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 40 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 $39, and the standard deviation for female consumers is assumed to be $20.

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.
   (  ,  )

Answer 1

Solution:

Given:

Male:

Sample size =n₁ = 40

Sample mean =

Population Standard Deviation =

Female:

Sample size =n₂ = 35

Sample mean =

Population Standard Deviation =

Part a) the point estimate of the difference between the population mean expenditure for males and the population mean expenditure for females

Part b) At 99% confidence, what is the margin of error

Z_c is z critical value for c = 0.99 confidence level.

Find Area = ( 1+c)/2 = ( 1 + 0.99 ) / 2 = 1.99 /2 = 0.9950

Thus look in z table for Area = 0.9950 or its closest area and find corresponding z critical value.

From above table we can see area 0.9950 is in between 0.9949 and 0.9951 and both are at same distance from 0.9950, Hence corresponding z values are 2.57 and 2.58

Thus average of both z values is 2.575

Thus Z_c = 2.575

Thus

Part c) Develop a 99% confidence interval for the difference between the two population means:

we have:

thus

Thus  a 99% confidence interval for the difference between the two population means is: