Question 536:
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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 they all occur at the same frequency.
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 6 flights that were classified as having low noise during the climb portion. 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*25)/61 = 5.74 as our expected value. Now we just plug and chug in the formula (6-5.74)2/5.74 = 0.0119
We now repeat this step for all values and then add them all up. This gets us 15.16, 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(15.16,4) where the second parameter is the degrees of freedom, found as (row count -1)*(column count-1) = 2*2=4. We get the p-value .00438 which is less than our alpha of .05 so we would reject the null hypothesis and say that there is independence between groups. Specifically, the amount of noise does differ by the portion of the flight
In looking at the counts, the biggest influence on the Chi-Square comes from the lack of high noise during Climbing (recorded in only 1 flight), its contribution to the Chi-Square Statistic was 5.512.