In Module Notes 5.2 we presented material for estimating and testing a population proportion from a single sample. This set of notes extends the methodology to the case where we want to estimate and test for the difference between two proportions, then test for the difference between multiple proportions. We conclude with a test for the relationship between two categorical variables.


Tests for the Difference in Two Proportions

Z-Test for the Difference in Two Proportions
Suppose we wanted to determine if a proportion in one population is equal or not to a proportion in another population. For example, suppose we surveyed 1,000 shoppers at both Sears and JCP and asked them to rate the shopping experience. Further suppose that in our two samples, 315 Sears shoppers rated the experience as Excellent (31.5%) and 323 shoppers at JCP rated the experience as Excellent (32.3%).

To determine if the two population proportions are equal or not, based on results from the sample, we set up the following hypotheses:

H₀: p_Sears = p_JCP (null hypothesis); note p_Sears - p_JCP = 0
H_a: p_Sears =/= p_JCP (alternative hypothesis)

The test statistic is the Z-Test statistic, and its computational formula is as follows:

Eq. 5.3.1: Z = (p_{Sears Sample} - p_{JCP Sample}) - (p_Sears - p_JCP) /

Sq Rt [ p_pooled * (1 - p_pooled) * (1/n_Sears + 1/n_JCP) ]
Where p_pooled = (x_Sears + x_JCP) / (n_Sears + n_JCP)

For this problem, first find p_pooled:

Eq. 5.3.2: p_pooled = (315 + 323) / (1,000 + 1,000) = 0.319

Next, compute the Z-Test statistic:

Eq. 5.3.3: Z = (0.315 - 0.323) - (0) /

Sq Rt [ 0.319 * (1 - 0.319) * (1 /1,000 + 1/1,000)] = -0.384

The final step is to find the p-value for a Z-score of -0.384. To do this, we put =NORMSDIST(-0.384) in an empty cell in an Excel Worksheet, and Excel returns 0.35. Since this is a two tail test, we multiple 0.35 by 2 and get a p-value of 0.70. Since the p-value is greater than an alpha threshold of 0.05, we do not reject the null hypothesis and conclude that the two proportions are equal. Any apparent difference in the two proportions in our samples is just do to random chance.

Worksheet 5.3.1 provides an Excel template for a general hypothesis test for p₁ = p₂.

Worksheet 5.3.1

Row 1	Column AF	AG
2	Hypothesis Test for (p1 = p2)
3	n1 (Sears and Excel)	1000
4	Successes	315
5	n2 (JCP & Excel)	1000
6	Successes	323
7	Confidence Level	0.95
8	Alpha	0.05
9	Ps1	0.315	= AG4 / AG3
10	Ps2	0.323	= AG6 / AG5
11	Pbar	0.319	= (AG4 + AG6)/(AG3 + AG5)
12	Null Hypothesis	p1 = p2
13	Z Test Statistic	-0.383800991	=(AG9-AG10)/SQRT(AG11*(1-AG11)*(1/AG3+1/AG5))
14	p-value (one-tail)	0.3506	= 2*(1-NORMSDIST(ABS(AG13)))
15	p-value (two-tail)	0.7011	= 2*AG14

Chi-Square (X²) Test for Difference Between Two Proportions
There is another test for the difference between two proportions - this one is a non parametric test based on the Chi-Square Distribution. This test is used for data set up in cross-classification tables.

To use the Chi-Square test for two proportions, we begin with the cross-classification table. I will use the data from the above example, and put it into a cross-classification table shown in rows 2 through 5 of Worksheet 5.3.2.

Worksheet 5.3.2

Row 1	Column AN	AO	AP	AQ	AR	AS
2		Excel	Not Excel	Total
3	Sears	315	685	1000
4	JCP	323	677	1000
5	Total	638	1362	2000
6
7	Cell	Oij	Eij	(Oij-Eij)	(Oij-Eij)^2	((Oij-Eij)^2)/Eij
8	Sears/Excel	315	319	-4	16	0.0502
9	Sears/Not Excel	685	681	4	16	0.0235
10	JCP/Excel	323	319	4	16	0.0502
11	JCP/Not Excel	677	681	-4	16	0.0235
12						0.1473	Total is X^2 Stat.
13						0.7011	=p-value for X^2
14							=CHIDIST (AS12,1)
15

The cross-classification table shows the joint events of Sears and Excellent ratings, and JCP and Excellent ratings, as well as the complements, Sears and not Excellent, and JCP and not Excellent. Recall that in the cross classification tables, we have to account for the entire sample space for the computation of the probabilities.

The hypotheses we are testing are identical to the hypotheses examined at the beginning of this section:

H₀: p_Sears = p_JCP (null hypothesis);
H_a: p_Sears =/= p_JCP (alternative hypothesis)

The Chi-Square statistic (note: text references use the symbol X² where X is the Greek symbol for Chi) assumes that the samples are randomly selected and independent, and that there is sufficient sample size. Sufficient sample size is defined as cell counts in the cross-classification table be at least 5. These requirements are met for this example.

