Test if proportion was the same across batches I have to solve the following question:
From each of six batches of seed, a random sample of $100$ seeds was selected for sowing. The numbers of seeds that failed to germinate in the six samples of $100$ seeds were:
$$
12,20,9,17,24,16.
$$
Test the hypothesis that the proportion of non-germinating seeds was the same for all batches.
My solution is the following:
We have a total of $98(=12+20+9+17+24+16)$ non-germinating seeds from our data, out of a total of $600$. So, the expected number of non germinating seeds would be $\frac{98}{600}\cdot100 = 16.3$.
Thus, from the Pearson-Chi-squared test we get that:
$$X^2 = \frac{(16.3-12)^2}{12} + \frac{(16.3-20)^2}{20}+\frac{(16.3-9)^2}{9} + \frac{(17-16.3)^2}{17}+ \frac{(24-16.3)^2}{24}+ \frac{(16-16.3)^2}{16} = 12.2795$$
Now, we know that $X^2\sim\chi^{2}_{6-\dim(\theta)-1}$. Now, my question is what is our $\dim(\theta)$ here. Here $\theta$ refers to the variable on which the null hypothesis depends. In this particular case, I assumed it would be just the mean, hence $\dim(\theta)=1$. Is my assumption correct?
Under this assumption, the problem would then imply that $X^2\sim\chi^2_{4}$. The $95\%$ point would thus be $9.488$. Since $12.2795>9.488$, we have that our point is outside the critical region, hence we reject $H_0$.
I would like to know if what I did is correct, or, if wrong, what I would need to change.
 A: Using prop.test in R:
prop.test(c(12,20,9,17,24,16), rep(100,6))

    6-sample test for equality of proportions 
    without continuity correction

data:  c(12, 20, 9, 17, 24, 16) out of rep(100, 6)
X-squared = 10.635, df = 5, p-value = 0.05912
alternative hypothesis: two.sided
sample estimates:
prop 1 prop 2 prop 3 prop 4 prop 5 prop 6 
  0.12   0.20   0.09   0.17   0.24   0.16 

Using chisq.test in R with a contingency table.
ng = c(12,20,9,17,24,16);  g = 100-ng
TBL = rbind(ng,g)
TBL
   [,1] [,2] [,3] [,4] [,5] [,6]
ng   12   20    9   17   24   16   
g    88   80   91   83   76   84

chisq.test(TBL)
    Pearson's Chi-squared test

data:  TBL
X-squared = 10.635, df = 5, p-value = 0.05912

Notice that the DF for the approximating chi-squared
distribution, based on a $6\times 2$ contingency table
is $\nu = (2-1)(6-1) = 5.$
P-value is the area under the density curve of $\mathsf{Chisq}(\nu=5)$ to the right of the chi-squared statistic.
1 - pchisq(10.63501, 5)
[1] 0.05911663

curve(dchisq(x,5), 0, 20, col="blue", lwd=2, 
      ylab="PDF", xlab="Chi-squared", main="")
 abline(v=0, col="green2"); abline(h=0, col="green2")
 abline(v=10.635, col="red", lwd=2, lty="dotted")


A: Consider using a  goodness-of-fit approach. The null hypothesis $H_{0}$ is that the data comes from a fully formed model vs $H_{1}$ that the data is unrestricted.
For $H_{0}$ assume that for each seed the probability of failure to germinate is $p$, ie each sample is a draw from a Binomial distribution $B(100,p)$. Here we can use the Poisson approximation $P(\lambda)$. This means the dimension of the null hypothesis is 1 (as it just depends on $\lambda$).
The MLE of $\lambda = \frac{1}{6}\sum_{i=1}^6 x_{i} = 98/6 = 16.33$.
$X^2 = \frac{(12-16.3)^2}{16.3} + \frac{(20-16.3)^2}{16.3}+\frac{(9-16.3)^2}{16.3} + \frac{(17-16.3)^2}{16.3}+ \frac{(24-16.3)^2}{16.3}+ \frac{(16-16.3)^2}{16.3} = 8.9$
$X^2\sim\chi^{2}_{6-\dim(\lambda)-1} =\chi^{2}_{4}$
The $90\%$ point is $7.78$ and the p-value of $8.9 = 6.3\%$
