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# An Example of Chi-Square Test for a Multinomial Experiment

To see how a chi-square hypothesis test works with a multinomial experiment, we will investigate the following two examples.

### Example 1: A Fair Coin

A fair coin has a equal probability of 1/2 of coming up heads or tails. We toss a coin 1000 times and record the results of a total of 580 heads and 420 tails. We want to test the hypothesis at a 95% level of confidence that the coin we flipped is fair. More formally, the null hypothesis H0 is that the coin is fair. Since we are comparing observed frequencies of results from a coin toss to the expected frequencies from an idealized fair coin, a chi-square test should be used.

### Compute the Chi-Square Statistic

We begin by computing the chi-square statistic for this scenario. There are two events, heads and tails. Heads has an observed frequency of f1 = 580 with expected frequency of e1 = 50% x 1000 = 500. Tails has an observed frequency of f2 = 420 with expected frequency of e1 = 500.

We now use the formula for the chi-square statistic and see that χ2 = (f1 - e1 )2/e1 + (f2 - e2 )2/e2= 802/500 + (-80)2/500 = 25.6.

### Find the Critical Value

Next we need to find the critical value for the proper chi-square distribution. Since there are two outcomes for the coin there are two categories to consider. The number of degrees of freedom is one less than the number of categories: 2 - 1 = 1. We use the chi-square distribution for this number of degrees of freedom and see that χ20.95=3.841.

### Reject or Fail to Reject?

Finally we compare the calculated chi-square statistic with the critical value from the table. Since 25.6 > 3.841, we reject the null hypothesis that this is a fair coin.

### Example 2: A Fair Die

A fair die has a equal probability of 1/6 of rolling a one, two, three, four, five or six. We roll a die 600 times and note that we roll a one 106 times, a two 90 times, a three 98 times, a four 102 times, a five 100 times and a six 104 times. We want to test the hypothesis at a 95% level of confidence that we have a fair die.

### Compute the Chi-Square Statistic

There are six events, each with expected frequency of 1/6 x 600 = 100. The observed frequencies are f1 = 106, f2 = 90, f3 = 98, f4 = 102, f5 = 100, f6 = 104,

We now use the formula for the chi-square statistic and see that χ2 = (f1 - e1 )2/e1 + (f2 - e2 )2/e2+ (f3 - e3 )2/e3+(f4 - e4 )2/e4+(f5 - e5 )2/e5+(f6 - e6 )2/e6 = 1.6.

### Find the Critical Value

Next we need to find the critical value for the proper chi-square distribution. Since there are six categories of outcomes for the die, the number of degrees of freedom is one less than this: 6 - 1 = 5. We use the chi-square distribution for five degrees of freedom and see that χ20.95=11.071.

### Reject or Fail to Reject?

Finally we compare the calculated chi-square statistic with the critical value from the table. Since the calculated chi-square statistic is 1.6

Courtney Taylor