# Chapter 16 Chi Squared Tests.

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Chapter 16 Chi Squared Tests

16.1 Introduction Two statistical techniques are presented, to analyze nominal data. A goodness-of-fit test for the multinomial experiment. A contingency table test of independence. Both tests use the c2 as the sampling distribution of the test statistic.

16.2 Chi-Squared Goodness-of-Fit Test
The hypothesis tested involves the probabilities p1, p2, …, pk.of a multinomial distribution. The multinomial experiment is an extension of the binomial experiment. There are n independent trials. The outcome of each trial can be classified into one of k categories, called cells. The probability pi that the outcome fall into cell i remains constant for each trial. Moreover, p1 + p2 + … +pk = 1. Trials of the experiment are independent

16.2 Chi-squared Goodness-of-Fit Test
We test whether there is sufficient evidence to reject a pre-specified set of values for pi. The hypothesis: The test builds on comparing actual frequency and the expected frequency of occurrences in all the cells.

The multinomial goodness of fit test - Example
Two competing companies A and B have enjoy dominant position in the market. The companies conducted aggressive advertising campaigns. Market shares before the campaigns were: Company A = 45% Company B = 40% Other competitors = 15%.

The multinomial goodness of fit test - Example
Example 16.1 – continued To study the effect of the campaign on the market shares, a survey was conducted. 200 customers were asked to indicate their preference regarding the product advertised. Survey results: 102 customers preferred the company A’s product, 82 customers preferred the company B’s product, 16 customers preferred the competitors product.

The multinomial goodness of fit test - Example
Example 16.1 – continued Can we conclude at 5% significance level that the market shares were affected by the advertising campaigns?

The multinomial goodness of fit test - Example
Solution The population investigated is the brand preferences. The data are nominal (A, B, or other) This is a multinomial experiment (three categories). The question of interest: Are p1, p2, and p3 different after the campaign from their values before the campaign?

The multinomial goodness of fit test - Example
The hypotheses are: H0: p1 = .45, p2 = .40, p3 = .15 H1: At least one pi changed. The expected frequency for each category (cell) if the null hypothesis is true is shown below: What actual frequencies did the sample return? 90 = 200(.45) 80 = 200(.40) 102 82 30 = 200(.15) 16

The multinomial goodness of fit test - Example
The statistic is The rejection region is

The multinomial goodness of fit test - Example
Example 16.1 – continued

The multinomial goodness of fit test - Example
Example 16.1 – continued c2 with 2 degrees of freedom Conclusion: Since 8.18 > 5.99, there is sufficient evidence at 5% significance level to reject the null hypothesis. At least one of the probabilities pi is different. Thus, at least two market shares have changed. Alpha P value 5.99 8.18 Rejection region

Required conditions – the rule of five
The test statistic used to perform the test is only approximately Chi-squared distributed. For the approximation to apply, the expected cell frequency has to be at least 5 for all the cells (npi ³ 5). If the expected frequency in a cell is less than 5, combine it with other cells.

16.3 Chi-squared Test of a Contingency Table
This test is used to test whether… two nominal variables are related? there are differences between two or more populations of a nominal variable To accomplish the test objectives, we need to classify the data according to two different criteria.

Contingency table c2 test – Example
In an effort to better predict the demand for courses offered by a certain MBA program, it was hypothesized that students’ academic background affect their choice of MBA major, thus, their courses selection. A random sample of last year’s MBA students was selected. The following contingency table summarizes relevant data.

Contingency table c2 test – Example
The observed values There are two ways to address the problem If each undergraduate degree is considered a population, do these populations differ? If each classification is considered a nominal variable, are these two variables dependent?

