Question 975169: The percentages of adults 25 years of age and older who have completed 4 or more years of college are 23.6% for females and 27.8% for males. A random sample of women and men who were 25 years old or older was surveyed with these results. Estimate the true difference in proportions with 95% confidence, and compare your interval with the Almanac statistics. The sample size is 350 for women and 400 for men with 100 women and 115 men completed 4 or more years respectively.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! This is a two sample proportion test.
The z value +/- the standard error will give the answer.
z= (phatw-phatm)/sqrt {[ (pw)(1-pw)/nw] + (pm)(1-pm)/pm]}
can do this by hand, but calculators are fine, so long as one knows what is going on.
We expect the difference to be 4.2%, and the confidence interval will contain that value should there be no statistical significance at the 5% level. Any result in the CI is considered equal to any other result.
It's worth looking at the point estimates: For women, it is 0.286 and men 0.288.
This is a difference of -0.002.
z=-0.053 p=0.478
standard error is (.286)(.714)/350=0.0005834
and (.288)(.712)/400 = 0.0005126
sum is 000110; sqrt(sum) is 0.0331=SE
multiply by z-value of 1.96, and we get 0.0649
CI (-0.0651,+0.0647)
The CI contains the difference of 0 for the estimation of the true difference in proportions; therefore, the null hypothesis of no difference cannot be rejected. Given the sample size, a percentage as large as shown in the Almanac could be found in a sample if there were no true difference in the parameters.
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