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Addendum: If you want a way to focus just on the tallest trees, you can compare the proportion of them in the logged forest with the the proportion of them in the undisturbed forest. That's $22/157$ vs. $46/189.$ Again you get a P-value about 2% (as for the chi-squared test), but without discussing observed and expected counts:

Addendum: If you want a way to focus just on the tallest trees, you can compare the proportion of them in the logged forest with the proportion of them in the undisturbed forest. That's $22/157$ vs. $46/189.$ Again, you get a P-value about 2% (as for the chi-squared test), but without discussing observed and expected counts:

Addendum: If you want a way to focus just on the tallest trees, you can compare the proportion of them in the logged forest with the the proportion of them in the undisturbed forest. That's $22/157$ vs. $46/185.$ Again you get a P-value about 2% (as for the chi-squared test), but without discussing observed and expected counts:

prop.test(c(22,46), c(157,185))

       2-sample test for equality of proportions 
       with continuity correction

data:  c(22, 46) out of c(157, 185)
X-squared = 5.6159, df = 1, p-value = 0.0178
alternative hypothesis: two.sided
95 percent confidence interval:
 -0.19703724 -0.02000528
sample estimates:
   prop 1    prop 2 
0.1401274 0.2486486 

Addendum: If you want a way to focus just on the tallest trees, you can compare the proportion of them in the logged forest with the the proportion of them in the undisturbed forest. That's $22/157$ vs. $46/189.$ Again you get a P-value about 2% (as for the chi-squared test), but without discussing observed and expected counts:

prop.test(c(22,46), c(157,189))

        2-sample test for equality of proportions 
        with continuity correction

data:  c(22, 46) out of c(157, 189)
X-squared = 5.1553, df = 1, p-value = 0.02318
alternative hypothesis: two.sided
95 percent confidence interval:
 -0.19088789 -0.01562982
sample estimates:
   prop 1    prop 2 
0.1401274 0.2433862 

Addendum: If you want a way to focus just on the tallest trees, you can compare the proportion of them in the logged forest with the the proportion of them in the undisturbed forest. That's $22/157$ vs. $46/185.$ Again you get a P-value about 2% (as for the chi-squared test), but without discussing observed and expected counts:

prop.test(c(22,46), c(157,185))

       2-sample test for equality of proportions 
       with continuity correction

data:  c(22, 46) out of c(157, 185)
X-squared = 5.6159, df = 1, p-value = 0.0178
alternative hypothesis: two.sided
95 percent confidence interval:
 -0.19703724 -0.02000528
sample estimates:
   prop 1    prop 2 
0.1401274 0.2486486