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I have the following issue: I have calculated a mean value of customer engagement over a sample of 30 individuals. We present them with material to consume and calculate the engagement as (number of videos/texts clicked on)/(number of videos/texts sent). My values are a mean of 88.9% of all content has been consumed, so on average, 88.9% of the content sent to the customer (videos, texts etc.) have been clicked on, but my sdev is +/- 28.9%. Now I am wondering

a) if I have an error in my calculations since in my head that does not make that much sense, you can't consume more than 100% of the content provided, and

b) how would I show this as a graph? Again, in my head it does not make sense to show error bars that go beyond 100%?

Thank you all for your guidance!

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    $\begingroup$ It would be useful to give a few more details about what your data looks like. For example, it is surprising to me that the standard deviation is so high. Specifically, even a single draw from a binomial distribution with $p = 0.889$ has a standard deviation of $0.31$, which is only a bit larger than the $0.289$ you report. $\endgroup$
    – stats_model
    Commented Apr 6, 2021 at 15:27
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    $\begingroup$ What is your sample size $n?$ You say your 'standard deviation` is $0.289,$ How did you obtain that? Maybe your're mixing up standard deviation, standard error, and margin of error. @stats_model (+1) asks for more details. // If you're using a Wald 95% CI it is possible to get an interval with endpoints outside $(0,1),$ But my guess is you're doing something wrong, and details would help us make sense of this. $\endgroup$
    – BruceET
    Commented Apr 6, 2021 at 19:27
  • $\begingroup$ Why not use my Answer, to the extent it may be helpful, to provide more detail in your Question--and possibly suggest a solution. Then maybe your question can be re-opened. $\endgroup$
    – BruceET
    Commented Apr 7, 2021 at 2:44
  • $\begingroup$ Hi @BruceET, I have added the details that hopefully clear that issue up and qualify the question for re-opening. I have not used your answer since I am located in an European country and I was, quite frankly, asleep when you posted it. $\endgroup$
    – P.Weyh
    Commented Apr 7, 2021 at 6:56

1 Answer 1

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Example: Suppose you have $n = 1000$ randomly chosen subjects of whom $x = 889$ are 'engaged.' Then the estimated proportion engaged in the population is $p = x/n = 0.889.$

Wald confidence intervals are intended for use with such large samples. A 95% Wald CI is of the form $\hat p \pm 1.96\sqrt{\frac{\hat p(1-\hat p)}{n}}.$ Using R as a calculator, we get the 95% CI $(0.870, 0.908),$ rounded to three places.

n = 1000;  x = 889
p.est = x/n;  p.est
[1] 0.889
se = sqrt(p.est*(1-p.est)/n);  se
[1] 0.00993373
ci = p.est + qnorm(c(.025,.975))*se;  ci
[1] 0.8695302 0.9084698

If this does not answer your question, please say what part of it seems different from what is being done in your course.

Note: There are many styles of CIs for a binomial proportion (google confidence interval binomial proportion; perhaps see the Wikipedia article, in particular). One additional style, called the Jeffries CI, relies on a Bayesian derivation, but has very good properties as a frequentist CI. For the data above the 95% Jeffreys CI is $(0.868, 0.907),$ easily computed in R as shown below. This style of confidence interval never gives endpoints outside of $(0,1).$

n = 1000;  x = 889
qbeta(c(.025,.975), x+.5, n-x+.5)
[1] 0.8684107 0.9073418
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  • $\begingroup$ Hi @BruceET, first of all, thank you for the answer. It is, however, not something that I am doing in a course but at work. We have 30 customers that have, on average, consumed 89% of the material that we have provided them with, so 11% of the materials have been ignored or not clicked on. From that average, there has been a 29% sdev. Hope that clears it up $\endgroup$
    – P.Weyh
    Commented Apr 8, 2021 at 7:00
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    $\begingroup$ @P.Weyh Neither the standard deviation of a bunch of numbers, nor the standard error of the mean, impose any kind of restriction on what values can realistically be included in a confidence interval (or a prediction interval for that matter). Therefore, you can't state that 29 percentage points is in any way unrealistic. Just forget about the number 29 and try this alternative approach. The answer by BruceET gives you a better way to construct an interval based on your type of data. The second part even shows an alternative that restricts the interval to $(0, 1)$, ensuring realistic values. $\endgroup$
    – Frans Rodenburg
    Commented Apr 12, 2021 at 12:59
  • $\begingroup$ @FransRodenburg thank you for your comment. I am not saying that they are, I would like to know if they are, and am hoping to get an idea on how to interpret a sdev of ">100%" and how to go about it if I have a sample size of 30, which I added after edit. And I unfortunately do not know which interval would work for me, since I am not well versed in statistics. Therefore what I was hoping to get was an insight in how I can find out which interval to use, and what data I need to provide for someone to tell me which interval to use. Thank you for your help. $\endgroup$
    – P.Weyh
    Commented Apr 13, 2021 at 14:05

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