Testing the hypothesis with level of significance

Question

How can we test the hypothesis whether the following die is fair with level of significance α = 0.o5? We cast it n = 1200 times and get the frequencies of the sides:$ν_1 = 184, ν_2 = 212, ν_3 = 190, ν_4 = 208, ν_5 = 212, ν_6 = 194$. Thanks.

Answer 1

I think you want to do a chi-squared goodness-of-fit (GOF) test. You have observed frequencies $X = (184, 212, 190, 208, 212, 194)$ for the various faces. The expected frequency of each face of a fair die is $E = n/6 = 1200/6 = 200.$

Intuitively, the question is whether the frequencies $X$ are collectively close enough to 200 to believe the die is fair, or so far away from 200 that you suspect the die is biased.

In this case, the usual GOF statistic is $$Q = \sum_{i=1}^6 \frac{(X_i - E)^2}{E}$$ and for a fair die $Q \stackrel{aprx}{\sim} CHISQ(df = 5).$

Big discrepancies from 200 counts per face yield big values of $Q,$ and we want to reject $H_0: \text{fair die}$ if the observed value of $Q$ exceeds the 95th percentile $q^* = 11.07$ of $CHISQ(df = 5),$ which can be found in printed tables of the chi-squared distribution or using software.

 qchisq(.95, 5)
 ## 11.0705

[Notice that you have $k = 6$ 'categories' (faces) and degrees of freedom $\nu = k-1 = 5.$ The value 11.07 is the 'critical value' for a test of the null hypothesis at the 5% level of significance.]

What remains is for you to compute the observed value of $Q$ based on the given face counts, correct to two or three decimal places. As a check, I got $Q \approx 3.7.$

Note: The mean of the distribution $CHISQ(df = \nu)$ is $\nu.$ So for your data, you might guess, even without consulting a chi-squared table, that such a small value as $Q \approx 3.7$ will not lead to rejection.

Below is a graph of the density function of $CHISQ(5).$ The area under the curve to the right of the broken red line is 5%. The rejection region lies to the right of this line. The solid black line shows the observed value of $Q$.