Example problem re: Goodness of Fit test:

You are given a six-faced die and asked to determine with 95% confidence if it is “fair.”

• What are the hypotheses?

Answer: the null H₀: the die is fair, p₁ = p₂ = … = p₆ = 1/6; the alternative Ha: the die is not fair, at least one p_i ≠ 1/6.

• What is the critical value of chi-square?

Answer: degrees of freedom df = 6 – 1 = 5. From a table or calculator, χ_0.05² =11.071.

Common student mistake #1:

using an incorrect value to enter table or calculator. Some tables give the area under the curve to the right of the critical value; others to the left. The same is true for calculators. You must check the table or calculator description and/or graph of the Chi-square distribution associated with it. Hint: remember that we want the area under the curve to the right of the critical value to be equal to alpha = 0.05 which is 1 – the 95% confidence level.

You toss the die 120 times and record the numbers of 1’s, 2’s, …, and 6’s (shown in the table in the column labeled “Observed”). Determine the chi-square statistic.

• Calculate the expected frequencies and complete the following table:

Solution:

Common student mistake#2: Incorrect “Expected values.” Hint: remember the expected frequency of each die face assuming the die is fair is 1/6 times the total number of tries.

The completed table:

• What is your conclusion?
  1. There is insufficient evidence at the 0.05 level to reject the null hypothesis the die is fair. [feedback: Incorrect. Because the chi-square statistic of 18.6 is greater than the critical value of 11.071, we should reject the null.]
  2. There is sufficient evidence at the 0.05 level to accept the null hypothesis the die is fair. [feedback: Incorrect for two reasons. One, we never can accept the null, only fail to reject it. Two, because the chi-square statistic of 18.6 is greater than the critical value of 11.071, we should reject the null.]
  3. There is insufficient evidence to accept the alternative hypothesis that the die in not fair. [feedback: Incorrect. Because the chi-square statistic of 18.6 is greater than the critical value of 11.071, we should reject the null.]
  4. There is sufficient evidence to accept the alternative hypothesis that the die is not fair. [Feedback: Correct, because the chi-square statistic of 18.6 is greater than the critical value of 11.071, we should reject the null.]