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## Chapter 14

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**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