For autosomal recessive inheritance what is the chi-square value for these results?

Aa X Aa should yield 75% dominant and 25% recessive.
Thus, out of 100 individuals, we should obtain 75 as dominant and 25 as recessive.
? = degree of freedom (n-1) = 1 for this case.

?² = ? [(o-e)²/e] = ? [d²/e], where d=deviation; o= observed; e=expected

First of all, we calculate for the dominant character.
d = o-e = 80-75=5
d²= 5²=25
d²/e = 25/75 = 0.33

Now for the recessive character,
d = o-e = 20-25= -5
d²= (-5)² = 25
d²/e = 25/25 = 1.0

Now, ?² = ? [d²/e] = 0.33 + 1.0
?² = 1.33

Now we have chi-square value (1.33) and degree of freedom (1). Now use the chart to determine the p-value.
http://passel.unl.edu/Image/Namuth-CovertDeana956176274/chi-sqaure%20distribution%20table.PNG
Look in the first horizontal line (for ?=1), you are getting chi square as 1.32 at p value = 0.25.

Since the p value (0.25) is greater than 0.05, the deviation is not of statistical significance and impies that chance deviations are occuring.
Thus, we accept the chance hypothesis..!!

This was my first chi-square solving question!! Hope i made it clear to you! I need to learn it even more.