Question

Bernard Associates conducted a survey in March 2004 of 1021 workers who held white-collar jobs and had changed jobs in the previous twelve months. Of these workers, 46% of the men and 37% of the women were paid more for their positions when they changed jobs. Suppose that these percentages are based on random samples of 510 mean 511 woman white-collar workers.

Construct a 95% confidence interval for the difference between the two population proportions.

Using a 1% significance level can you conclude that the two populations are different?

Answer 1

To Test :-

H0 :-  

H1 :-  

Test Statistic :-

is the pooled estimate of the proportion P
= ( x1 + x2) / ( n1 + n2)
= ( 234.6 + 189.07 ) / ( 510 + 511 )
= 0.415

Z = 2.9183   2.92

Test Criteria :-
Reject null hypothesis if

= 2.9183 > 2.5758, hence we reject the null hypothesis
Conclusion :- We Reject H0

Decision based on P value
P value = 2 * P ( Z < 2.9183 )
P value = 0.0036
Reject null hypothesis if P value <
Since P value = 0.0036 < 0.01, hence we reject the null hypothesis
Conclusion :- We Reject H0

There is sufficient evidence to support the claim that  the two populations are different.

Lower Limit =
upper Limit =
95% Confidence interval is ( 0.0298 , 0.1502 )
( 0.0298 < ( P1 - P2 ) < 0.1502 )