What you want to know is if these frequencies of true or false are different from chance. If so, the frequncies would be 50%-50%. In SAS:

data a;

input tf $ count;

cards;

true 140

false 216

;

proc freq data=a;

weight count;

tables tf / tesp=(.5 .5);

run;

This is yields a sign. result (see attachment): chi-square = 16.2247, df=1, p < .0001

I do n't understand the problem.

Thus is a one-sample proportional test, with theoretical-numm value for testing being coef=.50.

Jos Feys

Thank you very much.

In SPSS:

https://www3.nd.edu/~rwilliam/stats1/OneSample-Spss.pdf

also:

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

The actual chi-square value should be

Test Statistics

Chi-Square 33.980

df 1

Asymp. Sig. .000

The minimum expected cell frequency is 198.0.

Hence the differences observed in proportions of true and false is significant

Looks like you have the solution regarding computation - but can you tell us a little more about your research question/s? I am always interested in what other folks are doing, etc. Thank you-

Atikhom Thienthong , you might clarify what you are trying to determine. The answers proposed so far test a null hypothesis that counts of True and of False follow a 50% / 50% distribution. That is, that it is equally probable that students use words from the list or don't use those words. Is this a hypothesis you want to test?

* * *

A chi-square goodness-of-fit test ( Jos Feys ) or a binomial test ( I think that's what Matthew Kerry is proposing).

Either make sense.

For the chi-square g-o-f, the result is chi-square = 16.225, df=1, p ≈ 5.6e-05 ( Jos Feys ).

I don't know where the chi-square values reported by Ravishankar Suryanarayana or Atikhom Thienthong are coming from.

* * *

I think for what you are doing, probably the most useful result is the 95% confidence interval from the binomial test output below. The proportion of True words is 0.39 with the 95% confidence interval: 0.34 to 0.47. This gives you a meaningful estimate of this proportion.

* * *

R code:

chisq.test(c(140, 216))

### Chi-squared test for given probabilities

###

### data: c(140, 216)

### X-squared = 16.225, df = 1, p-value = 5.626e-05

binom.test(140, (140+216))

### Exact binomial test

###

### number of successes = 140, number of trials = 356, p-value = 6.636e-05

### alternative hypothesis: true probability of success is not equal to 0.5

### 95 percent confidence interval:

### 0.3421870 0.4461209

### sample estimates:

### probability of success

### 0.3932584

SPSS software where chisquare test for good news of fit