## Question 782:

1## Answer:

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First I'll walk through the calculations and I've provided an excel file with them as well.Part a

We want to perform a test of two proportions here. The sample sizes are large enough that we can use the normal approximation to the binomial. If the samples were small we'd need to use the exact probabilities of the binomial (more tedious calculations). The general check on the sample size is that both np and nq > 5 where n is the sample size and p is the proportion and q is 1-p. That gets us 65, 45, 435 and 455 all of which are well above 5.

So what you want to know is if the two proportions: 65/500= .13 and 45/500 =.09 come from different populations (meaning there is a significant difference). The difference we're testing is .13-.09 = .04 Given your sample size of 500 can we conclude the difference of .04 is greater than chance?

To find out, you divide this difference by the square-root of a denominator which accounts for the chance which will provide you with a z-score. The denominator is

1 1

-- + -- * PQ

n1 n2

Where P = (x1 + x2)/(n1+n2) Q is 1-P. The x's are just the number of responses and the n's are the sample size.

P = (65+45)/(500+500) = .11

Q = 1-.11 = ..89

PQ = .11*.89 = .0979

1/n1 + 1/n2 = 1/500 + 1/500 = .004

So multiply .0979 * .004 = 0.000392

Now the square root of this is SQRT(0.000392) = .019789

So the equation is the observed difference .04/.019789 = 2.02134

That last result of 2.02134 is the z-score which is your test-statistic. You now look this value up using the z-score to percentile calculator using the 2-sided area. This gets us a 2-sided p-value of .0432. Since this is less than our alpha of .05 we reject the NULL and conclude there IS a statistically significant difference.

Part b

Next we want a confidence interval around the observed difference in proportions. To compute this you'll use much of the same information you used to compute the 2-proportion test. The formula is:

p1- p2 +/- za/2 SQRT ( p1q1/n1 + p2q2/n1)

Lets plug in the numbers for the second example.

p1= .13

p2 = .09

za/2 = 1.96

q1 = .87

q2 = .91

n1 = 500

n2 = 500

Plugging in we get .13-.09 +/- 1.96 * SQRT( (.13*.09)/500 + (.09*.91)/500 ) gets us .04 +/- .038706 and a 95% confidence interval between (.00129 and 0.07871).

Notice how the confidence interval does NOT cross 0 (since the low end is above zero and high end is above 0) so we can also conclude with 95% confidence the difference is greater than chance.