Just from what you say, I would suggest a chi-squared test
of independence. I assume you have 271 independently chosen
subjects sorted into row and column categories as shown.
TBL = cbind(c(68,10,11),c(138,30,14)); TBL
[,1] [,2]
[1,] 68 138
[2,] 10 30
[3,] 11 14
A chi-squared test of independence is computed in R statistical
software as shown below. The P-value is $0.2824 > 0.05,$ so the null hypothesis (independence) cannot be rejected at the 5% level.
chi.out = chisq.test(TBL); chi.out
Pearson's Chi-squared test
data: TBL
X-squared = 2.5291, df = 2, p-value = 0.2824
Observed counts are the counts in the six cells of TBL.
chi.out$obs
[,1] [,2]
[1,] 68 138
[2,] 10 30
[3,] 11 14
Expected counts are computed under the assumption of independence. Look in your textbook or online to see the rationale for how this is done. Briefly, the expected count
in the upper-left cell is found as row total times column total
divided by grand total. That is. $\frac{206 \cdot 89}{271}\approx 67.65.$ All six, expected counts are shown below:
chi.out$exp
[,1] [,2]
[1,] 67.653137 138.34686
[2,] 13.136531 26.86347
[3,] 8.210332 16.78967
The Pearson residuals are of the form $R_{ij}=\frac{X_{ij} - E_{ij}}{\sqrt{E_{ij}}}.$ Roughly speaking they measure how
well the observed counts $X_{ij}$ and the expected counts
$E_{ij}$ agree in each of the six cells. The chi-squared
statistic $Q = \sum_{i=1}^3\sum_{j=1}^2 R_{ij}^2 \approx 2.53.$
(Denoted X-squared in the output.)
In particular, $R_{11} = (68 - 67.653)/\sqrt{67.653}
\approx 0.042.$
chi.out$resi
[,1] [,2]
[1,] 0.04217107 -0.02948994
[2,] -0.86538484 0.60515774
[3,] 0.97358112 -0.68081866
If the null hypothesis is true, then $Q \stackrel{aprx}{\sim} \mathsf{Chisq}(\nu = 2),$ where the degrees of freedom is $\nu = (r-1)(c-1) = (3-1)(2-1) = 2,$ (provided that there is enough data for all expected counts to exceed 5). In order to be significant at
the 5% level, the test statistic $Q$ must exceed the critical value $c = 5.99.$
qchisq(.95, 2)
[1] 5.991465
If the null hypothesis were rejected (with $Q \ge 5.99),$
it would be worthwhile to look at the individual residuals. Residuals larger than about 2 in absolute value indicate
cells in which there is especially poor agreement between observed and expected counts.
