For the following situation, find the mean and standard deviation of the population. List all samples (with replacement) of the given size from that population and find themean of each. Find the mean and standard deviation of the sampling distribution and compare them with the mean and standard deviation of the population.The word counts of five essays are 504, 629, 555, 607, and 580. Use a sample size of 2.The mean of the population is(Round to two decimal places as needed.)The standard deviation of the population is(Round to two decimal places as needed.)Identify all samples of size 2 with the correct accompanying means below.A.580, 555, x = 567.5B. 504, 555, x 529.5C.555, 629, x = 592D.555, 580, x = 567.5E.555, 607, x = 581580, 623, x = 601.5F.G.607, 555, x 581H. 580, 629, x = 604.5stI.580, 580, x 580J.629, 580, x = 604.5K.504, 607, x = 555.5L. 607, 504, x= 555.5ilestioM. 580, 607, x - 593.5N. 629, 555, x = 592O. 607, 607, x = 607P.g Qu629, 629, x= 629Q. 504, 629, x = 566.5R. 504, 580, x = 542e-S. 607, 580, x = 593.5629, 607, x= 618T.U. 555, 504, x = 529.5ered V. 607, 629, x 618W. 555, 555, x 555Z. 629, 504, x 566.5X. 504, 504, x= 504580, 504, x = 542Y.607, 629, x 551ted QThe mean of the sampling distribution is(Round to two decimal nlaces as needed)Click to select your answer(s).eaway between two entries, use the z-score nairway between the corresponding z-scores.NOVTwX P3tv This Question: 1 pt7 of 14 (8 complete)This Quiz: 14 pts possibleFor the following situation, find the mean and standard deviation of the population. List all samples (with replacement) of the given size from that population and find themean of each. Find the mean and standard deviation of the sampling distribution and compare them with the mean and standard deviation of the population.The word counts of five essays are 504, 629, 555, 607, and 580. Use a sample size of 2.M. 580, 607, X = 593.5N.629, 555, x = 592O. 607, 607, x= 607P.629, 629, x = 629Q. 504, 629, x = 566.5R. 504, 580, x = 542S. 607, 580, x = 593.5T.629, 607, x = 618U. 555, 504, x = 529.5tio607, 629, x = 618W. 555, 555, x = 555V.X. 504, 504, x 504Y. 580, 504, x 542607, 629, x 551Z.629, 504, x = 566.5eft fonThe mean of the sampling distribution is(Round to two decimal places as needed.)tThe standard deviation of the sampling distribution isile(Round to two decimal places as needed.)tioChoose the correct comparison of the population and sampling distribution below.QO A. The means are not equal and the standard deviation of the sampling distribution is larger.O B. The means are equal but the standard deviation of the sampling distribution is smaller.edO C. The means are equal but the standard deviation of the sampling distribution is larger.O D. The means are not equal and the standard deviation of the sampling distribution is smaller.ed QO E. The means and standard deviations are equal.?eaClick to select your answer(s).Priveway between two entries, use the z-score nalrway between the corresponding z-scores.WXPNOVtv

Question
Asked Nov 4, 2019
For the following situation, find the mean and standard deviation of the population. List all samples (with replacement) of the given size from that population and find the
mean of each. Find the mean and standard deviation of the sampling distribution and compare them with the mean and standard deviation of the population.
The word counts of five essays are 504, 629, 555, 607, and 580. Use a sample size of 2.
The mean of the population is
(Round to two decimal places as needed.)
The standard deviation of the population is
(Round to two decimal places as needed.)
Identify all samples of size 2 with the correct accompanying means below.
A.
580, 555, x = 567.5
B. 504, 555, x 529.5
C.
555, 629, x = 592
D.
555, 580, x = 567.5
E.
555, 607, x = 581
580, 623, x = 601.5
F.
G.
607, 555, x 581
H. 580, 629, x = 604.5
st
I.
580, 580, x 580
J.
629, 580, x = 604.5
K.
504, 607, x = 555.5
L. 607, 504, x= 555.5
ile
stio
M. 580, 607, x - 593.5
N. 629, 555, x = 592
O. 607, 607, x = 607
P.
g Qu
629, 629, x= 629
Q. 504, 629, x = 566.5
R. 504, 580, x = 542
e-
S. 607, 580, x = 593.5
629, 607, x= 618
T.
U. 555, 504, x = 529.5
ered V. 607, 629, x 618
W. 555, 555, x 555
Z. 629, 504, x 566.5
X. 504, 504, x= 504
580, 504, x = 542
Y.
607, 629, x 551
ted Q
The mean of the sampling distribution is
(Round to two decimal nlaces as needed)
Click to select your answer(s).
ea
way between two entries, use the z-score nairway between the corresponding z-scores.
NOV
T
wX P
3
tv
help_outline

