Question

A study of 25 graduates of four-year public colleges revealed the mean amount owed by a student in student loans was $55,051. Assume we conducted research and based on historical data found that the standard deviation for the population was $7,568. Construct a 90% confidence interval for the population mean. Carry all calculations out 4 decimal places.

Answer 1

Question 1

Solution:

Confidence interval for Population mean is given as below:

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

From given data, we have

x̄ = 55051

σ = 7568

n = 25

Confidence level = 90%

Critical Z value = 1.6449

(by using z-table)

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

Confidence interval = 55051 ± 1.6449*7568/sqrt(25)

Confidence interval = 55051 ± 2489.6504

Lower limit = 55051 - 2489.6504 = 52561.3496

Upper limit = 55051 + 2489.6504 = 57540.6504

Confidence interval = (52561.3496, 57540.6504)

Question 2

The sample size formula is given as below:

n = (Z*σ/E)^2

We are given

Population standard deviation = σ = 7568

Confidence level = 88%

Critical Z value = 1.5548

(by using z-table/excel)

Margin of error = E = 1000

The sample size is given as below:

n = (Z*σ/E)^2

n = (1.5548*7568/1000)^2

n = 138.4559

n = 139

Required sample size = 139