Questions: Three randomly selected children are surveyed. The ages of the children are 1, 4, and 10. Assume that samples of size n=2 are randomly selected with replacement from the population of 1, 4, and 10. Listed below are the nine different samples. Complete parts (a) through (d). 1,1 1,4 1,10 4,1 4,4 4,10 10,1 10,4 10,10 distribution of the variances in the format of a table representing the probability distribution of the distinct variance values. s^2 Probability 00 1/3 4.5 2/9 18 2/9 40.5 2/9 (Type integers or simplified fractions.) c. Find the mean of the sampling distribution of the sample variance. 14.0 (Round to three decimal places as needed.) d. Based on the preceding results, is the sample variance an unbiased estimator of the population variance? Why or why not? A. The sample variance targets the population variance. As such, the sample variance is an unbiased estimator of the population variance. B. The sample variance does not target the population variance. As such, the sample variance is an unbiased estimator of the population variance. C. The sample variance does not target the population variance. As such, the sample variance is a biased estimator of the population variance. D. The sample variance targets the population variance. As such, the sample variance is a biased estimator of the population variance.

Solution Steps

For each of the nine samples, we calculate the sample variance \( s^2 \):

1. Sample \( [1, 1] \):

   • Mean \( \mu = \frac{1 + 1}{2} = 1.0 \)
   • Variance \( s^2 = \frac{(1 - 1)^2 + (1 - 1)^2}{2 - 1} = 0.0 \)

2. Sample \( [1, 4] \):

   • Mean \( \mu = \frac{1 + 4}{2} = 2.5 \)
   • Variance \( s^2 = \frac{(1 - 2.5)^2 + (4 - 2.5)^2}{2 - 1} = 4.5 \)

3. Sample \( [1, 10] \):

   • Mean \( \mu = \frac{1 + 10}{2} = 5.5 \)
   • Variance \( s^2 = \frac{(1 - 5.5)^2 + (10 - 5.5)^2}{2 - 1} = 40.5 \)

4. Sample \( [4, 1] \):

   • Mean \( \mu = \frac{4 + 1}{2} = 2.5 \)
   • Variance \( s^2 = \frac{(4 - 2.5)^2 + (1 - 2.5)^2}{2 - 1} = 4.5 \)

5. Sample \( [4, 4] \):

   • Mean \( \mu = \frac{4 + 4}{2} = 4.0 \)
   • Variance \( s^2 = \frac{(4 - 4)^2 + (4 - 4)^2}{2 - 1} = 0.0 \)

6. Sample \( [4, 10] \):

   • Mean \( \mu = \frac{4 + 10}{2} = 7.0 \)
   • Variance \( s^2 = \frac{(4 - 7)^2 + (10 - 7)^2}{2 - 1} = 18.0 \)

7. Sample \( [10, 1] \):

   • Mean \( \mu = \frac{10 + 1}{2} = 5.5 \)
   • Variance \( s^2 = \frac{(10 - 5.5)^2 + (1 - 5.5)^2}{2 - 1} = 40.5 \)

8. Sample \( [10, 4] \):

   • Mean \( \mu = \frac{10 + 4}{2} = 7.0 \)
   • Variance \( s^2 = \frac{(10 - 7)^2 + (4 - 7)^2}{2 - 1} = 18.0 \)

9. Sample \( [10, 10] \):

   • Mean \( \mu = \frac{10 + 10}{2} = 10.0 \)
   • Variance \( s^2 = \frac{(10 - 10)^2 + (10 - 10)^2}{2 - 1} = 0.0 \)

The distinct variances obtained are \( 0.0, 4.5, 18.0, \) and \( 40.5 \).

The probabilities of each variance are calculated based on their frequency:

• \( P(0.0) = \frac{3}{9} = \frac{1}{3} \)
• \( P(4.5) = \frac{2}{9} \)
• \( P(18.0) = \frac{2}{9} \)
• \( P(40.5) = \frac{2}{9} \)

Thus, the probability distribution is: \[ \begin{array}{c|c} s^2 & \text{Probability} \\ \hline 0.0 & \frac{1}{3} \\ 4.5 & \frac{2}{9} \\ 18.0 & \frac{2}{9} \\ 40.5 & \frac{2}{9} \\ \end{array} \]

The mean of the sampling distribution of the sample variance \( \mu \) is calculated as follows: \[ \mu = 0.0 \times \frac{1}{3} + 4.5 \times \frac{2}{9} + 18.0 \times \frac{2}{9} + 40.5 \times \frac{2}{9} = 14.0 \]

The population variance \( \sigma^2 \) is calculated from the population \( [1, 4, 10] \):

• Mean \( \mu = \frac{1 + 4 + 10}{3} = 5.0 \)
• Variance \( \sigma^2 = \frac{(1 - 5)^2 + (4 - 5)^2 + (10 - 5)^2}{3} = 14.0 \)

Comparing the mean of the sampling distribution of the sample variance \( 14.0 \) with the population variance \( 14.0 \):

• Since they are equal, the sample variance is an unbiased estimator of the population variance.

Final Answer