The Canadian government would like to test the hypothesis that the standard deviation for the hourly wage for men is equal to the standard deviation for the hourly wage for women. The following data summarizes the sample statistics for hourly wages for men and women.

Men Women
Sample mean $25.40 $21.20
Sample size 20 18
Sample standard deviation $6.20 $5.90

If Population 1 is defined as men and Population 2 is defined as women, and using α = 0.025, which one of the following statements is true?

a) Because the p-value is less than α, the Canadian government cannot conclude that the average hourly wage for men is more than $2.00 higher than the average hourly wage for women.
b) Because the p-value is less than α, the Canadian government can conclude that the average hourly wage for men is more than $2.00 higher than the average hourly wage for women.
c) Because the p-value is greater than α, the Canadian government can conclude that the average hourly wage for men is more than $2.00 higher than the average hourly wage for women.
d) Because the p-value is greater than α, the Canadian government cannot conclude that the average hourly wage for men is more than $2.00 higher than the average hourly wage for women.

Because the p-value is greater than α, the Canadian government can conclude that the average hourly wage for men is more than $2.00 higher than the average hourly wage for women.

Step-by-step explanation:

See attached picture.

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codiepienagoya codiepienagoya · Answer 2 · 2022-04-07T19:57:13+02:00

P-value is greater than α, the Canadian government can't include the average hourly wage for men as more than $2 higher than the average hourly wage for women.

P-value calculation:

Calculating the pooled value by the Standard deviation:

[tex]\to Sp=\sqrt{(\frac{(n_1-1)\times s^2_1+(n_2-1)\times s^2_2)}{(n_1+n_2-2)}[/tex]

by solving the value we get

[tex]=6.0602[/tex]

Calculating the point of estimation:

[tex]\to x_1-x_2= 4.2000[/tex]

Calculating the standard error (se):

[tex]\to S_p\times \sqrt{(\frac{1}{n_1}+\frac{1}{n_2})}= 1.9689[/tex]

Calculating the testing start (t):

[tex]\to \frac{(x_1-x_2-\Delta o)}{Se}= 1.1170[/tex]

Calculating the p-value:

[tex]\to 0.1357[/tex]

Using an excel table to calculate value: [tex]\text{tdist(1.117,36,1)}[/tex]

Find out more information about the p-value here:

brainly.com/question/19881297