Question

Submit Assignment for Grading Save Exit tor Exercise 10.05 Algorithmic Question 1 of 3 Check My Work (No more tries available) 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 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 $39, and the standard deviation for female consumers is assumed to be $17. a. 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)? 67.03 b. At 99% confidence, what is the margin of error (to 2 decimals)? 16.30 c. Develop a 99% confidence interval for the difference between the two population means (to 2 decimals). Use z-table. 50.73 83.33 Hide Feedback Partially Correct Check My Work (No more tries available) 0 Icon Key Question 1 of 3 a 6:21 PM Exercise 10 05 Algorithmic 11/5/2019 xD

Answer 1

Solution: Males: n = 40 x1 = $135.67 01 = $39 Females: n2 = 30 X2 = $68.64 02 = $17 Part-A: Point estimate of the difference between the population mean expenditure for males and population mean expenditure for females = (x1 - x2) = $135.67 – $68.64 = $67.03 Part-B: For 99% confidence interval, a = 1-0.99 = 0.01 a/2 = 0.01/2 = 0.005 From t-distribution table, the value of z having an area of (a/2 = 0.005) in its upper tail, is given by = 20.005 = 2.576 Margin of Error: MoE = 20.005 X oi o vn N 392 172 → MoE = 2.576 x 30 V 40 → MoE = 17.78

Part-C: 95% confidence interval for the difference between the two population means: (x1 - x2) +MOE =$135.67 - $68.64 + 17.78 → 67.03 + 17.78 Lower limit of 99% confidence interval = 67.03 - 17.78 = 49.25 Upper limit of 99% confidence interval = 67.03 + 17.78 = 84.81 Hence, 99% confidence interval = (49.25, 84.81)