Question

# The USA Today reports that the average expenditure on Valentine's Day is \$100.89. Do male and...

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.

1. 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)?
2. At 99% confidence, what is the margin of error (to 2 decimals)?
3. Develop a 99% confidence interval for the difference between the two population means (to 2 decimals). Use z-table.

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]

#### Earn Coins

Coins can be redeemed for fabulous gifts.