I understand that SPSS can create a contingency table, and at the same time perform the chi-square test.
However, is it possible that when we already have the contingency table, to have SPSS do the chi-square test?
Solved – Can SPSS perform chi-square test on an existing contingency table
contingency tablesspss
Related Solutions
In the case of 2x2 contingency tables, the $z$-test of whether two proportions are equal is equivalent to the Pearson $\chi^2$ test of independence: the quantity $z^2$ is the Pearson $\chi^2$ statistic (Agresti, Introduction to Categorical Data Analysis, p. 26).
Keep in mind that the $t$-test is, essentially, a heavier-tailed $z$-test, and is used to counterbalance the (inappropriately narrow) confidence bands in the case of smaller sample sizes; asymptotically, as one increases the $t$ distribution's degrees of freedom to infinity (which happens as sample size increases), the tails of the $t$ distribution diminish and it approaches the normal.
Your results are interpreted in the ordinary way as for a Pearson $\chi^2$ test.
The links that whuber provided, especially Why does independence test use the chi-squared distribution? give some mathematical justification, but maybe here's an intuitive explanation for you.
Suppose you conducted a survey to see if having blue eyes was related to knowledge of Stackexchange, and you surveyed 100 people at random at the mall. Let say you found a table like this: $$ \begin{array}{lll} & \mbox{Blue} & \mbox{Not blue} \\ \mbox{Stackexchange yes} & 20 & 30 \\ \mbox{Stackexchange no} & 30 & 20. \end{array} $$ The expected counts are 25, all around, and your deviations are $$ \begin{array}{lll} & \mbox{Blue} & \mbox{Not blue} \\ \mbox{Stackexchange yes} & -5 & 5 \\ \mbox{Stackexchange no} & 5 & -5. \end{array} $$
Now, suppose we conducted the survey again, this time asking 1000 people, and found the following contigency table: $$ \begin{array}{lll} & \mbox{Blue} & \mbox{Not blue} \\ \mbox{Stackexchange yes} & 245 & 255 \\ \mbox{Stackexchange no} & 255 & 245 \end{array} $$ which of course, also yields proportional expected values of 250, and the same deviations of $\pm 5$.
Now which experiment do you think provides stronger evidence against the null of no association of eye color? With ~250 people per cell, a deviation of 5 counts can more easily come about through the sampling variability, compared to ~25 people per cell. In fact, the adjustment in the denominator of the statistic is precisely what's needed to account for the additional variance (Binomial variance) present for larger cell sizes.
Best Answer
Yes, it is possible in SPSS.
The data set must have all the counts in one column and the identifying information in two other columns.
The variable that contains the counts needs to be identified by choosing "Weight Cases" from the "Data" menu.
Click "Weight cases by frequency variable" and move over the variable containing the counts.
Then use "Crosstabs" to get the results.