In order to prove that a process has been improved, you must measure the process capability before and after improvements are implemented. This allows you to quantify the process improvement (e.g., defect reduction or productivity increase) and translate the effects into an estimated financial result – something business leaders can understand and appreciate. If data is not readily available for the process, how many members of the population should be selected to ensure that the population is properly represented? If data has been collected, how do you determine if you have enough data?
Determining sample size is a very important issue because samples that are too large may waste time, resources and money, while samples that are too small may lead to inaccurate results. In many cases, we can easily determine the minimum sample size needed to estimate a process parameter, such as the population mean .
When sample data is collected and the sample mean is calculated, that sample mean is typically different from the population mean . This difference between the sample and population means can be thought of as an error. The margin of error is the maximum difference between the observed sample mean and the true value of the population mean :
is known as the critical value, the positive value that is at the vertical boundary for the area of in the right tail of the standard normal distribution.
is the population standard deviation.
is the sample size.
Let’s put all this statistical mumbojumbo to work. Take for example that we would like to start an Internet service provider (ISP) and need to estimate the average Internet usage of households in one week for our business plan and model.
Problem
We would like to start an ISP and need to estimate the average Internet usage of households in one week for our business plan and model. How many households must we randomly select to be 95 percent sure that the sample mean is within 1 minute of the population mean . Assume that a previous survey of household usage has shown = 6.95 minutes.
Solution
We are solving for the sample size .
A 95% degree confidence corresponds to = 0.05. Each of the shaded tails in the following figure has an area of = 0.025. The region to the left of and to the right of = 0 is 0.5 – 0.025, or 0.475. In the table of the standard normal () distribution, an area of 0.475 corresponds to a value of 1.96. The critical value is therefore = 1.96.


© Copyright iSixSigma 20002016. User Agreement. Any reproduction or other use of content without the express written consent of iSixSigma is prohibited. More »
Comments
Excellent example using the startup of an Internet Service Provider (ISP)! I have moved the module of “How to Determine Sample Size” into my Six Sigma Green Belt for Service Course and have been looking for some varied well written examples. Your example fits the bill. I am going to point my students towards this article as a resource.
I look forward to reading more articles.
This is quite easy to understand. But can this formular be used for a twotailed hypothesis as well?
This is an example of a 2tailed test. Za/2. With a stated mean xbar +/ (Za/2)(sigma/sqrt n) or (s/sqrt of n1), where sigma is not known, as s is a biased estimator of sigma. Thus, the confidence interval would be xbar +/ the sampling error or (Za/2)(sigma/sqrt n).
This is excellent thank you! Would the calculation for the onetailed test be the same just with a different zscore? So, instead of using the Z score of 1.96, the Z score 1.64 should be used?
I am electrical engineer involved in testing of relays and ehv equipments. During last 6 months some where i came across the word ‘Confidance Interval’. I have tried a lot by searching the web to get in undestood. finaly today came across this web page and got the idea of confidance interval. Also this also may relate to Cetral Limit Therom. Thanks again.
Good enough. But what happens when the population is 100 or 150 ( or less than 186 for that matter).
The formula does not cover finite population. Thus 186 sample size arrived at ,should be corrected /adjusted for finite population. If the population is N, then the corrected sample size should be = (186N)/( N+185).
Example : If N=100, then the corrected sample size would be =18600/285 (=65.26 or 66)
Hello Arvind,
Thanks for sharing this info…
I am Msc student this has help me conceptualize use of normal curve in determining sample size Thanks a lot
You have provided good calculators
You mentioned twice that this formula can be used even when we do not know the population standard deviation … but you giving no example of that situation and how it would be done without knowing the standard deviation.
Also … there being another formula for sample size which using proportions (phat) and (1 – phat). How does that formula relate and compare to this formula ?
Thanks!
GOD Bless.
thank you . useful information.