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Margin of error and statistical testing

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



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?

Solution Preview

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 ...

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