Industrial Engineering Homework

Sample Size Determination

One of the primary variables in any measurement procedure is the size of the sample, referred to as "n." To compute the value of n, we need three numerical values:

1. a Z value based on an alpha value (alpha is the chance of being wrong),
2. an estimate of the variability of the process, and
3. how much error we are prepared to accept in our answer.

We will examine each of the above three values one-at-a-time.

Alpha is the chance of being wrong. For example, if we wish to be right 19 times out of 20, then we would be wrong 1 time out of 20, or 5% of the time. Alpha would then be equal to 0.05.

Once we know alpha, we can use the simplified Z Table shown below to find our Z value. To use the table below, we simply need to know the total alpha value and we look that number up in the alpha column. We then get the correct Z value from the column beside the alpha value. Continuing with our example above, if alpha is 0.050 then Z is 1.960 from the table below.

We also need an estimate of the variability of the process.

If we are working with variable data (which is numerical measurement data, such as length, weight, diameter, time), this would be the standard deviation of the process (s).

If we are working with attribute data (which is pass or fail data, such as the percent wrong), this would be the sample proportion (p).

The error factor represents how much error we are prepared to tolerate in our answer. If we want our answer to be perfect, then we will have to look at everything, and the sample size will be equal to our entire population of interest, or n = N. If we can tolerate something less than perfection, then our value of "n" can be smaller. The more error we will accept, the smaller the "n" value will be. Remember that there are many costs associated with "n" and we would like for "n" to be as small as practical to minimize those expenses.

We will call the acceptable error term "e." It will be equal to how far away from the mean (or average or proportion) we can be, and still feel comfortable with our answer.

In many cases, the acceptable error will be specified or given. If the error is not given, then we can estimate a reasonable value as follows:

If we are working with variable data, the error can be set equal to approximately one-half of the standard deviation, or e = s/2.

If we are working with attribute data, the error can be set equal to approximately one-half of the proportion, or e = p/2.

The equations for computing the sample size (n) are as follows:

Variable data: n = [ (z) (s) / (e) ] (squared)

Attribute data: n = [ (z)squared ] [ ( p ) ( 1 - p ) ] / [ (e)squared ]

1. A manufacturing operation has an average time-studied time of 0.750 minutes. The standard deviation of this operation is 0.055 minutes, based on the time study data. If we are interested in obtaining a time-studied estimate that is within 0.020 minutes of the true average for this job, with an alpha value of 6%, how large a sample must be taken?

Alpha = 6% or 0.060
Therefore, from the Z Table, Z = 1.881
s = 0.055 minutes
e = 0.020 minutes

Time is variable data so the correct equation is:

n = [ (z) (s) / (e) ] (squared)

n = [ (1.881) (0.055) / (0.020) ] (squared)
n = [ (0.103455) / (0.020 ) ] (squared)
n = [ 5.17275 ] (squared)
n = 26.757 observations or 27 observations

When determining the final value of "n" we always round up to the next whole number, as in the above example.

2. If we have a job that has a P, F, and D percentage of 25%, and we want our answer to be within 2% of the true percentage, with an alpha value of 10%, how large a sample must be taken?

p = 25% or 0.25
e = 2% or 0.02
alpha = 10% or 0.100
Therefore, from the Z Table, Z = 1.645

Percentages are attribute data. Therefore, the correct equation is:

n = [ (z)squared ] [ ( p ) ( 1 - p ) ] / [ (e)squared ]

n = [ (1.645)squared ] [ ( 0.25 ) ( 1 - 0.25 ) ] / [ (0.02)squared ]
n = [ 2.706 ] [ 0.1875 ] / [ 0.0004 ]
n = 1,268.4375 or 1,269 observations

When determining the final value of "n" we always round up to the next whole number, as in the above example.

The reason we have such a large sample size in the above example is due to the fact that we specified our acceptable error to be very small at 2%. If we change our acceptable error value to 8%, then answer would be as follows:

p = 25% or 0.25
e = 8% or 0.08
alpha = 10% or 0.100
Therefore, from the Z Table, Z = 1.645

Percentages are attribute data. Therefore, the correct equation is:

n = [ (z)squared ] [ ( p ) ( 1 - p ) ] / [ (e)squared ]

n = [ (1.645)squared ] [ ( 0.25 ) ( 1 - 0.25 ) ] / [ (0.08)squared ]
n = [ 2.706 ] [ 0.1875 ] / [ 0.0064 ]
n = 79.277 or 80 observations

When determining the final value of "n" we always round up to the next whole number, as in the above example.

THE ASSIGNMENT

Determine the correct sample size for each of the following problems. You must show the equation you used, the values you substituted into that equation, and your final value of "n" to three decimal places before you round the value of "n" to a whole number. In other words, follow the examples shown on the lecture page for this assignment.

1. After a brief time study on a delivery job, the data showed an average time to do the delivery job of 5.7 minutes with a standard deviation of 0.35 minutes. If we use an alpha value of 2%, and we want our final estimate to be within 0.1 minutes of the true value, how large a sample should be taken?

Your Answer:

2. A time study analyst has collected some preliminary data on a manufacturing job. The average time to do the job is 13.4 minutes with a standard deviation of 0.9 minutes. The analyst wants the final estimate to be within 0.5 minutes of the true time, and the alpha value has been established at 7%. How large a sample should be studied?

Your Answer:

3. The office operations of the company currently permits a personal factor of 22%. If the alpha value is equal to 6% and the acceptable error is 8%, how large a sample must be taken?

Your Answer:

4. The manufacturing area has had a historical delay percentage of 10.5%. Management wants to verify that percentage using an alpha value of 9% and they are willing to tolerate an acceptable error of 4%. How large a sample will be necessary?

Your Answer: