In: Statistics and Probability
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?
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 )