The average expenditure on Valentine's Day was expected to be $100.89 (USA Today, February 13, 2006). 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. The z value is 2.576 .

Answer:a) 67.03b) 17.02c) (50.01, 84.05)Step-by-step explanation:Given that:the male average expenditure [tex]\bar{x_1} = 135.67[/tex]the female average expenditure [tex]\bar{x_2} = 68.64[/tex]sample survey of the male  [tex]n_1 = 40[/tex]sample survey of the female  [tex]n_2 = 30[/tex]standard deviation of the male [tex]\sigma_1 = 35[/tex]standard deviation of the female [tex]\sigma_2 = 20[/tex]The Z-score is not given but it is meant to be =2.576a) the point estimate of the difference between the population mean expenditure for males and the population mean expenditure for females is:[tex]\bar {x1} - \bar{x_2}[/tex] = 135.67 - 68.64= 67.03b) At 99% confidence, the margin of error is calculated as:[tex]E = Z \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} }[/tex][tex]E = (2.576) \sqrt{\frac{35^2}{40} + \frac{20^2}{30} }[/tex][tex]E = (2.576) \sqrt{30.625 + 13.33} }[/tex][tex]E = 2.576*6.629[/tex]E =   17.02c) the  99% confidence interval for the difference between the two population means  is as follows:[tex]{( \bar {x_1} - \bar {x_2}) \pm z\sqrt{ \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} }[/tex]= [tex]{( 135.67- 68.64}) \pm (2.576) \sqrt{ \frac{35^2}{40} + \frac{20^2}{30} }[/tex]=  [tex]67.03 \pm 17.02[/tex]= (50.01, 84.05)