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 46 male consumers was $135.67, and the average expenditure in a sample survey of 40 female consumers was $68.64. Based on past surveys, the standard deviation for male consumers is assumed to be $32, and the standard deviation for female consumers is assumed to be $22. 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)? At 99% confidence, what is the margin of error (to 2 decimals)? Develop a 99% confidence interval for the difference between the two population means (to 2 decimals). Use z-table. (  ,   )

Answer 1

Here, we assume that the samples are taken from approximately normal distribution and the samples are random.

Given :

Sample mean for male consumer =

Sample mean for female consumer =

Standard deviation for male consumer =

Standard deviation for female consumer =

Sample size for male consumer = n1 = 46

Sample size for female consumer = n2 = 40

Significance Level = = 1 - 0.99 = 0.01

Critical Value =   ( From normal probability table )

(a) The point estimate of the difference between the two population means =

(b) The Margin of error E is given by ,

Therefore , margin of error = E = 15.1235

(c) The 99% confidence interval for the difference between the two population means is given by ,

Lower Limit =

Upper Limit =

Therefore , the 99% confidence interval using Z-distribution is