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 30 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 $20. 1. What is the point estimate of the difference between the population mean expenditure for males and the population mean expenditure for females?
2. At 99% confidence, what is the margin of error?
3. Develop a 99% confidence interval for the difference between the two population means.

- The following data is given for the average expenditures made on Valentine's Day. Samples from both gender populations were taken and the data obtained is given:

- The sample from male population, n1 = 40

- The average expenditure by males, x_bar1 = $135.67

- The standard deviation of expenditure made by male population, σ1 = $35

- The sample from female population, n2 = 30

- The average expenditure by females, x_bar2 = $68.64

- The standard deviation of expenditure made by female population, σ2 = $20

- The point estimation of difference in average expenditure by males and females on Valentine's day is:

x_bar1 - x_bar2 = $135.67 - $68.64

= $67.03

- The difference in average expenditures between both genders is estimated to be ( x_bar1 - x_bar2 ) = $ 67.03.

- The margin of error associated with the Confidence = 99% ( 0.99 ) is determined by the finding the Z-critical value associated with the significance level ( α ):

α = 1 - confidence

α = 1 - 0.99 = 0.01

Z-critical = Z_α/2 = Z_0.005

- Use the Z-score (standardized) look-up table and determine to the Z-score associated with:

P ( Z < Z-critical ) = α/2 = 0.005

Z-critical = 2.575

- Now determine the margin of error (ME) for different sample sizes n1 and n2 with known population standard deviations σ1 and σ2:

[tex]ME = Z_\alpha_/_2 *\sqrt{\frac{s1^2}{n1} + \frac{s2^2}{n2} } \\\\\\ME = 2.575 *\sqrt{\frac{35^2}{40} + \frac{20^2}{30} } \\\\\\ME = 17.0725[/tex]

- The margin of error (ME) is = 17.0725

- The interval estimate of the difference between two population means is calculated by:

[tex](x_1 - x_2) +/- ME[/tex]

- Hence,

point estimate ± ME

[ 67.03 ± 17.0725 ]

[ 67.03 - 17.0725 , 67.03 + 17.0725 ]

[ 49.9575 , 84.1025 ]