Question

In: Statistics and Probability

Use the following population data for the next question(s). At a bird watching center, the number...

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.

Solutions

Expert Solution

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


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