# Chi square goodness of fit test

The table below displays the number of accidents recorded at a particular intersection during each of the four seasons last year:

    Season           Spring Summer Fall Winter
no. of accidents 13      24     18  25


We would like to conduct a chi-square goodness-of-fit test to determine whether accidents are uniformly distributed over the four seasons.The value of the test statistic for the appropriate test of significance is?

I'm not quite sure how to proceed with this one.. Binomial distribution would not work, I think, since it's asking for a uniform distribution. How would I compute p-hat?

• Jack, the count within each cell (considered on its own, and coniditioned on the total number of accidents, $n$) would have a binomial distribution ($Y_i\sim \text{Bin}(n,p_i)$; though considering all cells at once, they're multinomial). The expected proportion across cells is uniform (i.e. $p_i=p_j \forall i,j$). If this is homework, could you please tag it as such? (I realize it's not the convention where this was previously posted, but it is here) – Glen_b Dec 7 '12 at 10:15
• Do you mean chi-square goodness of fit test? That sounds like what you are looking for. – Peter Flom Dec 7 '12 at 12:24

Since we are testing for a uniform distribution, the values of each season should be equal. The most likely value for each season assuming a uniform distribution is the average of all the seasons.

(13+24+18+25)/4 = 80/4 = 20


The formula for the Chi-Squared Statistic is given here.

http://en.wikipedia.org/wiki/Pearson's_chi-squared_test

((13-20)^2)/20 = 2.45
((24-20)^2)/20 = 0.80
((18-20)^2)/20 = 0.20
((25-20)^2)/20 = 1.25


Chi-Squared Statistic

2.45 + 0.80 + 0.20 + 1.25 = 4.7


The degrees of freedom are

df=4-1=3


You should be able to figure out the rest.

• (-1) This answer was looking good until it completely misinterpreted the p-value and therefore came to the wrong conclusion. (It also obtained an incorrect p-value.) – whuber Dec 7 '12 at 14:23