Chapter 14 Analysis of Count Data
Example • You buy a bag of grass seed from Big-K and it produced the following lawn.
Example • You are concerned because it doesn’t seem to be the same Varity of grasses that were advertised on the bag. You take a representative sample of the lawn to Jeff Houge (CR’s botany instructor) and he gives you the following break down.
Example • Annual Ryegrass 1300 pieces • Creeping Red Fescue 525 pieces • Perennial Ryegrass 275 pieces • Other Grasses 25 pieces • Weed & Other 375 pieces
How To Tell if Significant Difference? • We use the Chi-Squared Goodness-of-fit test. • Data consists of observed counts. • We calculate what we expect to see based on the nulls distribution. Calculate np for each category
The Big Idea • 1. The data consist of observedcounts—that is, how many of the items or subjects fall into each category. • 2. We will compute expected counts under Ho, that is, the counts that we would expect to see for each category if the corresponding null hypothesis were true. • 3. We will compare the observed and expected countsto each other via a test statistic that will be a measure of how close the observed counts are to the expected counts under Ho. So if this “distance” is large, we have some support for rejecting Ho.
Properties of the Chi-Square Distribution • The distribution is not symmetric and is skewed to the right. • The values are non-negative. • There is a different chi-square distribution for different degrees of freedom. • The mean of the chi-square distribution is equal to its degrees of freedom and is located to the right of the mode. • The variance of the chi-square distribution is 2(df).
Do the Test Page 933 • We could calculate the p-value by hand using the TI. To do that we’ll need to know our degrees of freedom which is K – 1 where K is the number of categories. • I wrote a program MULTNOM that is in your calculator to do this test.
Let’s Do It • LDI: 14.2, 14.3 • Page 936: Exercises 14.6, 14.10, 14.12, 14.13