Math 225. Chi-Square Tests. 1. A Chi-Square Goodness-of-Fit Test. Conditions. Chi-Square Distribution. The Test Statistic

Transcription

1 Chi-Square Tests Math 5 Chapter 11 In this chapter, you will learn about two chi-square tests: Goodness of fit. Are the proportions of the different outcomes in this population equal to the hypothesized proportions? Homogeneity of proportions. Are the proportions of the different outcomes in one population equal to those in another population? The procedure in each case is almost the same. The difference is in the type of question asked and the kinds of conclusions that you can draw. 1. A Chi-Square Goodness-of-Fit Test Conditions Definition: a Chi-Square Goodness-of-Fit Test is used to test whether a observed frequency distribution fits an expected distribution. The hypotheses: Ho: the frequency distribution fits the specified distribution Ha: the frequency distribution DOS NOT fits the specified distribution 1. The sample should be random.. All expected counts are larger than The Test Statistic Chi-Square Distribution The single number which summarizes the overall difference between observed and expected counts is the chi-square statistic χ, which tells us in a standardized way how far what we observed (data) is from what would be expected under Ho

2 Test Statistic The test statistic for chi-square tests is χ = ( Observed Count xpected Count) ( O ) = xpected Count Where O is the observed frequency in each category, and is the corresponding expected frequency. ( i =np i ) d.f.=n-1 Suppose the radio station claims that the distribution of music preferences for listeners in the broadcast region is shown here. Classical: 4% Country: 36% Gospel: 11% Oldies: % Pop:18% Rock: 9% xpected 7 8 A marketing executive randomly selects 500 music listeners from the broadcast region and asks each whether he or she prefers classical, country, gospel, oldies, pop, or rock music. The results are shown here: Classical:8 Country: 10 Gospel: 7 Observed Oldies: 10 Pop: 75 Rock: 15 The hypotheses: Ho: the frequency distribution of music preferences fits the specified distribution Ha: the frequency distribution of music preferences DOS NOT fits the specified distribution 9 10 Type of music % of listeners claimed by the radio stations Observed xpected Classical 4% 8 500(0.04)=0 Country 36% (0.36)=180 Gospel 11% 7 500(0.11)=55 Oldies % (0.0)=10 Conditions 1. The sample is random.. all expected counts are larger than 5. Pop 18% (0.18)=90 Rock 9% (0.9)=145 Total 100%

3 Type of music Observed (O) xpected () O- (O-) (O-) / Classical 8 500(0.04)= Country (0.36)= Gospel 7 500(0.11)= Oldies (0.0)= Pop (0.18)= Rock (0.9)= Total Test Statistic χ = ( O ) =. 7 Degrees of freedom (df) = number of categories -1 k 1 = 6 1 = 5 χ P-value from a TI-83/84: Conclusion p-value: DISTR 7:χ cdf(χ, , df) In our example: χ cdf(.7, , 5)= Since p-value<α=5%, we reject the null hypothesis that the frequency distribution of music preferences fits the specified distribution. We can conclude that the frequency distribution of music preferences in this region is different from what the radio station claims : M&M Activity : M&M Activity On average, the mix of colors for M&M s Plain Chocolate Candies will contain 13% browns 14% yellows 13% reds 0% oranges 16% greens 4% blues. Open your M&M bags and count each color. Do you have enough evidence that the distribution of M&M s is different from the advertised distribution? State the hypotheses, fill the tables, and check the conditions for the Chi-Square Goodness-of-Fit Test. Find the test statistic and the p-value, and state your conclusion

4 . The Chi-Square Test for Homogeneity Step 1: Stating the hypotheses In conducting a test of homogeneity, we can use the same requirements and procedure we used for the test of independence, with one exception: instead of testing the null hypothesis of independence between the row and column variables, we test the null hypothesis that the different populations have the same proportions of some characteristic. The null and alternative hypotheses are: Ho: The proportion of some characteristic in one population is the same as the proportion of the same characteristic in another population. Ha: The proportions are different Step 1: Stating the hypotheses When flipping a penny or spinning a penny, is the probability of getting heads the same? Here are the experimental data: Flipping: heads:048 tails: 199 Spinning: heads:953 tails: 1047 Use a 5% significance level to test the claim that the proportion of heads is the same with flipping as with spinning. The null and alternative hypotheses are: Ho: The proportion of heads is the same for flipping and spinning. Ha: The proportions are different. 1 Coin xperiment: Observed Counts For All Cells Heads Tails Total Flipping Spinning This will always be the case, and will help streamline our calculations: Column Total Row Total xpected Count = Table Total 3 4 4

5 Coin xperiment: xpected Counts Column Total Row Total xpected Count = = = Table Total 6040 Checking the conditions: 1. The data are random.. All the expected counts are above 5. Heads Tails Total Flipping Spinning We can therefore safely proceed with the chi-square test. 5 6 Observed and xpected values Results Observed Heads Tails Total Flipping Spinning xpected Heads Tails Total Flipping Spinning χ =4.96 df= (r-1)(c-1), where r is the number of rows for the data, and c is the number of columns for the data. In this example: df = (-1)(-1) = 1 P-value: χ cdf(4.96, ,1) = 0.06 Test statistic: χ = ( O ) = ( ) ( ) ( ) ( ) = Comparing the p-value with α=5%, we can reject the null hypothesis. There is enough evidence to reject the claim that the proportions are the same. It appears that flipping a penny and spinning a penny result in different proportions of heads