The formula for computing the Chi-Square Statistic begins with comparing the observed cell count to the expected. Let's take cell AO3 as an example. The observed cell count is 315. Next, we find the expected cell count, where the expected count is what is expected if there were no difference between the proportions. The expected count is obtained by taking the row total times the column total (the row and column marginal totals of the particular cell being studied) divided by the grand total of observations. The expected cell count for cell AO3 (Sears and Excellent) is:

Eq. 5.3.4: Expected_{Sears and Excellent} =

Row Total_Sears x Column Total_Excel ) / Total =
( 1,000 x 638 ) / 2,000 = 319.

This computation is shown in cell AP8 in Worksheet 5.3.2. Next, we find the difference between observed and expected (shown in cell AQ8), then we square the difference to remove the plus and minus signs, and convert to a relative frequency by dividing by the expected count for the cell of interest. Once this is done for each cell in the cross-classification table, as shown in Worksheet 5.3.2, sum the relative frequencies to get the Chi-Square statistic of 0.1473 for this example (cell AS12).

We next find the p-value for the Chi-Square statistic by using the Excel function =CHIDIST(0.1473,1) in an active worksheet cell. For this function, the first entry is the value of the Chi-Square statistic, and the second is the degrees of freedom for the Chi-Square distribution. Degrees of freedom are (number of rows - 1) times (number of columns - 1). For this example, we have two rows and two columns (not counting the total rows or columns). Thus, the degrees of freedom are (2 - 1) times (2 - 1) or 1.

The p-value returned by the Chi-Square function is 0.7011, which is identical to the two-tail p-value obtained from the Z-Score approach. Since the p-value is greater than 0.05, we fail to reject the null hypothesis, and conclude the proportions of interest are equal. Note that if the proportions Store and Excellent are equal, then the proportions for Store and not Excellent would also be equal.

If the Chi-Square and Z-Score test approaches are identical, why do we need to learn the Chi-Square? The main reason is that the Chi-Square approach may be used for cross-classification tables larger than 2 rows by 2 columns - multiple sample problems. The Z-Score approach is limited to comparing two proportions. We do the multiple sample problem next.

For the Chi-Square test to give accurate results for the 2 by 2 table, it is assumed that each expected frequency in the cross-classification table cells is at least five. References are provided (Levine, 1999) for small sample size problems beyond the scope of this course.


Test for the Difference in Multiple Proportions

Now suppose we were interested in comparing the proportion of shoppers who rated Kmart as Excellent, with those who rated Sears as Excellent, with those who rated JCP as Excellent, to those who rated Wards as Excellent with respect to their shopping experience. The hypothesis statements are:

H₀: p_{Kmart & Excel} = p_{Sears & Excel} = p_{JCP & Excel} = p_{Wards & Excel}
H_a: Not all proportions are equal

We gather our sample and prepare the cross-classification table as shown in rows 3 through 7 of Worksheet 5.3.3.

Worksheet 5.3.3

Row 1	AY	AZ	BA	BB	BC	BD
2		Excel	Good	Poor	Total
3	Kmart	272	477	251	1000
4	Sears	315	457	228	1000
5	JCP	323	470	207	1000
6	Wards	391	404	205	1000
7	Total	1301	1808	891	4000
8
9	Cell	Oij	Eij	(Oij-Eij)	(Oij-Eij)^2	((Oij-Eij)^2)/Eij
10	Kmart/E	272	325.25	-53.25	2835.5625	8.7181
11	Kmart/G	477	452	25	625	1.3827
12	Kmart/P	251	222.75	28.25	798.0625	3.5828
13	Sears/E	315	325.25	-10.25	105.0625	0.3230
14	Sears/G	457	452	5	25	0.0553
15	Sears/P	228	222.75	5.25	27.5625	0.1237
16	JCP/E	323	325.25	-2.25	5.0625	0.0156
17	JCP/G	470	452	18	324	0.7168
18	JCP/P	207	222.75	-15.75	248.0625	1.1136
19	Wards/E	391	325.25	65.75	4323.0625	13.2915
20	Wards/G	404	452	-48	2304	5.0973
21	Wards/P	205	222.75	-17.75	315.0625	1.4144
						35.8350	Total is X^2 Stat.
						6	= df = (row-1)(col-1)
						0.000	= p-value for X^2
							= CHIDIST(BD22,6)

Next, the Chi-Square statistic is again computed by working with computations involving the differences between the observed and expected frequencies in each cell. The resulting Chi-Square value of 35.8350 and 6 degrees of freedom has a p-value of 0.000. Since the p-value is less than a threshold alpha value of 0.05, we reject the null hypothesis and conclude that at least two of the proportions are not equal.

If you wanted to continue testing to determine which two proportions are not equal, you could do that with the Chi-Square test for Difference between two proportions. Caution: if you plan to conduct multiple tests with the data, you should adjust alpha down as done with multiple regression and ANOVA.

This information might be important if, for example, we were the marketing department at Wards and wanted to back up an advertising claim that shoppers rate their experience at Wards as more excellent than at the other three stores (note that Wards has the highest proportion of shoppers rating their experience as excellent).

