Question

ee randomly selected households are surveyed. The numbers of people in the households are 4,5, and 9. Assume that samples of size n 2 are randomly selected with replacement from the population of 4, 5, and 9. Listed below are the nine d Complete parts (a) through (c). 4,4 4,5 4,9 5,4 5,5 59 9,4 95 9,9p a. Find the variance of each of the nine samples, then summarize the sampling distribution of the variances in the format of a table representing the probability distribution of the distinct variance values. Probability Click to select your answer(s) iiee anduy sellcileu ioisuholus are surveyed. The humbers of people in the households are 4, 5, and 9. Assume that samp of sze n = 2 are ran only selected with replacement from the population of 4,5, and 9 Listed below are the nine dierent sample Complete parts (a) through (c) 4,4 4,5 4,9 5,4 5,5 5,9 9,4 9,5 9,9 Type an integer or a fraction. Use ascending order of the sample variances) b. Compare the population variance to the mean of the sample variances. Choose the correct answer below O A. The population variance is equal to the square root of the mean of the sample variances. 0 B. The population variance is equal to the mean of the sample variances. OC. The nonulation variance is eaual to the sauare of the mean of the samnle variances Click to select your answer(s). DELL F12 F11 F10 FO F8 Fo c. Do the sample variances target the value of the population variance? In general, do sample variances make good estimators df population variances? Why or why not? O A. The O B. The sample variances target the population variances,therefore, sample variances make good estimators of population O c. The sample varsances do not target the population varance, hermake good estimators of Click to select your answer(s) sample variances do not target the population variance, therefore, sample variances do not make good estimators of population variances variances e, therefore, sample vanances make good esti naors o

Answer 1

1. Find the variance of each of the nine​ samples, then summarize the sampling distribution of the variances in the format of a table representing the probability distribution.
Begin by finding all sample variances.

2. Use the formula below to find each sample​ variance, where x x-bar is the sample mean and n is the sample size.

s²=∑(x-x)²/n-1

3. Find the sample variance for (4,4​). Use the formula below to first solve for the sample mean x-bar ​, where n is the sample size.
x-bar =∑x/n
= 4+4/2
=8/2
=4/1
=4

4. Now solve for s². Substitute x-bar =4.0 and the sample values 4 and 4 for x into the sample variance formula and solve for s²

s²=∑(x-x)²/n-1
s₂=(4-4.0)²+(4-4)²/2-1
s²=(0)²+(0)²/1
s²=0+0/1
s²=0/1
s²=0

5. Use a similar process to find the sample variances for the other samples. Note that the probability of each sample is​ 1/9 because each of the nine samples are equally likely

6. Table will look like
Sample s₂ Probability
4,4 0 1/9
4,5 .5 1/9
4,9 12.5 1/9
5,4 .5 1/9
5,5 0 1/9
5,9 8 1/9
9,4 12.5 1/9
9,5 8 1/9
9,9 0 1/9

7. Now summarize the sampling distribution of the variances in the format of a table representing the probability distribution. Because each sample is equally likely to​ occur, the probability of a certain variance value is the number of occurrences of that variance value divided by the total number of​ samples, 9.

8. Complete the table for the probability distribution below
(count and add the number of different S²)
s² Probability
0 3/9
.5 2/9
8 2/9
12.5 2/9

9. Compare the population variance to the mean of the sample variances.
First find the mean of the sample variances. The mean of the sample variances is the sum of the nine sample variances divided by the number of samples.
0+.5+12.5+.5+0+8+12.5+8+0/9 = 4.667 rounded to the 3 decimal place

10. Find the population variance of 4​, 5​, 9 using the formula​ below, where muμ is the population mean and N is the population size.

σ²=∑(x-µ)²/N

11. First calculate the population mean.

µ = 4+5+9/3
µ=6

12. Now calculate the population variance. Substitute the values for x and µ=6 into the population variance formula.

σ²=∑(x-µ)²/N
σ²=(4-6)²+(5-6)²+(9-6)²/3
σ²=(-2)²+(-1)²+(3)²/3
σ²=4+1+9/3
σ²=14/3
σ²=4.667 rounded three decimal places

13. The population variance is 4.667 and the mean of the sample variances is 4.667. The values are equal.

14. Do the sample variances target the value of the population​ variance? In​ general, do sample variances make good estimators of population​ variances? Why or why​ not?

If the mean of the sample variances is equal to the population​ variance, then the sample variances target the population variance. Compare the values of the population variance and the mean of the sample variances to determine if the sample variances target the value of the population variance.

15. If the sample variances target the population​ variance, the sample variances are unbiased estimators. If the sample variances do not target the population​ variance, the sample variances are biased estimators. A biased estimator systematically underestimates or overestimates the parameter. Unbiased sample statistics make good​ estimators; biased sample statistics do not make good estimators.

Answers

a)

s^2	p
0	0.333333	(3/9)
0.5	0.222222	(2/9)
8	0.222222	(2/9)
12.5	0.222222	(2/9)

b)

option B) is correct

The population variance is 4.667 and the mean of the sample variances is 4.667. The values are equal.

c)

option B) is correct