Chi-Square Tests

Chi-Square Tests In this chapter, you will learn about two chi-square tests: Math

Chapter 11
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. 2 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:  
1. The sample should be random.
2. All expected counts are larger than 5. Ho: the frequency distribution fits the specified distribution Ha: the frequency distribution DOES NOT fits the specified distribution 3 The Test Statistic
4 Chi-Square Distribution The single number which summarizes the overall difference between observed and expected counts is the chi-square statistic χ2, which tells us in a standardized way how far what we observed (data) is from what would be expected under Ho. 5 6 1 Test Statistic
Example The test statistic for chi-square tests is χ2 = ∑ (Observed Count − Expected Count ) 2 = Expected Count
∑ (O − E ) 2 E Suppose the radio station claims that the distribution of music preferences for listeners in the broadcast region is shown here. 

 Where O is the observed frequency in each category, and E is the corresponding expected frequency. (Ei=npi) d.f.=n-1     Classical: 4% Country: 36% Gospel: 11% Oldies: 2% Pop:18% Rock: 29% Expected Frequency 7 Example

Example 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:       8 Classical:8 Country: 210 Gospel: 72 Oldies: 10 Pop: 75 Rock: 125
The hypotheses:   Observed Frequency Ho: the frequency distribution of music preferences fits the specified distribution Ha: the frequency distribution of music preferences DOES NOT fits the specified distribution 9 10 Conditions Type of music % of listeners claimed by the radio stations Observed Frequency Expected Frequency Classical 4% 8 500(0.04)=20 Country 36% 210 500(0.36)=180 Gospel 11% 72 500(0.11)=55 Oldies 2% 10 500(0.02)=10 Pop 18% 75 500(0.18)=90 Rock 29% 125 500(0.29)=145 Total 100% 500 500 11
1. The sample is random.
2. all expected counts are larger than 5. 12 2 O-E (O-E)2 (O-E)2/E Type of music Observed Frequency (O) Expected Frequency (E) Classical 8 500(0.04)=20 -12 144 7.2 Country 210 500(0.36)=180 30 900 5 Gospel 72 500(0.11)=55 17 289 5.25 Oldies 10 500(0.02)=10 0 0 0 Pop 75 500(0.18)=90 -15 225 2.5 Rock 125 500(0.29)=145 -20 400 2.76 Total 500 500 0 1958 22.7 Test Statistic 2 χ = ∑ (O − E ) 2 = 22.7 E Degrees of freedom (df) = number of categories -1 k–1=6–1=5 χ2 13 P-value from a TI-83/84:



14 Conclusion p-value: DISTR →7:χ2cdf(χ2, 9999999, df)
In our example: χ2cdf(22.7, 9999999, 5)=0.000385
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. 15 Example: M&M Activity Example: M&M Activity
On average, the mix of colors for“M&M’s”® Plain Chocolate Candies will contain       16

13% browns 14% yellows 13% reds 20% oranges 16% greens 24% blues.
17 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. 18 3 2. 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. 19 Step 1: Stating the hypotheses Example
When flipping a penny or spinning a penny, is the probability of getting heads the same? Here are the experimental data:  
Flipping: Spinning: 20 heads:2048 heads:953 tails: 1992 tails: 1047
The null and alternative hypotheses are:
Ho: The proportion of heads is the same for flipping and spinning. Ha: The proportions are different.
Use a 5% significance level to test the claim that the proportion of heads is the same with flipping as with spinning. 21 Coin Experiment: Observed Counts Heads Tails Total Flipping 2048 1992 4040 Spinning 953 1047 2000 Total 3001 3039 6040 22 For All Cells
This will always be the case, and will help streamline our calculations: Expected Count = 23 Column Total × Row Total Table Total 24 4 Checking the conditions: Coin Experiment: Expected Counts Expected Count = Column Total × Row Total 3001 ⋅ 4040 = = 2007.29 6040 Table Total

Heads Tails Total Flipping 2007.29 2032.71 4040 Spinning 993.71 1006.29 2000 Total 3001 3039 6040
1. The data are random. 2. All the expected counts are above 5. We can therefore safely proceed with the chi-square test. 25 Observed and Expected values Observed Heads Tails Total Expected Heads 26 Results Tails Total Flipping 2048 1992 4040 Flipping 2007.29 2032.71 4040 Spinning 953 1047 2000 Spinning 993.71 1006.29 2000 Total 3001 3039 6040 Total 3001 3039 6040 Test statistic:



(O − E ) 2 = E (2048 − 2007.29) 2 (1992 − 2032.71) 2 (953 − 993.71) 2 (1047 − 1006.29) 2 + + + = 4.96 2007.29 2032.71 993.71 1006.29 χ2 = ∑ 27
χ2=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 = (2-1)(2-1) = 1 P-value: χ2cdf(4.96, 9999999,1) = 0.026 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. 28 5