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Question on averages, margin of error

The problem statement is: "Five shops were randomly chosen as a sample in 3 regions A,B and C.
The monthly profits are given as follows:

Shops Profit in millions
----- A B C
1 7 3 5
2 6 9 6
3 2 4 4
4 5 3 9
5 4 4 2
Examine whether the average profit per shop is the same for three regions?"

The average of the samples are:
Average of Samples for Region A: 4.8
Average of Samples for Region A: 4.6
Average of Samples for Region A: 5.2

However, I think based on above, we cannot immediately conclude that the average profits of all 3 regions are same or different, as I vaguely remember that average of samples will have an error of sigma/(sqrt(n), where 'n' is the sample size.

So I am stumped as to how to proceed further to solve this problem.

Kindly help.

Comments for
Question on averages, margin of error

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Feb 06, 2011
 Compare Averages Using Margin of Error
by: Staff

------------------------------------------------------

Part II

6. Standard Deviation = Square root of result in step 5

Region A:
Standard Deviation = Sqrt(3.7) = 1.923538


Region B:
Standard Deviation = Sqrt(6.3) = 2.50998


Region C:
Standard Deviation = Sqrt(6.7) = 2.588436



The next step is to CHOOSE a “Z” VALUE and compute the margin of error for each region.

Your question did not state what the confidence level you wish to use. Therefore, I am going to arbitrarily choose a confidence level of 90%.


Confidence level, z* value
(ref: http://people.richland.edu/james/lecture/m170/ch08-int.html)

90% 1.645
95% 1.96
98% 2.33
99% 2.58

*The Z value for a 90% confidence limit = 1.645


Compute the Margin of Error for each region:

Margin of Error formula = (Z value)*(Standard Deviation)/( square root of sample size)

Region A:
Z (90% confidence level) = 1.645
σ (standard deviation) = 1.923538
N (sample size) = 5


Margin of Error, Region A = (1.645)*( 1.923538)/[sqrt(5)]

Margin of Error, Region A = 1.415 samples

Region B:
Z (90% confidence level) = 1.645
σ (standard deviation) = 2.50998
N (sample size) = 5


Margin of Error, Region B = (1.645)*( 2.50998)/[sqrt(5)]

Margin of Error, Region B = 1.846508 samples



Region C:

Z (90% confidence level) = 1.645
σ (standard deviation) = 2.588436
N (sample size) = 5


Margin of Error, Region C = (1.645)*( 2.588436)/[sqrt(5)]

Margin of Error, Region C = 1.904225 samples




Compute the Range of Possible Values for each region:

Region A:

Actual Average, Region A = calculated average ± margin of error

Actual Average, Region A = 4.8 ± 1.415083


Actual Average, Region A falls between the values of 3.38 and 6.22




Region B:
Actual Average, Region B = calculated average ± margin of error

Actual Average, Region B = 4.6 ± 1.846508


Actual Average, Region B falls between the values of 2.75 and 6.45





Region C:
Actual Average, Region C = calculated average ± margin of error

Actual Average, Region C = 5.2 ± 1.904225



Actual Average, Region C falls between the values of 3.30 and 7.10





Compare the range of values for the three regions:

Region A: 3.38 to 6.22
Region B: 2.75 to 6.45
Region C: 3.30 to 7.10



The final answer: The ranges of values overlap for all three regions. Therefore, you can say with a 90% confidence level that the average profits for all three regions could be the same.




Thanks for writing.


Staff
www.solving-math-problems.com

Feb 06, 2011
 Compare Averages Using Margin of Error
by: Staff


The question:

The problem statement is: "Five shops were randomly chosen as a sample in 3 regions A,B and C.

The monthly profits are given as follows:

Shops Profit in millions
-----

. ABC
1 7 3 5
2 6 9 6
3 2 4 4
4 5 3 9
5 4 4 2

Examine whether the average profit per shop is the same for three regions?"

The average of the samples are:

Average of Samples for Region A: 4.8
Average of Samples for Region B: 4.6
Average of Samples for Region C: 5.2

However, I think based on above, we cannot immediately conclude that the average profits of all 3 regions are same or different, as I vaguely remember that average of samples will have an error of sigma/(sqrt(n), where 'n' is the sample size.

So I am stumped as to how to proceed further to solve this problem.

Kindly help.



The answer:

Part I


An excellent question.

You are correct.

Actual Average = calculated average ± margin of error




Margin of Error formula = (Z value)*(Standard Deviation)/( square root of sample size)

In order to continue, we need to calculate the standard deviation.

The problem doesn’t state whether the standard deviation is the same for all three regions, so I will compute a separate value for each region.



Calculate the Standard Deviation for each region:

1. Calculate the sample average
You have already done this:

Sample average for Region A: 4.8
Sample average for Region B: 4.6
Sample average for Region C: 5.2

2. Compute each deviation (for each sample) by subtracting the mean from each number in the sample

Region A:
7 - 4.8 = 2.2

6 - 4.8 = 1.2

2 - 4.8 = -2.8

5 - 4.8 = .2

4 - 4.8 = -.8

Region B:

3 - 4.6 = -1.6

9 - 4.6 = 4.4

4 - 4.6 = -.6

3 - 4.6 = -1.6

4 - 4.6 = -.6

Region C:

5 – 5.2 = -.2

6 – 5.2 = .8

4 – 5.2 = -1.2

9 – 5.2 = 3.8

2 – 5.2 = -3.2

3. Square each deviation
Region A:

(2.2)^2 = 4.84

(1.2)^2 = 1.44

(-2.8)^2 = 7.84

(.2)^2 = .04

(-.8)^2 = .64

Region B:

(-1.6)^2 = 2.56

(4.4)^2 = 19.36

(-.6)^2 = .36

(-1.6)^2 = 2.56

(-.6)^2 = .36

Region C:

(-.2)^2 = .04

(.8)^2 = .64

(-1.2)^2 = 1.44

(3.8)^2 = 14.44

(-3.2)^2 = 10.24


4. Add the squares
Region A:

4.84
1.44
7.84
0.04
0.64
------

14.8

Region B:

2.56
19.36
0.36
2.56
0.36
-------

25.2

Region C:

0.04
0.64
1.44
14.44
10.24
-------

26.8


5. Divide each total by the number of samples minus 1 (There are 5 samples for each region. Therefore, divide each total by 4 since 5 – 1 = 4.)

Region A:

14.8 ÷ 4 = 3.7

Region B:

25.2 ÷ 4 = 6.3

Region C:

26.8 ÷ 4 = 6.7

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