Image Transcriptionclose

For the following situation, find the mean and standard deviation of the population. List all samples (with replacement) of the given size from that population and find the mean of each. Find the mean and standard deviation of the sampling distribution and compare them with the mean and standard deviation of the population. The word counts of five essays are 504, 629, 555, 607, and 580. Use a sample size of 2. The mean of the population is (Round to two decimal places as needed.) The standard deviation of the population is (Round to two decimal places as needed.) Identify all samples of size 2 with the correct accompanying means below. A. 580, 555, x = 567.5 B. 504, 555, x 529.5 C. 555, 629, x = 592 D. 555, 580, x = 567.5 E. 555, 607, x = 581 580, 623, x = 601.5 F. G. 607, 555, x 581 H. 580, 629, x = 604.5 st I. 580, 580, x 580 J. 629, 580, x = 604.5 K. 504, 607, x = 555.5 L. 607, 504, x= 555.5 ile stio M. 580, 607, x - 593.5 N. 629, 555, x = 592 O. 607, 607, x = 607 P. g Qu 629, 629, x= 629 Q. 504, 629, x = 566.5 R. 504, 580, x = 542 e- S. 607, 580, x = 593.5 629, 607, x= 618 T. U. 555, 504, x = 529.5 ered V. 607, 629, x 618 W. 555, 555, x 555 Z. 629, 504, x 566.5 X. 504, 504, x= 504 580, 504, x = 542 Y. 607, 629, x 551 ted Q The mean of the sampling distribution is (Round to two decimal nlaces as needed) Click to select your answer(s). ea way between two entries, use the z-score nairway between the corresponding z-scores. NOV T wX P 3 tv

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This Question: 1 pt
7 of 14 (8 complete)
This Quiz: 14 pts possible
For the following situation, find the mean and standard deviation of the population. List all samples (with replacement) of the given size from that population and find the
mean of each. Find the mean and standard deviation of the sampling distribution and compare them with the mean and standard deviation of the population.
The word counts of five essays are 504, 629, 555, 607, and 580. Use a sample size of 2.
M. 580, 607, X = 593.5
N.
629, 555, x = 592
O. 607, 607, x= 607
P.
629, 629, x = 629
Q. 504, 629, x = 566.5
R. 504, 580, x = 542
S. 607, 580, x = 593.5
T.
629, 607, x = 618
U. 555, 504, x = 529.5
tio
607, 629, x = 618
W. 555, 555, x = 555
V.
X. 504, 504, x 504
Y. 580, 504, x 542
607, 629, x 551
Z.
629, 504, x = 566.5
eft f
on
The mean of the sampling distribution is
(Round to two decimal places as needed.)
tThe standard deviation of the sampling distribution is
ile
(Round to two decimal places as needed.)
tio
Choose the correct comparison of the population and sampling distribution below.
Q
O A. The means are not equal and the standard deviation of the sampling distribution is larger.
O B. The means are equal but the standard deviation of the sampling distribution is smaller.
ed
O C. The means are equal but the standard deviation of the sampling distribution is larger.
O D. The means are not equal and the standard deviation of the sampling distribution is smaller.
ed Q
O E. The means and standard deviations are equal.
?
ea
Click to select your answer(s).
Prive
way between two entries, use the z-score nalrway between the corresponding z-scores.
WXP
NOV
tv
help_outline

