In: Statistics and Probability
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 48 male consumers was $135.67, and the average expenditure in a sample survey of 34 female consumers was $68.64. 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 $23.
Develop a 99% confidence interval for the difference between the two population means (to 2 decimals).
Solution:
Given:
Male:
Sample mean =
Population standard deviation =
Sample size = n1 = 48
Female:
Sample mean =
Population standard deviation =
Sample size = n2 = 34
We have to find a 99% confidence interval for the difference between the two population means .
Formula;
where
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
Thus
Thus we are 99% confident that the true difference between the two population means of male and female expenditure on Valentine's Day is between ( $50.53 , $83.53 )