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Submitted by Jdpt3 on Fri, 2012-06-15 23:14
due on Fri, 2012-06-15 22:39
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Statistics Assignment

6. A population consists of the following five values: 2,2,4,4 and 8.

A. List all samples of size 2 and compute the mean of each sample.

B. compute the mean of distribution of sample means and the population mean. Compare the two values.

C. Compare the dispersion in the population with that of the sample means.

10. There are five sales associates at mid-motors ford. The five representatives and the number of cars they sold last week are:

Sales. Cars Sold

Representative.

Peter Staller. 8

Connie Stallter. 6

Juan Lopez. 4

Ted Barnes. 10

Peggy Chu 6

A. How many different samples of size 2 are possible?

B. List all possible samples of size 2, and compute the mean of each sample

C. Compare the mean of the sampling distribution of sample means with that of the population

D. On a chart compare the dispersion in sample means with that of the population.

12. Scrapper elevator company has 20 sales representatives who sell its product through out the US and Canada. The number of units sold last month by each representative is listed below. Assume these sales figures to be population values.

2 3 2 3 3 4 2 4 3 2 2 7 3 4 5 3 3 3 3 5

A. Draw a graph showing the population distribution

B. compute the mean of the population

C. Select five random samples of 5 each. Compute the mean of each sample. Use the methods described the methods to be used in order to determine the items to be included in the sample.

D. Compare the mean of the sampling distribution of the sample means to the population mean. Would you expect the two values to be about the same?

E. draw a histogram of the sample means. Do you notice the difference in the shape of the distribution of sample means compared to the shape of the population distribution?

16. A normal population has a mean of 75 and a standard deviation of 5. You select a sample of 40. Compute the probability the sample mean is:

A. Less than 74

B. between 74 and 76

C. Between 76 and 77

D. Greater than 77

18. According to an IRS study it takes a mean of 330 minutes for taxpayers to prepare, copy, and electronically file a 1040 tax form. This distribution of times follows the normal distribution and the standard deviation is 80 minutes. A customer watchdog agency selects a random sample of 40 tax payers.

A. What is the standard error of the mean in this example?

B. what is the likelihood that sample mean is greater than 320 minutes?

C. What is the likelihood the sample mean is between 320 and 350 minutes?

D. What is the likelihood the sample mean is greater than 350 minutes?

Answer
Submitted by StatSolver on Sat, 2012-06-16 02:44
teacher rated 56 times
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Please see the attachment for solution.

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Please xxx xxx attachment xxx xxxxxxxxx xx xxx need any clarification xxxxxx ask xxx xxxxxxx

[Note: The normal probability P(Z < xx can be calculated using xxx Excel xxxxxxx

=NORMSDIST(z) See the xxxxx xxxx for computation]

file1.doc preview (1049 words)

xx x xxxxxxxxxx consists of the xxxxxxxxx xxxx xxxxxxx xxxxxxx xxx xxxxxxx xxxx xxx samples of xxxx x xxx xxxxxxx xxx mean of each xxxxxxxxxxxx compute the mean of xxxxxxxxxxxx xx sample means and the population xxxxx Compare the two xxxxxxxxxxxx xxxxxxx xxx xxxxxxxxxx xx xxx xxxxxxxxxx with that of the xxxxxx means.

xxxxxxxxx

A)

xxxxxxxxxxxxxxxxxxx

Sample

xxxxxx

Mean

x

x

x

2

2

x

x

x

x

x

4

x

4

x

8

x

x

2

x

3

x

x

4

3

x

2

x

5

x

x

4

4

x

x

x

6

10

x

x

x

xx The population xxxxx

( =(2 x 2 x x x x x xxxx x 20/5 = x

The xxxx xx xxxxxxxxxxxx of xxxxxx xxxxxx

x (2 x x x x + 5 x 3 + x + x + x x 6 x 6)/10 x 40/10 x xx

xxxxx xxx mean of xxxxxxxxxxxx of sample means and xxx xxxxxxxxxx mean are equal.

