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 49 male consumers was $135.67, and the average expenditure in a sample survey of 31 female consumers was $68.64. Based on past surveys, the standard deviation for male consumers is assumed to be $36, and the standard deviation for female consumers is assumed to be $18.

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

a)

point estimate = 135.67 - 68.64 = 67.03

b)

Given CI level is 0.99, hence α = 1 - 0.99 = 0.01                  
α/2 = 0.01/2 = 0.005, zc = z(α/2, df) = 2.576                  

Pooled Variance
sp = sqrt(s1^2/n1 + s2^2/n2)
sp = sqrt(1296/49 + 324/31)
sp = 6.0746
                  
Margin of Error                  
ME = zc * sp                  
ME = 2.576 * 6.0746                  
ME = 15.65

c)
                  
                  
CI = (x1bar - x2bar - tc * sp , x1bar - x2bar + tc * sp)                  
CI = (135.67 - 68.64 - 2.576 * 6.0746 , 135.67 - 68.64 - 2.576 * 6.0746                  
CI = (51.38 , 82.68)

### if you take z value upto 2 decimal i. e . 2.58 answer would be change