Contingency table c2 test – Example
Solution The hypotheses are: H0: The two variables are independent H1: The two variables are dependent Since ei = npi but pi is unknown, we need to estimate the unknown probability from the data, assuming H0 is true. k is the number of cells in the contingency table. The test statistic The rejection region

Estimating the expected frequencies
Undergraduate MBA Major Degree Accounting Finance Marketing Probability 60 BA 60 60/152 BENG 31 31/152 BBA 39 39 39/152 Other 22 22/152 61 44 152 152 61 44 47 152 Probability 61/152 44/152 47/152 Under the null hypothesis the two variables are independent: P(Accounting and BA) = P(Accounting)*P(BA) = [61/152][60/152]. The number of students expected to fall in the cell “Accounting - BA” is eAcct-BA = n(pAcct-BA) = 152(61/152)(60/152) = [61*60]/152 = 24.08 The number of students expected to fall in the cell “Finance - BBA” is eFinance-BBA = npFinance-BBA = 152(44/152)(39/152) = [44*39]/152 = 11.29

The expected frequencies for a contingency table
The expected frequency of cell of raw i and column j in the contingency table is calculated by eij = (Column j total)(Row i total) Sample size

Calculation of the c2 statistic
Solution – continued Undergraduate MBA Major Degree Accounting Finance Marketing BA 31 (24.08) 13 (17.37) 16 (18.55) 60 BENG 8 (12.44) (8.97) 7 (9.58) 31 BBA 12 (15.65) 10 (11.29) 17 (12.06) 39 Other (8.83) 5 (6.39) 7 (6.80) 22 61 44 47 152 The expected frequency c2= ( )2 24.08 ( )2 6.39 ( )2 6.80 = 14.70 +….+ +….+

Contingency table c2 test – Example
Solution – continued The critical value in our example is: Conclusion: Since c2 = > , there is sufficient evidence to infer at 5% significance level that students’ undergraduate degree and MBA students courses selection are dependent.

Using the computer Define a code to specify each nominal value. Input the data in columns one column for each category. Select the Chi squared / raw data Option from Data Analysis Plus under tools. See Xm16-02 Code: Undergraduate degree 1 = BA 2 = BENG 3 = BBA 4 = OTHERS MBA Major 1 = ACCOUNTING 2 = FINANCE 3 = MARKETING

Required condition Rule of five
The c2 distribution provides an adequate approximation to the sampling distribution under the condition that eij >= 5 for all the cells. When eij < 5 rows or columns must be added such that the condition is met. Example 18 (17.9) 23 (22.3) 12 (12.8) 4 (5.1) 7 (6.3) 4 (3.6) We combine column 2 and 3

16.5 Chi-Squared test for Normality
The goodness of fit Chi-squared test can be used to determined if data were drawn from any distribution. The general procedure: Hypothesize on the parameter values of the distribution we test (i.e. m = m0, s = s0 for the normal distribution). For the variable tested X specify disjoint ranges that cover all its possible values. Build a Chi squared statistic that (aggregately) compares the expected frequency under H0 and the actual frequency of observations that fall in each range. Run a goodness of fit test based on the multinomial experiment.

15.5 Chi-Squared test for Normality
Testing for normality in Example 12.1 For a sample size of n=50 (see Xm12-01) ,the sample mean was with standard error of Can we infer from the data provided that this sample was drawn from a normal distribution with m = and s = 38.83? Use 5% significance level.

c2 test for normality Solution
First let us select z values that define each cell (expected frequency > 5 for each cell.) z1 = -1; P(z < -1) = p1 = .1587; e1 = np1 = 50(.1587) = 7.94 z2 = 0; P(-1 < z< 0) = p2 = .3413; e2 = np2 = 50(.3413) = 17.07 z3 = 1; P(0 < z < 1) = p3 = .3413; e3 = 17.07 P(z > 1) = p4 = .1587; e4 = 7.94 The cell boundaries are calculated from the corresponding z values under H0. The expected frequencies can now be determined for each cell. e2 = 17.07 e3 = 17.07 z1 =(x )/38.83 = -1; x1 = .1587 .3413 421.55 e1 = 7.94 e4 = 7.94 460.38 499.21

c2 test for normality The test statistic c2= = 1.72 + + + (10 - 7.94)2
( )2 7.94 ( )2 17.07 ( )2 17.07 ( )2 7.94 = 1.72 + + + f3 = 19 e2 = 17.07 e3 = 17.07 f2 = 13 f1 = 10 f4 = 8 e1 = 7.94 e4 = 7.94

c2 test for normality The test statistic c2= The rejection region
( )2 7.94 ( )2 17.07 ( )2 17.07 ( )2 7.94 = 1.72 + + + The rejection region Conclusion: There is insufficient evidence to conclude at 5% significance level that the data are not normally distributed.