Question

Confidence Intervals – Means & Proportions

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?

Answer 1

Solution

Back-up Theory

Let X = starting salaries of recently graduated business majors

Let μ and σ be the mean and standard deviation of X.

100(1 - α) % Confidence Interval for μ, when σ is known: Xbar ± MoE…….. (1)

Where

MoE = (Z_{α /2})σ/√n ………………………………….......................………………………..... (2)

With

Xbar = sample mean,

Z_{α /2} = upper (α/2)% point of N(0, 1),

σ = population standard deviation and

n = sample size.

Now to work out the solution,

Part (a)

Vide (1), 90% confidence interval to estimate the average starting salary of a recently graduated business major is: [42059.16, 45540.84]   Answer 1

Details of calculations

Given	α	0.1	1 - (α/2) =	0.95
	n	60	SQRT(n)	7.74596669
	σ	8198
	Xbar	43800
	Zα/2	1.6449
95% CI for μ		43800	±	1740.8428
Lower Bound		42059.157
Upper Bound		45540.843

Part (b)

Vide (2), Margin of Error = (Z_{α /2})σ/√n. We want this to be 500. So, we should have:

(1.6449 x 8198)/√n = 500 [vide the above Excel output]

Or, n = {(1.6449 x 8198)/500}²

= 745.

Thus, number of recently graduated business majors to be sampled is 745 Answer 2

[Going beyond,

Answer 2 could be got directly from Answer 1 by using the result:

if M₁ is the MoE with n = n₁ and M₂ is the MoE with n = n₂, then, given M₁, M₂ = M₁ x √(n₁/n₂)]

DONE