Question

The *USA Today* reports that the average expenditure on
Valentine's Day is $100.89. Do male and female consumers differ in
the amounts they spend? The average expenditure in a sample survey
of 46 male consumers was $135.67, and the average expenditure in a
sample survey of 35 female consumers was $68.64. Based on past
surveys, the standard deviation for male consumers is assumed to be
$30, and the standard deviation for female consumers is assumed to
be $16.

- What is the point estimate of the difference between the
population mean expenditure for males and the population mean
expenditure for females (to 2 decimals)?
- At 99% confidence, what is the margin of error (to 2
decimals)?
- Develop a 99% confidence interval for the difference between
the two population means (to 2 decimals). Use
*z*-table.

Answer #1

**Solution:**

Given that,

n1= 46, n2 = 35

x̅1 = $135.67, x̅2 = $68.64

σ1 = $30, σ2 = $16

a) x̅1- x̅2 = 135.67 - 68.64 = 67.03

----------------------------------------------------

b) at 99% confidence level

Level of significance α = 1-0.99 = 0.01

Critical value Zα/2 = Z0.005 = 2.58

Margin of error E = Zα/2 * sqrt[σ1^2/n1 + σ2^2/n2]

= (2.58) * sqrt[30^2/46 + 16^2/35]

= (2.58) *5.1845 = 13.38

Margin of error is 13.38

-----------------------------------------------------

c) 99% confidence interval for the difference between the two
population means

=> (x̅1- x̅2) - E ≤ μ1- μ2 ≤ (x̅1- x̅2) + E

=> 67.03 - 13.38 ≤ μ1- μ2 ≤ 67.03 + 13.38

=> 53.65 ≤ μ1- μ2 ≤ 80.41

=>[53.65, 80.41]

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Valentine's Day is $100.89. Do male and female consumers differ in
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