Explore BrainMass

# Margin of error and statistical testing

Not what you're looking for? Search our solutions OR ask your own Custom question.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

Question: Do you have university education?

Yes=45% (n=450)
No=60% (n=800)

To find out whether 45% is significantly different from 60%, I use this:

=SQRT (45*(100-45)*1.96*1.96/450)

Then I use

=SQRT(+60*(100-60)*1.96*1.96/800)

Then I calculate the total margin of error 4.59+3.40=7.99

If the total margin of error is < than my % difference (60%-45%=15%), then I say that the difference between these two %s is significantly different. In this case, 7.99<15 therefore it is significant.

I need to understand this formula. Could you please explain it to me in layman terms. I don't need to understand the whole mathematical basis for 'why it works'. I just need to know in simple terms what I am doing.

Specific questions:

-Is there a name for this test? Or is it just comparing the margin of error at 95% confidence interval?

-Why do I multiply 1.96 twice?

© BrainMass Inc. brainmass.com September 28, 2022, 1:56 am ad1c9bdddf
https://brainmass.com/statistics/confidence-interval/margin-error-statistical-testing-38653

## SOLUTION This solution is FREE courtesy of BrainMass!

See attached file where formatting is conserved:

Whenever we estimate the population proportion from the sample proportion we get the estimate in terms of a confidence interval corresponding to the level of significance that we select.

Thus , suppose we get the sample proportion as
p= 45%
for a sample size of n= 450

Then the estimate of the population proportion will lie within 45% (+ or -) margin of error
The margin of error depends on the sample size n and the level of significance alpha (a)
Suppose we choose alpha (a) = 0.05 or 5%

margin of error = Z * sp
where z value comes from normal distribution table depending on alpha (a)
and sp=standard error of proportion=square root of (pq/n)=square root of (p(1-p)/n) as p+q=1 and therefore q=1-p
n is the sample size

Confidence limits for proportions when sample proportion = 45%
p= 45.00%
q=1-p= 55.00%
n=sample size= 450
sp=standard error of proportion=square root of (pq/n)= 2.35% =square root of ( 45.% * 55.% / 450)
Significance level=alpha (a) = 5%
No of tails= 2
This is a 2 tailed test because we are calculating upper and lower confidence level

Since sample size= 450 >= 30 use normal distribution
Z at the 0.05 level of significance 2 tailed test = 1.96

Margin of error= 4.596% =1.96*2.35%

SQRT (45*(100-45)*1.96*1.96/450)
=1.96 * SQRT (45% * (100-45%) / 450)
as square root of (1.96 x 1.96) = 1.96
(there is no need to multiply 1.96 twice , you can take it outside the square root sign once)
1.96 * SQRT (45% * (100-45%) / 450)
= 1.96 * SQRT (p%(100%-p%)/n)
= 1.96 * SQRT ((p%q%)/n)
= Z * sp , where Z is =1.96
and where sp =standard error of proportion=square root of (pq/n)

Upper confidence limit= p+z*sp= 49.596% =45.%+4.596%
Lower confidence limit= p-z*sp= 40.404% =45.%-4.596%

Thus if the sample proportion is 45.00%
then the population proportion will lie between 40.404% and 49.596%
at 5% significance level

Confidence limits for proportions when sample proportion = 60%
p= 60.00%
q=1-p= 40.00%
n=sample size= 800
sp=standard error of proportion=square root of (pq/n)= 1.73% =square root of ( 60.% * 40.% / 800)
Significance level=alpha (a) = 5%
No of tails= 2
This is a 2 tailed test because we are calculating upper and lower confidence level

Since sample size= 800 >= 30 use normal distribution
Z at the 0.05 level of significance 2 tailed test = 1.96

Margin of error= 3.395% =1.96*1.73%

SQRT (60*(100-60)*1.96*1.96/800)
=1.96 * SQRT (60% * (100-60%) / 450)
as square root of (1.96 x 1.96) = 1.96

1.96 * SQRT (60% * (100-60%) / 450)
= 1.96 * SQRT (p%(100%-p%)/n)
= 1.96 * SQRT ((p%q%)/n)
= Z * sp , where Z is =1.96
and where sp =standard error of proportion=square root of (pq/n)

Upper confidence limit= p+z*sp= 63.395% =60.%+3.395%
Lower confidence limit= p-z*sp= 56.605% =60.%-3.395%
Thus if the sample proportion is 60.00%
then the population proportion will lie between 56.605% and 63.395%

To check whether 45% is significantly different from 60%
we wish to know whether there is any overlap in the population confidence intervals for
the estimated population proportions

Upper limit for the population proportion when the sample proportion is 45% is
ie 45% + 4.596%

Lower limit for the population proportion when the sample proportion is 60% is
ie 60% - 3.395%

Since 56.605% is greater than 49.596% there is no overlap
There is a gap of 7.009% =56.61%-49.596%
Thus 45% is significantly different from 60%

This could also have been calcualted as such
difference between 60% and 45 % is= 15%

Because of sampling effects ( sample proportion cannot be exactly equal to population proportion)
there are errors of 4.596% and 3.395% in estimating the respective population proportions
The sum of these errors = 7.991% =4.6%-3.395%

The sum of these errors is not able to bridge the gap between 45% and 60%
as there is a gap of 7.009% =15.%-7.991%
Thus 45% is significantly different from 60%

-Is there a name for this test? Or is it just comparing the margin of error at 95% confidence interval?

Two tailed test for difference between proportions

-Why do I multiply 1.96 twice?
(there is no need to multiply 1.96 twice , you can take it outside the square root sign once)

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

© BrainMass Inc. brainmass.com September 28, 2022, 1:56 am ad1c9bdddf>
https://brainmass.com/statistics/confidence-interval/margin-error-statistical-testing-38653