Hi,

What is your null and alternative hypothesis.

I suppose your null hypothesis is like

H0 : the pi ’s arise from a fully specified model (or specified values) against H1: the

pi ’s are unrestricted (but we must still have pi ≥ 0,

This is a goodness of fit test of a fully-specified null distribution

For calculating the chi-square statistic and test p-value you may use SPSS.

For your data, see the sub section 15.3.2. of the following www link (insert 2 categories instead of 3 in the worked example ).

You may find the first part of the following webpage useful:

https://openpress.usask.ca/introtoappliedstatsforpsych/chapter/15-3-spss-lesson-13-proportions-goodness-of-fit-and-contingency-tables/

Please note you may use binomial test for you data (two categories) as well.

Good luck,

Hamid

Hello Lina,

As Hamid Ghorbani explains, the answer depends on what specific null hypothesis you're evaluating. With a two-category variable as you have, if you opt for a chi-square test, you'll be using the chi-square goodness of fit test (not the chi-square test of independence). However, for a goodness of fit test to be appropriately evaluated, it's critical to know how to compute the expected cell frequencies.

If your null hypothesis was: no difference in rate of infected and non-infected (p1 = p2; which doesn't sound very plausible to me),

Then expected cell values are N/2 for infected and non-infected, respectively (where N is your total sample size).

However, if your null hypothesis is that the rate of infected cases is no different from some benchmark rate (e.g, p1 = .20, p2 = .80 if these represent prior empirical rates observed in other samples), then the expected cell frequencies are 0.20 * N for infected and 0.80 *N for non-infected, respectively.

So, there is no single answer as to what the resultant computation should be, without knowing your specific hypothesis.

Good luck with your work.

This is the goodness of fit chi-square:

### --------------------------------------------------------------

### Chi-square goodness-of-fit, Both 50/50?

### --------------------------------------------------------------

## for infected

observed = c(61, 581) # observed frequencies

expected = c(0.50, 0.50) # expected proportions

chi_infected

Chi-squared test for given probabilities data: observed X-squared = 421.18, df = 1, p-value < 2.2e-16

R script in attachment

Chi-Square Goodness-of-Fit Test in SPSS Statistics

https://statistics.laerd.com/spss-tutorials/chi-square-goodness-of-fit-test-in-spss-statistics.php