Answer:
a) 67.03
b) 17.02
c) (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 =2.576
a) 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
= 67.03
b) 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 = 17.02
c) 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)
