A human resources HR director wants to investigate the natur

A human resources (HR) director wants to investigate the nature of pay in the modern workplace. She will use data from the 2014 General Social Survey (GSS), a comprehensive survey of American adults 18 and older. The data are shown below.

a). Find a 95% confidence interval for the difference in the proportion of females and males who are paid by the hour. Show your calculations. An answer without work shown is worth 0 points.

b). Interpret the interval above in the context of the problem.
   

c). The human resources director now wishes to study whether there is a significant difference in the mean number of hours spent at work per week between men and women in the U.S. who are paid by the hour. Analyzing the 2014 GSS data again, she obtains the table below.

Find a 95% confidence interval for the difference in the mean amount of time spent working per week between Females (Group 1) and Males (Group 2), assuming equal population standard deviations. Show all calculations.

d). Interpret the interval above in the context of the problem.

e). Based on your answer to (c), if the HR director were to test the null hypothesis that the mean working times are equal between males and females at the 0.05 level of significance, what her conclusion be, and why?

Solution

Q1.
a)
Confidence Interval for Diffrence of Proportion
CI = (p1 - p2) ± Z a/2 Sqrt(p1(1-p1)/n1 + p2(1-p2)/n2 )
Proportion 1
Probability Succuses( X1 )=60
No.Of Observed (n1)=203
P1= X1/n1=0.296
Proportion 2
Probability Succuses(X2)=46
No.Of Observed (n2)=207
P2= X2/n2=0.222
C.I = (0.296-0.222) ±Z a/2 * Sqrt( (0.296*0.704/203) + (0.222*0.778/207) )
=(0.296-0.222) ± 1.96* Sqrt(0.002)
=0.073-0.085,0.073+0.085
=[-0.011,0.158]

b)
we are 95% confident there is no significance b/w them as zero lies interval