Ch 15 - Chi-Squared _Categorical Data_

45 since we reject h0 for large values of x 2 and we

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: erved X 2 test statistic value = 1.44 df = 4 – 1 = 3 . Sketch distribution to find bounds… p‐value is > 0.50. Are the results statistically significant at the 5% significance level? NO Conclusion at a 5% level: It appears that .... the 4 booths are used equally often Aside: Using our frame of reference for chi‐square distributions. Recall that if we have a chi‐square distribution with df degrees of freedom, then the mean is equal to df , and the standard deviation is equal to 2(df ) So, if H0 were true, we would expect the X 2 test statistic to be about 3 give or take about sqrt(2*3) = 2.45 . Since we reject H0 for large values of X 2 , and we only observed a value of 1.44 , even less than expected under H0, we certainly do not have enough evidence to reject H0. Goodness of Fit Test Summary Assume: We have 1 random sample of size n . We measure one discrete response X that has k possible outcomes Test: H0: A specified discrete model for X p1 p10 , p 2 p 20 , , pk pk 0 Ha: The probabilities are not as specified in the null hypothesis. Test Statistic: X 2 observed - expected 2 expected where expected E i np i 0 If H0 is true, then X 2 has a 2 distribution with ( k 1) degrees of freedom, where k is the number of categories. The necessary conditions are: at least 80% of the expected counts are greater than 5 and none are less than 1. Be aware of the sample size (pg 598). 209 Try It! Crossbreeding Peas For a genetics experiment in the cross breeding of peas, Mendel obtained the following data in a sample from the second generation of seeds resulting from crossing yellow round peas and green wrinkled peas. n = 556 Yellow Round Yellow Wrinkled Green Round Green Wrinkled 315 101 108 32 312.75 104.25 104.25 34.75 556(9/16) = 312.75, etc. Do these data support the theory that these four types should occur with probabilities 9/16, 3/16, 3/16, and 1/16 respectively? Use = 0.01. H 0 : p1 9/16 , p 2 3/16 , p3 3/16 , p 4 1/16 . X2 315 312.752 101 104.252 108 104.252 32 34.75...
View Full Document

This document was uploaded on 02/25/2014 for the course STATS 250 at University of Michigan.

Ask a homework question - tutors are online