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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?
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]
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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)
xxxxxxxxxxxxxxxxxxxSample | 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
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6,10
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx | Values | Mean | Sample | xxxx sold | xxxx | Cars sold | No of Sales | xx of xxxxx | ||
x | x | x | 2 | 1 | x | 6 | 7 | 4 | x | x |
2 | x | 4 | x | 2 | x | 4 | x | x | x | 2 |
x | x | x | 3 | x | 8 | 10 | 9 | x | 2 | x |
x | x | 8 | 5 | x | x | 6 | x | 7 | 0 | 3 |
5 | x | x | x | x | x | x | 5 | 8 | 1 | x |
6 | x | x | 3 | x | 6 | xx | 8 | x | x | 1 |
7 | x | x | 5 | 7 | x | x | x | 10 | 1 | x |
x | x | 4 | x | x | 4 | xx | x | |||
x | 4 | x | x | 9 | x | x | x | |||
xx | 4 | 8 | x | xx | 10 | 6 | 8 |
xxxx
xx
xxxxxx
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxData | Units sold | Frequency | Sample | xxxxxx Values | xxxx | ||||
2 | x | x | 1 | x | x | x | x | 3 | 2.8 |
3 | 2 | x | x | 2 | x | 2 | x | 3 | 3.2 |
x | 3 | x | 3 | x | 4 | 3 | x | x | 3.4 |
3 | 4 | 3 | 4 | 4 | x | x | x | x | xxx |
x | 5 | 2 | 5 | x | 3 | x | 4 | 3 | 3.2 |
4 | x | 0 | |||||||
x | x | 1 | |||||||
x | |||||||||
3 | |||||||||
x | |||||||||
x | xxxxxx Mean | Frequency | |||||||
7 | xxx | x | |||||||
3 | 3.2 | x | |||||||
4 | 3.4 | 1 | |||||||
x | 3.8 | x | |||||||
x | |||||||||
3 | |||||||||
3 | |||||||||
x | |||||||||
x |
16,18
xxxx0 |
x |
x |
3 |
2 |
0 |
x |
x |
x |
1 |
x |
Question 16 | xxxxxxxx xx | ||||
x | x | xxx < z) | x | z | P(Z < xx |
74 | xxxxxxx | xxxxxx | xxx | -0.7906 | 0.2146 |
xx | 1.2649 | xxxxxx | xxx | 1.5811 | xxxxxx |
77 | xxxxxx | 0.9943 |
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Answer file is attached with all calculations and with easy printing format. Feel free to contact for any further assistance
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xx
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx | |||||||||||
xxxxxxxxxx size | 2 | x | x | , | x | x | x | x | x | ||
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 | C | n | x | x | x | 2 | xxxx xxxxxxx xxxxxxxxxxx | |||
x | 10 | ||||||||||
xxxxxx xxx | xxx Possible xxxxxxx | xxxxxx mean xxx | |||||||||
1 | x | , | 2 | x | |||||||
2 | x | , | x | 3 | |||||||
x | 2 | , | x | 3 | |||||||
x | x | , | x | x | |||||||
5 | 2 | , | 4 | 3 | |||||||
6 | x | , | x | x | |||||||
Sample No. | x | x | , | 8 | x | ||||||
x | 4 | x | x | x | |||||||
9 | 4 | , | x | x | |||||||
xx | x | , | x | x | |||||||
Total | 40 | ||||||||||
xxxxxx xxxxxxxxxxxx xx x is | x | x | xx | ||||||||
2 | 1 | 2 | |||||||||
x | 4 | 12 | |||||||||
x | x | 4 | |||||||||
5 | 2 | xx | |||||||||
6 | x | xx | |||||||||
Total | xx | 40 | |||||||||
Sample xxxx x | 4 | ||||||||||
Population xxxx = x | x | x | 2 | 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
xxxxxxxxxxxxxxxxxxxxxxxxxxRequired: | |||||||||
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 xxxxxxx | x | |||||||
x | Connie Stallter | x | |||||||
x | Juan Lopez | x | |||||||
x | Ted Barnes | xx | |||||||
5 | Peggy Chu | x | |||||||
Total | xx | ||||||||
Population size | x | , | 6 | x | 4 | , | xx | , | 6 |
x am xxxxxxxxx a xxx xxxxxx xxxxxx from the list xx 5 persons | |||||||||
No. xx xxxxxxxx sample | x | C | n | = | x | x | x | xxxx xxxxxxx replacement) | |
x | xx | ||||||||
Sample No. | xxx Possible samples | Sample xxxx |
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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
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx | 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
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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
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxmean | 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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