In: Math
A newspaper 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 30 female consumers was $61.64. Based on past surveys, the standard deviation for male consumers is assumed to be $35, and the standard deviation for female consumers is assumed to be $20.
(a)
What is the point estimate (in dollars) of the difference between the population mean expenditure for males and the population mean expenditure for females? (Use male − female.)
$
(b)
At 99% confidence, what is the margin of error (in dollars)? (Round your answer to the nearest cent.)
$
(c)
Develop a 99% confidence interval (in dollars) for the difference between the two population means. (Use male − female. Round your answer to the nearest cent.)
$ to $
Solution:
Given:
Sample 1 Male:
n1 = 40
Sample 2 Female:
n2 = 30
Population standard deviations of male and female expenditures is:
and
Part a)
The point estimate (in dollars) of the difference between the population mean expenditure for males and the population mean expenditure for females is given by:
Part b)
The margin of error at 99% confidence level is given by:
Zc 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 Zc = 2.575
Thus we get:
Part c)
A 99% confidence interval (in dollars) for the difference between the two population means is given by:
From part a) we have and from part b) we have
Thus
Thus A 99% confidence interval (in dollars) for the difference between the two population means is between :