## Question 795:

1

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1. The Null Hypothesis is that there is no difference between the male and femal proportions. The alternative hypothesis is that there IS a difference between proportions.  We would conduct a 1-tailed test since we're just curious to know if there is a difference (not whether men have a higher proportion than women). The test statistic will be the Z-score which we can use from the Normal Approximation to the Binomial.
2.  In conducting the 2-proprtion test, we will be subtracting the proportions to find a difference of  .74-715 = .02815. We now want to know if this difference is greater than what we'd expect from chance alone. We divide the difference by the standard error of the difference between proportions (SE). The SE is found as SQRT(PQ/n1+n2) where P is the sum of the proportions and Q is the sum of 1-the proportions and the n's are the sample sizes for each proportion.
1. P = (432+562)/(604+756) = .73088
2. Q = 1-P = 1-.73088 = .269117
3. PQ = .269117*.73088 = .1966693
4. n1+n2 = 1360
5. We get SQRT(.1966693/1360) = .02420. This gets us the fraction as the difference between proportions / SE = .02815/.02420 = 1.16322.
6. 1.16322 is the z-score and we look this value up in a normal table and get the 2-sided p-value 0.244
7. Since .244 is above .05 we fail to reject the null hypothesis and conclude there is no difference in the proportions.

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