## Question 537:

1

1. Q537_1_chiSquare2PQ537.xls

We will use the Chi-Square test of independence here to see if the difference in cock-pit noise is different at different levels of the flight. We'd need these differences to be greater than what we'd expect to see from chance alone. The null hypothesis of the Chi-Square test is that the categories are not independent from each other--meaning there is no difference in accuracy between diet and regular.

The Chi-Square test uses the formula

Sum of       (O-E)2
---------
E

• O means the observed frequencies
• E means the expected frequencies

We will need to do this for each cell. I'll walk through the first cell example. We have 7 who got regular correct. This is the observed count. The expected count is found by multiplying the row total and column totals together and then multiplying this by the grand total. We get (14*19)/46 = 5.78 as our expected value. Now we just plug and chug in the formula (7-5.78)2/5.78 = 0.2563

We now repeat this step for all values and then add them all up. This gets us .627, which is our test statistic (the Chi-Square Statistic). To see if this is significant, we can use a table of Chi-Square values in a statistics textbook or use the excel formula =CHIDIST(.627,1) where the second parameter is the degrees of freedom, found as (row count -1)*(column count-1) = 1*1=1. We get the p-value .428 which is GREATER than our alpha of .05 so we can NOT reject the null hypothesis. There is not sufficient evidence for us to say there is independence between groups. Specifically, the different correct guesses are not greater than what we'd expect from chance alone.

The second part of the question asks us to complete a 2-sample Z-test. By this, I believe this means a 2-proportion test using the Normal Approximation to the Binomial. We can  use this test since it is a 2x2 table and there is one proportion to compare to another proportion.

The two proportions we are calculating are 7/19 = .368 and 7/27 =.259 and the difference is .109. Can we conclude this difference is greater than chance using this test?

We 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 correct guesses and the n's are the sample size.

P = (7+7)/(19+27) = .3043
Q = 1-.3043 = .6956
PQ = .3043*.6956 = .2117

1/n1 + 1/n2 = 1/19 + 1/27 = .08966

So multiply .2117 * .08966 = 0.01895

Now the square root of this is SQRT(0.01895) = .1377

So the equation is the observed difference .259/.1377 = .7922

That last result of .7922 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 p-value of .428, which is the same as we got with the Chi-Square test.  So again, not enough evidence with this sample size that correct guesses between diet and regular are better than chance alone.

We are using the normal approximation to the binomial, which greatly simplifies the calculations. It's usually Ok to use the normal approximation to the binomial when n*p and n*q > 5, where p is the proportion of getting the correct answer which we defined above. We get 7 and 13.2, so we meet that test. For 2x2 tables I prefer using the z-test on the difference in proportions as the computation is a bit more straight forward. Both tests have limitations for small samples. We just saw the np > 5 test, for Chi-Square, we really would like to see no cell counts less than 5.