C) xxx dispersion xxx xxx population is xxxxxxx than xxxx xxx the sample xxxxxx xxx xxxxxxxxxx varies from 2 xx xx xxxxxxx xxx xxxxxx means only vary from 2 to 6.

xxx There xxx five xxxxx associates xx xxxxxxxxxx xxxxx xxx xxxx xxxxxxxxxxxxxxx xxx the

- - - more text follows - - -

file2.xls preview (62 words)

6,10

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
xxxxxxValues Mean Samplexxxx soldxxxxCars sold No of Sales xx of xxxxx
x x x 2 1x 6 74 x x
2 x4 x 2 x 4 xxx2
xx x 3 x8 109x 2 x
x x8 5 x x 6x7 03
5 x xxx x x5 8 1 x
6 x x 3x 6 xx 8 x x 1
7xx 57 xx x10 1 x
x x 4 xx4xx x
x4xx 9 x xx
xx4 8xxx106 8

xxxx

Cars sold
xxxxxx xx Sales
xxxxxxxxxx mean

xx

x
xxxx sold
Number xx Means
Sample mean

xxxxx

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
DataUnits sold FrequencySample xxxxxx Values xxxx
2xx1 xx xx32.8
32x x 2x2 x 3 3.2
x3x3x43 xx 3.4
34 344x x xx xxx
x5 2 5x 3x 4 33.2
4x 0
xx 1
x
3
x
x xxxxxx Mean Frequency
7xxxx
33.2 x
43.41
x3.8x
x
3
3
x
x

16,18

xxxx
0
x
x
3
2
0
x
xxxxx sold
xxxxxxxxx
x
x
1
x
xxxxxx xxxxx
xxxxxxxxx
xxxxxxxxxxxxxxxxxx
Question 16 xxxxxxxx xx
x xxxx < z)x zP(Z < xx
74 xxxxxxxxxxxxxxxx -0.79060.2146
xx1.2649 xxxxxx xxx1.5811xxxxxx
77xxxxxx 0.9943


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Answer
Submitted by Kumail Raza on Sat, 2012-06-16 07:02
teacher rated 24 times
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price: $50.00

Answer file is attached with all calculations and with easy printing format. Feel free to contact for any further assistance

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file1.xlsx preview (796 words)

xx

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
xxxxx
xxxxxxxxxx size 2 xx, xx x xx
xxxxxxxxx
xx xxxx xxx samples of xxxx x and compute the xxxx xx each sample.
xx compute xxx mean xx xxxxxxxxxxxx of xxxxxx xxxxx and xxx xxxxxxxxxx mean. xxxxxxx the xxx xxxxxxx
Solution:
No. xx Possible sample x Cnx xx 2xxxx xxxxxxx xxxxxxxxxxx
x 10
xxxxxx xxxxxx Possible xxxxxxx xxxxxx mean xxx
1 x, 2x
2x, x3
x 2 , x3
xx ,xx
52,4 3
6 x,x x
Sample No. xx, 8x
x 4 x xx
9 4 , xx
xxx,x x
Total40
xxxxxx xxxxxxxxxxxx xx x is xx xx
21 2
x 412
xx4
5 2xx
6xxx
Total xx 40
Sample xxxx x4
Population xxxx = xxx2 x 4 x 4 x x) x x
x
Population xxxxx xxx xxxxxx xxxxx xxx same.
C. xxxxxxx xxx xxxxxxxxxx xx xxx population with that xx the xxxxxx means.
=8 - 2
Range xx xxxxxxxxxx mean =6
= 6-x
Range of sample xxxx = 4
The sample xxxx xx less dispersed than xxxxxxxxxx mean.

Q10

xxxxxxxxxxxxxxxxxxxxxxxxxx
Required:
A. List all samples xx size 2 xxx compute the xxxx of xxxx sample.
xx xxxxxxx xxx xxxx xx xxxxxxxxxxxx of sample xxxxx and xxx population xxxxx xxxxxxx the two xxxxxxx
No. Sales Representative xxxx xxxx
x Peter xxxxxxxx
xConnie Stallter x
x Juan Lopez x
x Ted Barnesxx
5 Peggy Chu x
Totalxx
Population size x, 6x 4,xx ,6
x am xxxxxxxxx a xxx xxxxxx xxxxxx from the list xx 5 persons
No. xx xxxxxxxx samplex C n=xx x xxxx xxxxxxx replacement)
xxx
Sample No. xxx Possible samples Sample xxxx

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Answer
Submitted by shahimermaid on Sat, 2012-06-16 10:34
teacher rated 387 times
4.322995
price: $30.00

the answer is attached

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xx x xxxxxxxxxx consists xx xxx following xxxx values: xxxxxxx xxx 8. A. List xxx samples of xxxx 2 and xxxxxxx xxx mean xx each xxxxxxx

xxxxxxx xx xxxx 2 are

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

xxxx

Difference xxxx mean (4)

