Question

In: Statistics and Probability

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 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).

Solutions

Expert Solution

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 )


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