The assumption for application of the Chi-Square test to cross-classification tables larger than 2 by 2 is that the expected cell frequencies are at least equal to five, although some sources suggest the expected cell frequencies can be as low as 1 as long as the total sample size is large (number of cells times 5) (Sheskin, 2000). If the expected cell frequencies are less than five (or one with very large total samples), then one or more rows or one or more columns may have to be collapsed.

Test for Independence

The final test we can perform with categorical data in cross-classification tables is the Chi-Square test for independence. The hypotheses for this test are:

H₀: The two categorical variables are independent (there is no

relationship between the two variables)

H_a: The two categorical variables are dependent (there is a

relationship between the two variables)

Suppose we are interested in determining if the amount of money one spends on a shopping experience is related to the store where one shops. The following cross-classification table illustrates data collected (rows 3 through 7) for a scenario such as this:

Worksheet 5.3.4

1	BH	BI	BJ	BK	BL	BM	BN
2		>$100	$50-100	<$50	Total
3	Kmart	50	325	478	853
4	Sears	457	315	50	822
5	JCP	250	250	250	750
6	Wards	275	200	100	575
7	Total	1032	1090	878	3000
8
9	Cell	Oij	Eij	(Oij-Eij)	(Oij-Eij)^2	((Oij-Eij)^2)/Eij
10	Kmart/E	50	220.074	-170.074	28925.165	131.4338
11	Kmart/G	325	232.4425	92.5575	8566.8908	36.8560
12	Kmart/P	478	187.2335	290.7665	84545.158	451.5493
13	Sears/E	457	212.076	244.924	59987.766	282.8598
14	Sears/G	315	223.995	91.005	8281.9100	36.9736
15	Sears/P	50	180.429	-130.429	17011.724	94.2849
16	JCP/E	250	193.5	56.5	3192.25	16.4974
17	JCP/G	250	204.375	45.625	2081.6406	10.1854
18	JCP/P	250	164.625	85.375	7288.8906	44.2757
19	Wards/E	275	148.35	126.65	16040.223	108.1242
20	Wards/G	200	156.6875	43.3125	1875.9727	11.9727
21	Wards/P	100	126.2125	-26.2125	687.09516	5.4440
22						1230.46	Total is X^2 Stat.
23						6	= df = (row-1)(col-1)
24	Since p-value is < 0.01, reject Ho; the two variables					0.000	= p-value for X^2
25	are related. The strength of relationship is:						= CHIDIST( BM22,6)
26
27	Cramer's Phi = SQRT(BM22/(3000*(3-1))) =				0.453

The test statistic is the Chi-Square, and its computation is identical to the Chi-Square test for differences in multiple samples. Here, the Chi-Square computed of 1230.46 with 6 degrees of freedom has a p-value of 0.000. Since the p-value is less than a threshold value of alpha of 0.05, we reject the null hypothesis and conclude that dollars spent during the shopping event is related to the store at which one shops (Sears and Wards attracting shoppers who spend more than $100, whereas Kmart attracting shoppers who spend less than $50). This information might be of use to someone planning a target market ad campaign.

The last item concerning the relationship between two categorical variables is strength. There is a coefficient called Cramer's Phi Coefficient which is similar to the correlation coefficient. It is shown in Row 27 of Worksheet 5.3.4. The formula is:

Eq. 5.3.5: Cramer's Phi = Square Root {Chi-Square / [ ( n * (k-1) ] }

where k - 1 is the smaller of (rows - 1) or (columns - 1).

For this problem, the Chi-Square value is 1230.46, the sample size is 3,000, and since the number of columns is less than the number of rows, we use columns - 1. The resulting computations are:

Eq. 5.3.6: Cramer's' Phi = Square Root {1230.46 / [ 3000 * (3-1) ] }

= 0.453

While there is a significant statistical relationship, we would have to say it is relatively weak at 0.453 (interpreted just as a correlation coefficient in regression analysis).


That's it!!! We are finished with the notes for QMB 6305. You may want to review the notes in Module 5 by working through the "Airline Satisfaction Survey" assignment in the Main Module 5 Overview. After this set of notes, you can answer questions 3 and 4.


References:


Anderson, D., Sweeney, D., & Williams, T. (2001). Contemporary Business Statistics with Microsoft Excel. Cincinnati, OH: South-Western, Chapter 11.

Levine, D., Berenson, M. & Stephan, D. (1999). Statistics for Managers Using Microsoft Excel (2nd. ed.). Upper Saddle River, NJ: Prentice-Hall, Chapter 11.

Mason, R., Lind, D. & Marchal, W. (1999). Statistical Techniques in Business and Economics (10th. ed.). Boston: Irwin McGraw Hill, Chapter 14.

Sheskin, D. (2000). Handbook of Parametric and Nonparametric Statistical Procedures (2nd ed.). Boca Raton, FL: Chapman & Hall/CRC, Test 16 -- The Chi-Square Test for r x c Tables