According to United Nations Population Division, the age distribution of the Commonwealth of Australia is: 21% less than 15 years of age, 67% between 15 and 65 years of age, and 12% over 65. A random sample of 210 Canberra residents revealed 40 under 15, 145 between 15 and 65, and 25 over 65. Are the ages of Canberra residents unusual in any way?

According to United Nations Population Division, the age distribution of the Commonwealth of Australia is: 21% less than 15 years of age, 67% between 15 and 65 years of age, and 12% over 65. A random sample of 210 Canberra residents revealed 40 under 15, 145 between 15 and 65, and 25 over 65. Are the ages of Canberra residents unusual in any way?

• A. Chi-Square Goodness of Fit Test
• B. 1-Proportion Z-Test
• C. ANOVA
• D. Kolmogorov-Smirnov Test

Answer: (A)

Testing whether the observed age distribution in Canberra fits a specified distribution across several categories is a classic case for a chi-square goodness-of-fit test. You use this when you have counts in several categories and a hypothesized proportion for each category, and you want to know if the observed counts deviate from those expectations. Compute the expected counts from the given proportions and the sample size: under 15 should be 0.21 × 210 = 44.1, ages 15–65 should be 0.67 × 210 = 140.7, and over 65 should be 0.12 × 210 = 25.2. The observed counts are 40, 145, and 25 respectively. Calculate the chi-square statistic by summing (O − E)² / E for each category: for under 15, (40 − 44.1)² / 44.1 ≈ 0.38; for 15–65, (145 − 140.7)² / 140.7 ≈ 0.13; for over 65, (25 − 25.2)² / 25.2 ≈ 0.0016. The total is about 0.51. With three categories, the degrees of freedom are 3 − 1 = 2. A chi-square value of 0.51 with 2 degrees of freedom corresponds to a p-value around 0.8, which is far above standard significance levels like 0.05. Therefore, there’s no evidence that Canberra’s ages deviate from the specified distribution; the observed counts are consistent with the expected distribution. This test is the right choice because it directly compares observed counts in each category to what would be expected under a specified distribution. Other options don’t fit: a 1-proportion z-test looks at a single category proportion, ANOVA compares means of continuous outcomes across groups, and the Kolmogorov-Smirnov test is used for comparing continuous distributions rather than counts in discrete categories.

Testing whether the observed age distribution in Canberra fits a specified distribution across several categories is a classic case for a chi-square goodness-of-fit test. You use this when you have counts in several categories and a hypothesized proportion for each category, and you want to know if the observed counts deviate from those expectations.

Compute the expected counts from the given proportions and the sample size: under 15 should be 0.21 × 210 = 44.1, ages 15–65 should be 0.67 × 210 = 140.7, and over 65 should be 0.12 × 210 = 25.2. The observed counts are 40, 145, and 25 respectively.

Calculate the chi-square statistic by summing (O − E)² / E for each category: for under 15, (40 − 44.1)² / 44.1 ≈ 0.38; for 15–65, (145 − 140.7)² / 140.7 ≈ 0.13; for over 65, (25 − 25.2)² / 25.2 ≈ 0.0016. The total is about 0.51. With three categories, the degrees of freedom are 3 − 1 = 2. A chi-square value of 0.51 with 2 degrees of freedom corresponds to a p-value around 0.8, which is far above standard significance levels like 0.05. Therefore, there’s no evidence that Canberra’s ages deviate from the specified distribution; the observed counts are consistent with the expected distribution.

This test is the right choice because it directly compares observed counts in each category to what would be expected under a specified distribution. Other options don’t fit: a 1-proportion z-test looks at a single category proportion, ANOVA compares means of continuous outcomes across groups, and the Kolmogorov-Smirnov test is used for comparing continuous distributions rather than counts in discrete categories.