Image Transcriptionclose

This Question: 1 pt 7 of 14 (8 complete) This Quiz: 14 pts possible For the following situation, find the mean and standard deviation of the population. List all samples (with replacement) of the given size from that population and find the mean of each. Find the mean and standard deviation of the sampling distribution and compare them with the mean and standard deviation of the population. The word counts of five essays are 504, 629, 555, 607, and 580. Use a sample size of 2. M. 580, 607, X = 593.5 N. 629, 555, x = 592 O. 607, 607, x= 607 P. 629, 629, x = 629 Q. 504, 629, x = 566.5 R. 504, 580, x = 542 S. 607, 580, x = 593.5 T. 629, 607, x = 618 U. 555, 504, x = 529.5 tio 607, 629, x = 618 W. 555, 555, x = 555 V. X. 504, 504, x 504 Y. 580, 504, x 542 607, 629, x 551 Z. 629, 504, x = 566.5 eft f on The mean of the sampling distribution is (Round to two decimal places as needed.) tThe standard deviation of the sampling distribution is ile (Round to two decimal places as needed.) tio Choose the correct comparison of the population and sampling distribution below. Q O A. The means are not equal and the standard deviation of the sampling distribution is larger. O B. The means are equal but the standard deviation of the sampling distribution is smaller. ed O C. The means are equal but the standard deviation of the sampling distribution is larger. O D. The means are not equal and the standard deviation of the sampling distribution is smaller. ed Q O E. The means and standard deviations are equal. ? ea Click to select your answer(s). Prive way between two entries, use the z-score nalrway between the corresponding z-scores. WXP NOV tv

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check_circleExpert Solution
Step 1

Calculation of population mean and population standard deviation

Enter the given data in excel as shown below

A
1
504
2
629
3
555
607
5
580
Population mean (u) 575 (From Excel -AVERAGE (A1:A5))
=
Population standard Deviation (o) = 43.3728 (From Excel -STDEV.P(A1:A5))
Ln
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A 1 504 2 629 3 555 607 5 580 Population mean (u) 575 (From Excel -AVERAGE (A1:A5)) = Population standard Deviation (o) = 43.3728 (From Excel -STDEV.P(A1:A5)) Ln

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Step 2

Taking a samples of two and finding the mean of that

Enter the data in excel as shown below and the following for...

A
В
A
В
C
1 x1
x2
1 х1
X-bar
x2
504 2 504
2
504
-AVERAGE(A2:B2)
AVERAGE(A3:B3)
504
629 3 504
3
504
629
555 4 504
4
504
AVERAGE(A4:B4)
AVERAGE(A5:B5)
555
607 5 504
504
607
580 6 504
504
6
AVERAGE(A6:B6)
AVERAGE(A7:B7)
580
629 7 629
7
629
629
555 8 629
AVERAGE(A8:B8)
8
629
555
607 9 629
629
-AVERAGE(A9:B9)
607
580 10 629
10
629
AVERAGE(A10:B10)
580
555 11 555
555
11
AVERAGE(A11:B11)
555
607 12 555
12
555
AVERAGE(A12:B12)
607
580 13 555
-AVERAGE(A13:B13)|
13
555
580
607 14 607
14
607
AVERAGE(A14:B14)
607
580 15 607
15
607
AVERAGE(A15:B15)
-AVERAGE(A16:B16)
580
580 16 580
580
16
580
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Image Transcriptionclose

A В A В C 1 x1 x2 1 х1 X-bar x2 504 2 504 2 504 -AVERAGE(A2:B2) AVERAGE(A3:B3) 504 629 3 504 3 504 629 555 4 504 4 504 AVERAGE(A4:B4) AVERAGE(A5:B5) 555 607 5 504 504 607 580 6 504 504 6 AVERAGE(A6:B6) AVERAGE(A7:B7) 580 629 7 629 7 629 629 555 8 629 AVERAGE(A8:B8) 8 629 555 607 9 629 629 -AVERAGE(A9:B9) 607 580 10 629 10 629 AVERAGE(A10:B10) 580 555 11 555 555 11 AVERAGE(A11:B11) 555 607 12 555 12 555 AVERAGE(A12:B12) 607 580 13 555 -AVERAGE(A13:B13)| 13 555 580 607 14 607 14 607 AVERAGE(A14:B14) 607 580 15 607 15 607 AVERAGE(A15:B15) -AVERAGE(A16:B16) 580 580 16 580 580 16 580

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