Square of difference

xxx

2

xx

4

2,4

3

xx

1

xxx

3

-1

1

xxx

x

x

x

xxx

x

xx

x

2,4

3

-1

x

xxx

x

xx

x

2,8

5

1

x

xxx

x

-1

x

xxx

x

xx

x

4,4

4

0

x

xxx

6

2

4

4,2

3

-1

x

xxx

x

-1

x

4,4

x

0

x

xxx

x

2

4

xxx

x

1

x

8,2

x

1

x

xxx

6

2

x

xxx

x

x

x

xx computes xxx xxxx of xxxxxxxxxxxx of sample means xxx the population mean. xxxxxxx xxx two values.

xxxx xx xxxxxxxxxxxx xx sample means xx xxxxxx expected value xx X. it xx xxxxx xx the table.

∑xxxxxx xxxxx xxxxx =4

Population mean is 2+2+4+4+8= xxxxx 4

C. xxxxxxx the xxxxxxxxxx xx xxx population xxxx xxxx xx the xxxxxx means.

The xxxxx xx xxx xxxxxxxx measure xx xxxxxxxxxx

xxxxxx mean xxxxx x xxxx 4

Population xxxxxx 8-2= 6

Variance is the xxxxxxx squared difference xx scores from xxx mean xxxxx xx a distribution. 

Variance xx xxxxxx 36/20=1.8

xxxxxxxx in xxxxxxxxxxx ∑xxxxxxx xxxxxxxxx xxxxxxx xxx -4)^2 x (8-4)^2= 4+4+0=0+16 =24/5 x xxx

Standard xxxxxxxxx is xxx square root of the variance.

xxxxxx xxxxxxx

Population SD=2.19

10. There are five xxxxx associates at xxxxxxxxxx ford. xxx five representatives xxx xxx xxxxxx xx cars they sold last week xxxxxxxxxxx xxxx xxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxx x - xxxxxxxxx Stallter. xx xxxxxxx

- - - more text follows - - -


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Answer
Submitted by shahimermaid on Sat, 2012-06-16 10:34
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the answer is attached

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file1.docx preview (769 words)

xx x xxxxxxxxxx consists of the xxxxxxxxx five values: 2,2,4,4 xxx xxxxx List all samples of xxxx x and xxxxxxx the mean xx xxxx xxxxxxx

Samples xx size x are

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

mean

xxxxxxxxxx xxxx mean (4)

xxxxxx xx difference

xxx

x

-2

4

2,4

x

-1

1

xxx

x

xx

1

2,8

5

1

1

xxx

2

-2

4

xxx

x

-1

1

2,4

x

-1

1

xxx

5

x

x

4,2

x

-1

1

4,2

x

xx

1

4,4

x

x

0

4,8

6

2

4

4,2

3

-1

1

4,2

x

xx

x

xxx

x

x

x

xxx

6

2

x

xxx

x

1

x

8,2

5

x

x

xxx

6

x

x

8,4

x

x

x

B. xxxxxxxx the xxxx xx xxxxxxxxxxxx of sample means and the population xxxxx Compare xxx two xxxxxxx

Mean of distribution xx sample means is called expected value xx xx xx is xxxxx xx xxx table.

∑sample xxxxx xxxxx =4

xxxxxxxxxx mean xx xxxxxxxxxx xxxxx x

xx Compare the xxxxxxxxxx in xxx xxxxxxxxxx xxxx that of xxx sample xxxxxx

xxx range xx xxx xxxxxxxx measure xx dispersion

Sample mean xxxxx x 6-2= 4

xxxxxxxxxx range= 8-2= x

Variance is xxx xxxxxxx squared difference of scores xxxx the xxxx score xx x xxxxxxxxxxxxx 

xxxxxxxx xx sample 36/20=1.8

Variance xx population= ∑xxxxxxx +(2-4)^2+ (4-4)^2 xxx xxxxx + (8-4)^2= xxxxxxxxxx xxxxx x 4.8

Standard deviation xx xxx square xxxx of xxx xxxxxxxxx

Sample SD=1.34

Population SD=2.19

10. xxxxx are five sales xxxxxxxxxx xx xxxxxxxxxx ford. The xxxx representatives and xxx xxxxxx xx cars they sold last xxxx xxxxxxxxxxx xxxx xxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxx 8 - PS Connie xxxxxxxxx 6- xxxxxxx

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