Question
In: Statistics and Probability
The following data represent the number of days absent per year in a population of six...
The following data represent the number of days absent per year in a population of six employees in a small company: 1 3 6 7 9 10
a. Compute the population mean.
b. Compute the population standard deviation.
c. Assuming that you sample without replacement, select all possible samples of size n=2 and construct the sampling distribution of the mean.
d. Compute the mean of all the sample means. How does the mean of the sample means and the population mean compare?
e. Compute the standard deviation of all the sample means. How does the standard deviation of the sample means and the population standard deviation compare?
f. Repeat parts (c) – (e) for all possible samples of size n=3. How do the results compare with those of sample size n=2? Comment on what happened as the sample size increased (from 2 to 3).
Solutions
Expert Solution
Following table shows the calculations for population mean and population standard deviation:
| X | (X-mean)^2 | |
| 1 | 25 | |
| 3 | 9 | |
| 6 | 0 | |
| 7 | 1 | |
| 9 | 9 | |
| 10 | 16 | |
| Total | 36 | 60 |
(a)
The population mean is
(b)
The population standard deviation is
(c)
Sine samples are drawn without replacement so possible number of samples are C(6,2) = 15
Since all samples are equally likely so probability of getting each sample is 1/15 = 0.06667
Following table shows all the possible samples and sample means:
| Samples | Sample mean, xbar | P(xbar) | |
| 1 | 3 | 2 | 0.06667 |
| 1 | 6 | 3.5 | 0.06667 |
| 1 | 7 | 4 | 0.06667 |
| 1 | 9 | 5 | 0.06667 |
| 1 | 10 | 5.5 | 0.06667 |
| 3 | 6 | 4.5 | 0.06667 |
| 3 | 7 | 5 | 0.06667 |
| 3 | 9 | 6 | 0.06667 |
| 3 | 10 | 6.5 | 0.06667 |
| 6 | 7 | 6.5 | 0.06667 |
| 6 | 9 | 7.5 | 0.06667 |
| 6 | 10 | 8 | 0.06667 |
| 7 | 9 | 8 | 0.06667 |
| 7 | 10 | 8.5 | 0.06667 |
| 9 | 10 | 9.5 | 0.06667 |
Now we need to sum the probabilities of samples that have same sample mean. Following table shows the sampling distribution of sample mean:
| Sample mean, xbar | P(xbar) |
| 2 | 0.06667 |
| 3.5 | 0.06667 |
| 4 | 0.06667 |
| 4.5 | 0.06667 |
| 5 | 0.13334 |
| 5.5 | 0.06667 |
| 6 | 0.06667 |
| 6.5 | 0.13334 |
| 7.5 | 0.06667 |
| 8 | 0.13334 |
| 8.5 | 0.06667 |
| 9.5 | 0.06667 |
d)
Following table shows the calculation for mean of sampling distribution of sample means:
| Sample mean, xbar | P(xbar) | xbar*P(xbar) |
| 2 | 0.06667 | 0.13334 |
| 3.5 | 0.06667 | 0.233345 |
| 4 | 0.06667 | 0.26668 |
| 4.5 | 0.06667 | 0.300015 |
| 5 | 0.13334 | 0.6667 |
| 5.5 | 0.06667 | 0.366685 |
| 6 | 0.06667 | 0.40002 |
| 6.5 | 0.13334 | 0.86671 |
| 7.5 | 0.06667 | 0.500025 |
| 8 | 0.13334 | 1.06672 |
| 8.5 | 0.06667 | 0.566695 |
| 9.5 | 0.06667 | 0.633365 |
| Total | 6.0003 |
So,
The mean of the sample means and the population mean are equal.
orchestra answered 2 years agoRelated Solutions
The following data represent the number of days absent (x) and the final grade (y) for...
The following data represent the number of days absent (x) and
the final grade (y) for a sample of college students in a
statistics course at a large university.
Number of absences (x):
0
1
2
3
4
5
6
7
8 9
Final grade
(y):
89.2 86.4 83.5
81.1 78.2 73.9
64.3 71.8 65.5 66.2
Draw a scatter diagram on graph paper. The number of days
absent is the explanatory variable and the final grade is the
response variable....
The following data represent the number of days absent (x) and the final grade (y) for...
The following data represent the number of days absent (x) and
the final grade (y) for a sample of college students in a
statistics course at a large university.
