# Thread: Margin of Error - Confidence Level

1. ## Margin of Error - Confidence Level

I sent out a questionnaire to hundreds of people and recieved 154 responses.
Some of the questions have a yes/no answer. So for one of them I received 69 yes and 85 no.
What I want to add to my answers/graph is a margin of error for those that misinterpreted the question.
This is where I am getting stumped. In excel, for confidence level, it asks for a standard deviation. I don't know how to calculate a standard deviation on yes/no responses.

help/suggestions???

2. Originally Posted by psrs0810
I sent out a questionnaire to hundreds of people and recieved 154 responses.
Some of the questions have a yes/no answer. So for one of them I received 69 yes and 85 no.
What I want to add to my answers/graph is a margin of error for those that misinterpreted the question.
This is where I am getting stumped. In excel, for confidence level, it asks for a standard deviation. I don't know how to calculate a standard deviation on yes/no responses.

help/suggestions???
The distribution is Bernoulli (i.e. yes = 1 or no = 0).

Thus, the Standard Deviation is Sqrt[p*q] = Sqrt[(69/154)*(85/154)].

3. Dragon - thanks for the quick response!

would you check me?
so if the standard deviation is: = Sqrt[154*(69/154)*(85/154)]
stdev= 6.171257

and the confidence level = (.05, 6.171257, 154)
confidence = .9747

4. Originally Posted by psrs0810
Dragon - thanks for the quick response!

would you check me?
so if the standard deviation is: = Sqrt[154*(69/154)*(85/154)]
stdev= 6.171257

and the confidence level = (.05, 6.171257, 154)
confidence = .9747

I adjusted my post ( I thought too quickly ).

Leave the 154 out like I did above.

5. I now have a confidence level of .0785

does that sound right?

6. Originally Posted by psrs0810
I now have a confidence level of .0785

does that sound right?

Yes, it looks good i.e. the standard error X 1.96.

I applied the stdev of Sqrt[(A1/A3)*(A2/A3)] to all of my Yes/No questions
A1 = Total Yes
A2 = Total No
A3 = Total of Yes and No

My concern is that my standard deviation is so low that my confidence level (Margin of error) is below 1. Therefore when I apply my +- (error bar) to my Total Yes and Total No there is no fluctuation. It's basically saying that the data is very confident and no potential adjustments.

I tried ajusting my Alpha in the confidence level and the end result is minute.

HELP!!!

8. You need to know which parameter you are interested in to construct the
confidence interval. For example, if you want to construct for the
proportion of Yes, then the parameter of interest is
and the approximate confidence interval is
, where ,
i.e. is the number of observed Yes.

However, if you want to construct for the number of expected Yes,
then the parameter of interest is
The point estimate is
and thus the estimated variance is
So the approximate confidence interval is

9. Originally Posted by psrs0810
...What I want to add to my answers/graph is a margin of error for those that misinterpreted the question....
The question I have -- going back to the original post -- is how do you know who the people are (and how many) that misinterpreted the question??

These people are from other organizations that are in the same position that I am in. I am trying to get a feel for how they are handling their processes. That is how this questionnaire derived.

The total number of responders was 154.
As for the number of who misinterpreted the question(s), I am not sure. but I was going with a high confidence level for most of the questions.

I was trying to structure the confidence level around how polling results work, if that makes sense.

You guys have been a big help thus far - thank you.

Oh, BGM - I REALLY appreciate the statistical formulas, but I need to see the numbers working to understand the formula. I am a visual person, so I need to see the numbers being applied.

Thanks

11. If I apply 1.96 * SQRT((p * (1-p))/N)
would this be accurate?
P=69 (number that answered yes) would that be correct?
N=154 (total number who returned the survery)

(69*(-68))/154
(-4692)/154
-30.4673
SQRT (30.4673)
5.51974
*1.96
Margin of Error = 10.81869

12. Originally Posted by psrs0810
If I apply 1.96 * SQRT((p * (1-p))/N)
would this be accurate?
P=69 (number that answered yes) would that be correct?
N=154 (total number who returned the survery)

(69*(-68))/154
(-4692)/154
-30.4673
SQRT (30.4673)
5.51974
*1.96
Margin of Error = 10.81869
No. is the proportion of the sample that answered yes (ie. 69/154)

13. OK - if I correct P to be 69/154 or .44805

I am getting a margin of error of ±7.85%

(1-.44805) = .551948

.44805*.551948 = .247301

.247301/154 = .001606

Sqrt .001606 = .040073

.040073 * 1.96 = .078543 or ±7.85%
???

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