Ch 26 – Comparing Counts Day 1 - The Chi-Square Distribution Part VII – Exploring and Understanding Data Ch 26 – Comparing Counts Day 1 - The Chi-Square Distribution

The Chi-Square Test The Chi-square test has two uses: Goodness of fit: to determine if data fits a given distribution Independence: to determine if there is a relationship between two categorical variables

The Chi-square test statistic The basic idea behind chi-square is to see if what really happened is significantly different from what you expected to happen The formula for a chi-square test statistic is:

Decisions using Chi-square We use chi-square only for hypothesis tests – no confidence intervals Once we get a test statistic, we use the chi-square table to make a decision as to whether or not to reject H0 The table works in a similar way to the t-table

Goodness of Fit Test A botanist wants to know if a certain insect has a preference in the color of roses it eats. She records the number of insects who infest each color in her study. Red White Yellow Pink 53 71 46 30

Hypotheses For this type of test there are no symbols in the hypotheses, only words The null hypothesis basically states that there is nothing unusual going on In this case: H0: the insects have no color preference Ha: the insects do have a color preference

Type of Test Chi-square test for goodness of fit α = .05 df = 3 For goodness of fit tests: df = # of categories – 1

Conditions for Chi-square Check Random Sample Read the problem – assume, if no description of sample is given All expected counts are ≥ 5 Verify – if no, make a note of this and go on with the test Make sure you show your expected counts somewhere in your problem

Expected Values The expected values are based on your null hypothesis In this problem, if our null hypothesis was true, all cells would have the same value Red White Yellow Pink Observed 53 71 46 30 Expected

Conditions for Chi-square (This problem) Check Random Sample Assume All expected counts are ≥ 5 yes

The test statistic

Conclusion Since p < .05, reject H0. There is enough evidence to conclude that the insects do have a color preference.

Another example… An anatomy teacher hypothesizes that the final grades in her classes are distributed as 10% A’s, 23% B’s, 45% C’s, 14% D’s and 8% F’s. At the end of the semester, a random sample of her students has the following grades. Was her hypothesis correct? A B C D F 4 13 24 7 5

Expected Values 10% A’s, 23% B’s, 45% C’s, 14% D’s and 8% F’s Total in sample = 53 A B C D F Observed 4 13 24 7 5 Expected 5.3 12.19 23.85 7.42 4.24

Χ2 test for goodness of fit, α=.05, df =4 H0: The distribution does fit the teacher’s hypothesis Ha: The distribution doesn’t fit the teacher’s hypothesis Χ2 test for goodness of fit, α=.05, df =4 Condition Check Random Sample All expected counts are ≥ 5 Stated Since p >.05, fail to reject H0. There is not enough evidence to conclude that the teacher’s grades do not fit her hypothesis. No – one count is < 5 this concerns us, but we will proceed

Homework 26-1 GOF WS