Question 1071804
Our sampling statistic is the proportion, namely
:
N1 = 600 and N2 = 400
:
women(p1) is 200/600 = 1/3 and men(P2) is 150/400 = 3/8
:
A) At 0.05 significance level
Ho: P1 = P2 and Ha: P1 not = P2
:
P = (200 + 150) / (600 + 400) = 350 / 1000 = 0.35
:
standard error(SE) = square root(0.35 * 0.65) = 0.477
:
test statistic(z-value) = ((1/3) - (3/8)) / SE = -0.0874
:
since our Ho has an =, we use a two tailed test
:
Therefore we split the 0.05 between the the high and low probabilities, that is 0.05/2 = 0.025
:
The associated z-scores for 0.025 is -1.96 and 1.96
:
Our test statistic(-0.0874) is not less than -1.96 and
0.0874 is not greater than 1.96
:
Therefore we can not reject our Ho
:
B) Since we have a two-tailed test, the P-value is the probability that the z-score is less than -0.0874 or greater than 0.0874.
:
Use z-tables or a Normal Distribution Calculator to find P(z < -0.0874) = 0.4641, and P(z > 0.0874) = 0.4641. Thus, the P-value = 0.4641 + 0.4641 = 0.9282.
:
Since the P-value (0.9282) is greater than the significance level (0.05), we accept the null hypothesis.