# Thread: Chi square test

1. ## Chi square test

According to a theory, four categories of strawberries are supposed to occur in the ratio 9:3:3:1. For these four categories, 124, 30, 43 and 11 were counted, respectively. Are these compatible with the theory?Let $\alpha=0.05$.

Work so far-
124+30+43+11=208

Expected values are therefore 117, 39, 39, 13 for the ratio 9:3:3:1

(4 columns - 1)=3 degrees of freedom

I calculated the Chi-test statistic by doing (124-117)^2/117+....+(11-13)^2/13 and got 3.21, but I am not sure what to compare it to or what to do with the $\alpha=0.05$. Thanks for your help!

2. Originally Posted by larz
According to a theory, four categories of strawberries are supposed to occur in the ratio 9:3:3:1. For these four categories, 124, 30, 43 and 11 were counted, respectively. Are these compatible with the theory?Let $\alpha=0.05$.

Work so far-
124+30+43+11=208

Expected values are therefore 117, 39, 39, 13 for the ratio 9:3:3:1

(4 columns - 1)=3 degrees of freedom

I calculated the Chi-test statistic by doing (124-117)^2/117+....+(11-13)^2/13 and got 3.21, but I am not sure what to compare it to or what to do with the $\alpha=0.05$. Thanks for your help!
df = 4 - 1 = 3.

Critical value for 3 df at 0.05 significance level is 7.81 (got from tables).

3.21 < 7.81 therefore ......

3. Therefore the observations are compatible with the theory! Thanks. I found the table you are talking about. Now it makes sense.