Number of absences (x): 0 1 2 3 4 5 6 7 8 9
Final grade (y): 89.2 86.4 83.5 81.1 78.2 73.9 64.3 71.8 65.5
66.2
a) Draw a scatter diagram on graph paper. The number of days
absent is the explanatory variable and the final grade is the
response...
The data below represent the number of days absent, x, and the final grade, y, for...
The data below represent the number of days absent, x, and the
final grade, y, for a sample of college students at a large
university. Complete parts (a) through (e) below.
Number of absences, x
Final grade, y
0
88.5
1
85.6
2
82.5
3
79.9
4
76.9
5
72.4
6
62.6
7
67.1
8
64.1
9
61.1
(a) Find the least-squares regression line treating the number
of absences, x, as the explanatory variable and the final grade,
y, as...
The data below represent the number of days absent, x, and the final grade, y, for...
The data below represent the number of days absent, x, and the
final grade, y, for a sample of college students at a large
university. Complete parts (a) through (e) below.
No. of absences, x
00
11
22
33
44
55
66
77
88
99
Final grade, y
88.188.1
85.285.2
82.282.2
79.879.8
76.876.8
72.372.3
62.862.8
67.267.2
64.364.3
61.461.4
(a) Find the least-squares regression line treating the number
of absences, x, as the explanatory variable and the final grade,
y, as...
The data below represent the number of days absent, x, and the final grade, y, for...
The data below represent the number of days absent, x, and the
final grade, y, for a sample of college students at a large
university. Complete parts (a) through (e) below.
0
1
2
3
4
5
6
7
8
9
89.0
86.1
83.0
80.5
77.4
73.0
63.3
67.8
64.8
61.8
(a) Find the least-squares regression line treating the number
of absences, x, as the explanatory variable and the final grade,
y, as the response variable.
(b) Interpret the slope...
The accompanying data represent the number of days absent, x, and the final exam score, y,...
The accompanying
data represent the number of days absent, x, and the final exam
score, y, for a sample of college students in a general education
course at a large state university. Complete parts (a) through (e)
below.
Click
the icon to view the absence count and final exam score
data.
Absences and Final
Exam Scores
No. of
absences, x
0
1
2
3
4
5
6
7
8
9
Final
exam
score,
y
88.3
87.3
84.2
80.9
77.6
73.8...
7. The following data represent the number of workdays absent during the past year, y, and...
7. The following data represent the number of workdays absent
during the past year, y, and the number of years employed by the
company x, for seven employees randomly selected from a large
company. Assume data is normally distributed.
Y 2 0 5 6 4 9 2
X 7 8 2 3 5 3 7
The slope estimate (b1) was found to be = -1.09 That is:
b 1= nExy - (Ex) (EY)
---------------------------- = - 1.09
nEx2 - (EX)2...
Consider the number of days absent from a random sample of six students during a semester:...
Consider the number of days absent from a random sample of six
students during a semester: A= {2, 3, 2, 4, 2, 5}
Compute the arithmetic mean, geometric mean, median, and mode
by hand and verify the results using R
Arithmetic Mean:
X=i=1nXin=2+3+2+4+2+56=3
mean(data2$absent)
[1] 3
Geometic Mean:
GMx=Πi=1nX11n=2∙3∙2∙4∙2∙516=2.79816
>gmean <-
prod(data2$absent)^(1/length(data2$absent))
> gmean
[1] 2.798166
Median:
X=12n+1th,
Xi2,2,2,3,4,5,
n=6=126+1th
ranked value=3.5, value=2.5 days absent
>median(data2$absent)
[1] 2.5
Mode: Most frequent value=2
> mode <-
names(table(data2$absent))
[table(data2$absent)==max(table(data2$absent))]
> mode
[1]...
1. T The following data represents the number of days absent from school in one school...
1. T The following data represents the
number of days absent from school in one school year for a sample
of 40 students in Ms. Jinn’s fourth grade class.
0
1
2
2
2
4
5
5
7
7
7
7
8
8
8
8
8
8
10 10
12
12 12
13
14
...
Based on the following data for the current year, what is the number of days' sales...
Based on the following data for the current year, what is the
number of days' sales in receivables? Assume 365-Day year.
Sales on account during year
$570,068
Cost of goods sold during year
219,238
Accounts receivable, beginning of year
46,044
Accounts receivable, end of year
51,623
Inventory, beginning of year
91,672
Inventory, end of year
116,124
Round your answer up to the nearest whole day.
a.31
b.74
c.67
d.140
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