 95

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

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

Confidence interval = 3.7 ± 1.96*5/sqrt(95)

Confidence interval = 3.7± 1.0054

Lower limit = 3.7- 1.0054 = 2.6946

Upper limit = 3.7+ 1.0054 = 4.7054

Confidence interval = (2.69%, 4.71%)

a) Calculate the lower bound for the 95% confidence interval for the population mean forecast.

Answer: 2.69%

b) Calculate the upper bound for the 95% confidence interval for the population mean forecast.

Answer: 4.71%