3. The following table lists the frequency distribution for 60 rolls of a die.
Outcome 1-spot 2-spot 3-spot 4-spot 5-spot 6-spot
Frequency 7 12 8 15 11 7
Using the “Goodness of Fit” test at the 5% level of significance, test whether the null hypothesis that the given die is fair is true. (Hint: If the die is fair, each “spot” has the same chance of appearing). Use the back of the page for work if needed.
Use the Chi-square (X^2) method.
X^2 = ∑ (O-E)^2/E, where O = observed frequency and E = expected frequency.
∑ = sum of all the cells.
E = (column total * row total)/grand total
df = n - 1, where n = number of cells
Look up value in X^2 table in the back of your textbook.
To perform the "Goodness of Fit" test, we need to calculate the expected frequencies for each outcome under the assumption that the die is fair. Then, we will compare the observed frequencies with the expected frequencies using a statistical test.
Step 1: Calculate the expected frequencies:
Since each "spot" has an equal chance of appearing on a fair die, the expected frequency for each outcome can be calculated by dividing the total number of rolls (60) by the number of possible outcomes (6).
Expected Frequency = Total Rolls / Number of Outcomes = 60 / 6 = 10
So, the expected frequencies for each outcome are as follows:
Outcome Expected Frequency
1-spot 10
2-spot 10
3-spot 10
4-spot 10
5-spot 10
6-spot 10
Step 2: Calculate the test statistic:
To test whether the observed frequencies significantly differ from the expected frequencies, we will use the chi-square test statistic. The formula for the chi-square test statistic is:
χ² = Σ((Observed Frequency - Expected Frequency)² / Expected Frequency)
We calculate the test statistic by plugging in the observed and expected frequencies:
χ² = ((7-10)²/10) + ((12-10)²/10) + ((8-10)²/10) + ((15-10)²/10) + ((11-10)²/10) + ((7-10)²/10)
= (9/10) + (4/10) + (4/10) + (25/10) + (1/10) + (9/10)
= 5.2
Step 3: Determine the critical value:
To determine whether the calculated test statistic is significant at the 5% level of significance, we need to compare it with the critical value from the chi-square distribution table. The degrees of freedom for this test are the number of categories minus 1. Since we have 6 categories, the degrees of freedom are 6-1 = 5. Looking up this value in the chi-square distribution table at the 5% significance level, we find the critical value to be approximately 11.07.
Step 4: Compare the test statistic with the critical value:
Since the calculated test statistic (χ² = 5.2) is less than the critical value (11.07), we do not have enough evidence to reject the null hypothesis that the given die is fair. In other words, the observed frequencies do not significantly differ from the expected frequencies, suggesting that the die being tested is fair.
Note: If the calculated test statistic were greater than the critical value, we would reject the null hypothesis and conclude that the die is unfair, as the observed frequencies differ significantly from the expected frequencies.