In: Statistics and Probability
Use the following population data for the next question(s). At a bird watching center, the number of yellow throated warblers observed over Memorial Day Weekend was: three on Saturday, 10 on Sunday, and eight on Monday. Assume sample sizes of two are randomly selected with replacement from this population.
10) List the nine different possible samples of yellow throated warblers and find the mean of each of them.
11) Identify the probability of each sample of yellow throated warblers and describe the sampling distribution of sample means.
12) Find the mean of the sampling distribution of yellow throated warblers.
13) Compare the mean of the sampling distribution of yellow throated warblers to the mean of the population of yellow throated warblers.
Solution
Population values: 3, 10, 8
Back-up Theory
Mean = Σx.p(x) …………………………………………………………………………………… (1)
Now, to work out the solution,
Q10
i) List of nine different possible with replacement samples of 2 yellow throated warblers
ii) Mean of each sample
Sample # |
Sample Values |
Sample Mean |
|
1 |
3 |
3 |
3 |
2 |
3 |
10 |
6.5 |
3 |
3 |
8 |
5.5 |
4 |
10 |
3 |
6.5 |
5 |
10 |
10 |
10 |
6 |
10 |
8 |
9 |
7 |
8 |
3 |
5.5 |
8 |
8 |
10 |
9 |
9 |
8 |
8 |
8 |
Answer 1 & 2 DONE
Q11)
i) Probability of each sample of yellow throated warblers
Sample # |
Sample Values |
Probability |
|
1 |
3 |
3 |
1/9 |
2 |
3 |
10 |
1/9 |
3 |
3 |
8 |
1/9 |
4 |
10 |
3 |
1/9 |
5 |
10 |
10 |
1/9 |
6 |
10 |
8 |
1/9 |
7 |
8 |
3 |
1/9 |
8 |
8 |
10 |
1/9 |
9 |
8 |
8 |
1/9 |
Total |
1 |
Answer 3 DONE
ii) Sampling distribution of sample means
Sample Mean |
Frequency |
Probability |
3 |
1 |
1/9 |
5.5 |
|| |
2/9 |
6.5 |
|| |
2/9 |
8 |
| |
1/9 |
9 |
|| |
2/9 |
10 |
| |
1/9 |
Total |
10 |
`1 |
Answer 4 DONE
Q12)
Mean of the sampling distribution of yellow throated warblers = 7. Answer 5
Sample Mean - xibar |
Probability - p(xibar) |
xibar x p(xibar) |
3 |
1/9 |
0.333333333 |
5.5 |
2/9 |
1.222222222 |
6.5 |
2/9 |
1.444444444 |
8 |
1/9 |
0.888888889 |
9 |
2/9 |
2 |
10 |
1/9 |
1.111111111 |
Total |
`1 |
7 |
Q13)
i) Population mean, µ = (3 + 10 + 8)/3 = 7 Answer 6
ii) Mean of the sampling distribution, X double bar = 7 Answer 7
iii) Comparison: Population mean = Mean of the sampling distribution of sample means. Answer 8
DONE