Standard Error of The Mean

Standard Error of the Mean Definition

Standard deviation and standard error of the mean are important topics in statistical studies of various subjects such as finance, engineering, medicine, etc. The standard error of the mean is a method used to determine the standard deviation of a sampling distribution provided for a population. It measures the extent by which the sample mean or average differs from the true population mean. The standard error of the mean gives an accurate idea of the sample mean by analyzing the various samples from the sampling distribution. 

Standard Error of Mean 

Let's say for a particular instance, we pick a sample space from a population distribution and the estimated mean or sample mean is denoted by x. We pick another sample from the same population distribution and we find that the estimated mean or the sample mean may have a different value like y and not x. Hence, different samples from the same population distribution may end up giving us different variance and mean. Here the standard error of mean gives us the standard deviation of the sample means collected from all possible samples from the sampling distribution. The standard error of mean is sometimes abbreviated as SEM. 

Standard Error of the Mean Formula

How to Calculate Standard Error of Mean

Now let us learn how to calculate the standard error of the mean. 

Standard deviation represents the dispersion or variation in the set of values for a given data.

Standard deviation \[\sigma\] = \[\sqrt{\frac{\sum_{i=1}^{n}(x_{i} - \bar{x})^{2}}{(n-1)}}\] 

Variance = \[\sigma\]\[^{2}\]

The standard error of the mean  \[\sigma\]\[_{\bar{x}}\] = \[\frac{\sigma}{\sqrt{n}}\]

(where \[\bar{x}\] = the sample mean and n=the sample size)

Hence, the standard error of mean mathematically is the ratio of the standard deviation to the square root of the sample size.

The standard error of the mean (SEM)=Standard deviation \[\frac{SD}{\sqrt{samplesize(n)}}\]  

The steps required to find the standard deviation of a given data is-

• The first step involves taking the square root of the difference between the data entries and sample means and further adding them up.

• The second step involves calculating the variance by dividing the sum by one less than the sample size.

• The final step involves taking the square root of the variance to arrive at the standard deviation.

After the calculation of standard deviation, apply the formula \[\sigma\]\[_{\bar{x}}\] = \[\frac{\sigma}{\sqrt{n}}\].

From the formula, it is evident that the value of the standard error of mean decreases with an increase in the sample size (n). This can be easily explained because if we calculate the sample at each stage consisting of several data entries, a small variation in one of the values will not explicitly affect the mean. 

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Solved Examples

1. Calculate the Standard Error of the Mean for the Given Data: 5,10,15,20,25.

Given, x=5,10,15,20,25.

Sample size (n) =5

Mean (x\[_{m}\]) = Sum of individual elements/Sample size

Mean = (5+10+15+20+25)/5

Mean = 75/5

Mean (x\[_{m}\]) = 15

We know that,

Standard deviation \[\sigma\] = \[\sqrt{\frac{\sum_{i=1}^{n}(x_{i} - \bar{x})^{2}}{(n-1)}}\] 

The standard error of the mean \[\sigma\]\[_{\bar{x}}\] = \[\frac{\sigma}{\sqrt{n}}\]

\[\sigma\] = \[\sqrt{\frac{(x_{1} - x_{m})^{2} + (x_{2} - x_{m})^{2} + (x_{3} - x_{m})^{2} + (x_{4} - x_{m})^{2} + (x_{5} - x_{m})^{2}}{n-1}}\]

\[\sigma\] = \[\sqrt{\frac{(5-15)^{2} + (10 - 15)^{2} + (15 - 15)^{2} + (20 - 15)^{2} + (25 - 15)^{2} }{5-1}}\]

\[\sigma\]  =  \[\sqrt{\frac{100+25+0+25+100}{4}}\]

\[\sigma\]  =  \[\sqrt{\frac{250}{4}}\]

\[\sigma\] = 7.91 

Now standard error of the mean \[\sigma\]\[_{\bar{x}}\] = \[\frac{7.91}{\sqrt{5}}\] = \[\frac{7.91}{2.24}\] = 3.53 

Hence, the standard error of the mean for the given data is 3.53.

2. For a Given Sample: 20, 22, 24, 26, 28, Find Out the Standard Error of the Mean.

Given, x=20,22,24,26,28.

Sample size (n) =5

Mean(x\[_{m}\]) = \[\frac{\text{Sum of individual elements}}{\text{Sample Size}}\]

Mean(x\[_{m}\]) = \[\frac{20+22+24+26+28}{5}\]

Mean(x\[_{m}\]) = \[\frac{120}{5}\]

Mean(x\[_{m}\]) = 24

We know that,

Standard deviation \[\sigma\] = \[\sqrt{\frac{\sum_{i=1}^{n}(x_{i} - \bar{x})^{2}}{(n-1)}}\] 

The standard error of the mean \[\sigma\]\[_{\bar{x}}\] = \[\frac{\sigma}{\sqrt{n}}\] 

\[\sigma\] = \[\sqrt{\frac{(x_{1} - x_{m})^{2} + (x_{2} - x_{m})^{2} + (x_{3} - x_{m})^{2} + (x_{4} - x_{m})^{2} + (x_{5} - x_{m})^{2}}{n-1}}\]

\[\sigma\] = \[\sqrt{\frac{(20-24)^{2} + (22-24)^{2} + (24-24)^{2} + (26-24)^{2} + (28-24)^{2}}{5-1}}\]

\[\sigma\] = \[\sqrt{\frac{16+4+0+4+16}{4}}\]

\[\sigma\] = \[\sqrt{\frac{40}{4}}\]

\[\sigma\] = 3.16

Now standard error of the mean  \[\sigma\]\[_{\bar{x}}\] = \[\frac{3.16}{\sqrt{5}}\] = \[\frac{3.16}{2.24}\] = 1.41 

Hence, the standard error of the mean for the given data is 1.41.