Executing z-tests by using chi-square tests in SPSS

Imagine that a researcher is interested in testing the claim that vitamin C helps prevent one

from catching a cold. 500 subjects are recruited just before cold season and given a bottle of 120

pills that are either a placebo (n = 250) or a substantial dose of vitamin C (n = 250). They are

instructed to take one pill per day. They are further instructed that they should eat, exercise, and

so on as usual. At the end of the four-month period, each subject is asked yes or no if they

caught at least one cold in the interim. Of the placebo group, 75 (75 / 250 = .30) report having

caught a cold, and in the vitamin C group, 65 (65 / 250 = .26) report having caught a cold. Is

there enough evidence to conclude that vitamin C reduces the incidence of the common cold?

One way to address this is with a z-test for to compare proportions for independent groups. The

formula1 for this is below (where p1 is the proportion of subjects in group 1 who have a cold, and

p2 is the same for group 2, and n1 and n2 are sample sizes):

𝑧 = (𝑝1 − 𝑝2)

√𝑝1(1 − 𝑝1)

𝑛1+

𝑝2(1 − 𝑝2)𝑛2

=. 3 − .26

√. 3(.7)250

+. 26(.74)

250

= 0.997

This formula is a bit of work to use, and has a pretty high probability of leading to (human) error.

SPSS allows you to do test the same hypothesis using a different statistic, chi-square, which is

closely related to z (z2 = χ2) when the DV is measured in a binary fashion. To carry out this chi-

square test of independence2, enter the data as illustrated at left. I recommend creating value

labels for group and cold so that the data look like at right.

From here, go to Data > Weight Cases, select Weight cases by and move the count variable (f) to

the Frequency Variable box.

Then click OK. Now go to

Analyze > Descriptive

Statistics > Crosstabs. Put

one variable (e.g., group)

into Row(s) and the other

into Column(s). Click the

Statistics button and make

sure Chi-square is checked.

You'll get some output like

that at right. Of particular

interest is the row labeled Pearson Chi-Square. The value of chi-square is 0.992 (which is equal to

z2), and it's significance value is .319, which is not less than the conventional .05 level, and

therefore there is no significant difference between the two groups.

1 If you look around, you will find that there is an alternative version of this formula that uses a pooled proportion (p = p1*n1 + p2*n2) / (n1 + n2) in place of p1 and p2. 2 This is called a test of independence because we are interested in seeing whether the status of having a cold (or not) is dependent on having taken vitamin C or not.

Chi-Square Tests

.992b 1 .319

.804 1 .370

.993 1 .319

.370 .185

.990 1 .320

500

Pearson Chi-Square

Continuity Correctiona

Likelihood Ratio

Fisher's Exact Test

Linear-by-Linear

Association

N of Valid Cases

Value df

Asy mp. Sig.

(2-sided)

Exact Sig.

(2-sided)

Exact Sig.

(1-sided)

Computed only f or a 2x2 tablea.

0 cells (.0%) hav e expected count less than 5. The minimum expected count is 70.

00.

b.

Now give this a try on your own to answer the following question: Howell and Huessy (1981)

used a rating scale to classify children in a second-grade class as showing or not showing

behavior commonly associated with attention deficit disorder (ADD). The researchers then

classified the same children again when they were in fourth and fifth grades. When the children

reached the end of the ninth grade, the researchers examined school records and noted which

children were enrolled in remedial English. In the following data, all children who were ever

classified as exhibiting behavior associated with ADD have been combined into one group

(labeled ADD):

remedial English non-remedial English proportion in

remedial English

no ADD 22 187 .105

ADD 19 74 .204

Use the chi-square test of independence to examine whether behavior during elementary school

is related to later placement in remedial English classes. Please export the Chi-Square Tests table

into a Word file and add a brief conclusion that looks something like this: "There was (not) a

significant difference in assignment to remedial English based on whether earlier ADD behavior

was present or not, χ2(1) = ???, p = ???." If there is a significant difference, be sure to add some

words to the conclusion that say something about which group had higher or lower assignment

to remedial English classes. You would never say there is merely a significant effect without

telling your reader what the effect is, would you? After you finish with you conclusion, please

email the file to me.