Standard deviation for proportions

[Zues]

Hi,,,can u please explain me how to calculate standard deviation and standard error for a binomial distribution when you have several samples?

For exapmple:
I don't know the population size. I take a sample of 10 and check for a particular characteristic. Let's say number of successes for this sample is x. So the proportion of successes is x/n. Then I repeat this process 3 times. That means I take 3 samples. Then I'll calculate the mean of the x/n for these 3 samples. So how do I calculate standard deviation or standard error for this mean value?

Eg: Sample 1 => x/n = x/10 =3/10 =30%

When this is done to all three samples,

Sample 1 => 30% +- a
Sample 2 => 32% +-b
Sample 3 => 32% +- c
Mean = 31.33% +-d

How do I calculate a,b,c and d? And what if I have different sample sizes for the three occasions? (having 10, 15, 8 instead of 10,10,10).

Thank you very much for your help

 

[I like Serena]

Hi,,,can u please explain me how to calculate standard deviation and standard error for a binomial distribution when you have several samples?

For exapmple:
I don't know the population size. I take a sample of 10 and check for a particular characteristic. Let's say number of successes for this sample is x. So the proportion of successes is x/n. Then I repeat this process 3 times. That means I take 3 samples. Then I'll calculate the mean of the x/n for these 3 samples. So how do I calculate standard deviation or standard error for this mean value?

Eg: Sample 1 => x/n = x/10 =3/10 =30%

When this is done to all three samples,

Sample 1 => 30% +- a
Sample 2 => 32% +-b
Sample 3 => 32% +- c
Mean = 31.33% +-d

How do I calculate a,b,c and d? And what if I have different sample sizes for the three occasions? (having 10, 15, 8 instead of 10,10,10).

Thank you very much for your help

Hi Zues! Welcome to MHB! :)

The standard deviation of the proportion of a binomial distribution is estimated with:
$$\hat \sigma = \sqrt{\frac{\hat p (1 - \hat p)}{n}}$$
where $\hat p$ is the estimated proportion.

When combining N measurements with different standard deviations, you'll need a weighted average.
The weights are:
$$w_i = \frac{1}{\hat\sigma_i^2}$$
The weighted mean is then:
$$\bar{p} = \frac{ \displaystyle\sum_{i=1}^N \hat p_i w_i}{\displaystyle\sum_{i=1}^N w_i}$$
And the standard error $\sigma_{\bar{p}}$ of the weighted mean is:
$$\sigma_{\bar{p}} = \sqrt{\frac{ 1 }{\sum_{i=1}^N w_i}}$$

 

[Zues]

Hi Zues! Welcome to MHB! :)

The standard deviation of the proportion of a binomial distribution is estimated with:
$$\hat \sigma = \sqrt{\frac{\hat p (1 - \hat p)}{n}}$$
where $\hat p$ is the estimated proportion.

When combining N measurements with different standard deviations, you'll need a weighted average.
The weights are:
$$w_i = \frac{1}{\hat\sigma_i^2}$$
The weighted mean is then:
$$\bar{p} = \frac{ \displaystyle\sum_{i=1}^N \hat p_i w_i}{\displaystyle\sum_{i=1}^N w_i}$$
And the standard error $\sigma_{\bar{p}}$ of the weighted mean is:
$$\sigma_{\bar{p}} = \sqrt{\frac{ 1 }{\sum_{i=1}^N w_i}}$$

Thank you very very much Serena. I spent a whole day trying to find this. Thank you very much (Smile)

 

[Zues]

Hi,, I have another question regarding this. I would like to report my results as percentages. Then how should I report the standard errors and standard deviations? I'm asking this because we use the proportion (instead of the percentage value) to calculate the SD an SE

Thank you very much and I'm so sorry for bothering. Thank you

 

[I like Serena]

Hi,, I have another question regarding this. I would like to report my results as percentages. Then how should I report the standard errors and standard deviations? I'm asking this because we use the proportion (instead of the percentage value) to calculate the SD an SE

Thank you very much and I'm so sorry for bothering. Thank you

A proportion and a percentage represent the same thing.
The only difference is a factor of a 100.

So, suppose you have a weighted mean of $\bar p = 0.31$ and an estimated standard error of $\hat \sigma_{\bar p}=0.12$, then you might also say that $\bar p = 31\%$ and $\hat \sigma_{\bar p}=12\%$.
Or for short:
$$\bar p = 31 \pm 12 \%$$

 

[Zues]

Thank you very much. This means a lot. You are so kind, Thank you(Smile)