Question

1.    You would like to estimate the starting salaries of recently graduated business majors (B.S. in any business degree). You randomly select 60 recently graduated business majors and get a sample mean of $43,800 and the population standard deviation is known to be $8,198

A.    Construct a 90% confidence interval to estimate the average starting salary of a recently graduated business major (Round to the nearest penny and state the answer as an interval – for example $351.89 to $728.14).

B.    Using the same confidence level, you would like the margin of error to be within $500, how many recently graduated business majors should you sample?

2.    You would like to estimate the amount of student loan debt a graduating senior will have at the time of repayment which begins in November. You randomly select 72 graduating seniors and get a sample mean of $31,172 with a standard deviation of $6,423. Construct a 98% confidence interval for the amount of debt a graduating senior will have. (Make sure you are careful selecting the correct values and that you round to the nearest penny. You will not get any credit if you are off by more than 1 cent.) (Round to the nearest penny and state the answer as an interval – for example $351.89 to $728.14).

3.    You would like to estimate the proportion of student loan debts that are in default. You randomly select 211 people who have student loans and find that 25 are in default.

A.    Construct a 95% confidence interval to estimate the proportion of student loans that are in default. (Round your sample proportion to 4 decimal points as well as round your margin of error to 4 decimal points. For example: .23916 would be .2392 and this represents 23.92%. State the answer as an interval – for example 27.36% to 31.43%).

B.    Using the same confidence level, you would like the margin of error to be with 3%, how many people with student loans should you sample?

Answer 1

1) A) At 90% confidence level, the critical value is z_0.05 = 1.645

The 90% confidence interval is

= 42059.002, 45540.998

= 42059.00, 45541.00

B) Margin of error = 500

or, n = 728

2) df = 72 - 1 = 71

At 98% confidence level, the critical value is t^* = 2.380

The 98% confidence interval is

= 29370.44, 32973.56

3) = 25/211 = 0.1185

At 95% confidence level, the critical value is z_0.025 = 1.96

The 95% confidence interval is

+/- sqrt((1 - )/n)

= 0.1185 +/- 1.96 * sqrt(0.1185(1 - 0.1185)/211)

= 0.1185 +/- 0.0436

= 0.0749, 0.1621

= 7.49%, 16.21%

B) Margin of error = 0.03

or, sqrt(p(1 - p)/n) = 0.03

or, sqrt(0.1185(1 - 0.1185)/n) = 0.03

or, n = (sqrt(0.1185(1 - 0.1185))/0.03)^2

or, n = 117