Wednesday, May 28, 2014

Min Sample Size of Confidence Interval of Proportion in Excel 2010 and Excel 2013

This is one of the following five articles on Confidence Intervals in Excel

z-Based Confidence Intervals of a Population Mean in 2 Steps in Excel 2010 and Excel 2013

t-Based Confidence Intervals of a Population Mean in 2 Steps in Excel 2010 and Excel 2013

Minimum Sample Size to Limit the Size of a Confidence interval of a Population Mean

Confidence Interval of Population Proportion in 2 Steps in Excel 2010 and Excel 2013

Min Sample Size of Confidence Interval of Proportion in Excel 2010 and Excel 2013

 

Min Sample Size to Limit

Width of a Confidence

Interval of a Population

Proportion

The larger the sample taken, the smaller the Confidence Interval becomes. That makes intuitive sense because the more sample information that is gathered, the more tightly the position of the population mean can be defined. The Confidence Interval is an interval believed to contain the population mean with specific degree of certainty.

As sample size increase, the Confidence Interval shrinks because greater certainty has been attained. The margin of error, which is equal to half the width of the Confidence Interval, therefore shrinks as well.

During the design phase of a statistical experiment, sample size should be determined. Sampling has a cost and additional sampling beyond what is necessary to attain a desired level of certainty is often undesirable. One common objective of the design phase of a statistical test involving sampling is to determine the minimum sample size required to obtain a specified degree of certainty.

This minimum sample size, n, can be derived by the following equation:

(Half-width of C.I.) = z Valueα, 2-tailed * SQRT[ (pest * qest) / n]

Estimates of population parameters p and q must be used in this equation because sample statistics p_bar and q_bar are not available since a sample has not been taken.

(Half-width of C.I.) = z Valueα, 2-tailed * SQRT[ (pest * qest) / n]

(Half-width of C.I.) = NORM.S.INV(1 – α/2) * SQRT[ (pest * qest) / n]

Squaring both sides gives the following:

(Half-width of C.I.)2 = NORM.S.INV2 (1 – α/2) * pest * qest / n

Further algebraic manipulation provides the following:

n = [NORM.S.INV2 (1 – α/2) * pest * qest] / (Half-width of C.I.)2

or, equivalently

n = [NORM.S.INV2 (1 – α/2) * pest * qest] / (Margin of Error)2

The count of data observations in a sample, n, must be a whole number so n must be rounded up to the nearest whole number. This is implemented in Excel as follows:

n = Roundup( ( NORM.S.INV2 (1 – α/2) * pest * qest ) / (Half-width of C.I.)2 )

pest and qest are estimates of the actual population parameters p and q. The most conservative estimate of the minimum sample size would use pest = 0.50.

If pest = 0.05, then qest = 1 – p = 0.50

The product pest * qest has its maximum value of 0.25 when pest = 0.50. This maximum value of pest * qest produces the highest and therefore most conservative value of the minimum sample size, n.

If p is fairly close to 0.5, then pest should be set at 0.5. If p is estimated to be significantly different than 0.5, pest should be set to its estimated value.

 

Example 1 of Calculating Min Sample Size in Excel

Min Number of Voters Surveyed to Limit Poll Error Margin

Two candidates are running against each other in a national election. This election is considered fairly even. What is the minimum number of voters who should be randomly surveyed to obtain a survey result that has 95 percent certainty of being within 2 percent of the nationwide preference for either one of the candidates?

pest should be set at 0.5 since the election is considered even.

pest = 0.5

qest = 1 – pest = 0.5

Half-width of the confidence interval = Margin of Error = 2 percent = 0.02

n = Roundup( ( NORM.S.INV2 (1 – α/2) * pest * qest ) / (Half-width of C.I.)2 )

n = Roundup( ( NORM.S.INV2 (1 – 0.05/2) * 0.50 * 0.50 ) / (0.02)2 )

n = Roundup( ( NORM.S.INV2 ( 0.975) * 0.250 ) / (0.02)2 )

n = Roundup(2400.912)

n = 2401

The preferences of at least 2,401 voters would have to be randomly surveyed to obtain a sample proportion that has 95 percent certainty of being within 2 percent of the national voter preference for one of the candidates.

 

Example 2 of Calculating Min Sample Size in Excel

Min Number of Production Samples to Limit Defect Rate Estimate Error Margin

A production line is estimated to have a defect rate of approximately 15 percent of all units produced on the line. What would be the minimum number of completed production units that should be randomly sampled for defects to obtain a sample proportion of defective units that has 95 percent certainty to being within 1 percent of the real defect rate of all unites produced on that production line?

pest should be set more conservatively than its estimate. The more conservative that p is, the higher will be the minimum sample size required. The most conservative setting for pest would 0.5. pest should be set between its estimate of 0.15 and 0.5. A reasonable setting for pest would be 0.25.

pest = 0.25

qest = 1 – pest = 0.75

Half-width of the confidence interval = Margin of Error = 1 percent = 0.01

n = Roundup( ( NORM.S.INV2 (1 – α/2) * pest * qest ) / (Half-width of C.I.)2 )

n = Roundup( ( NORM.S.INV2 (1 – 0.05/2) * 0.25 * 0.75 ) / (0.01)2 )

n = Roundup( ( NORM.S.INV2 ( 0.975) * 0.1875 ) / (0.01)2 )

n = Roundup(7202.735)

n = 7203

At least 7,203 completed units should be randomly sampled from the production line to obtain a sample proportion defective that has 95 percent certainty of being within 1 percent of the actual proportion defective of all units produced on that production line. If pest were set at 0.15 instead of the more conservative 0.25, the minimum sample size would have been reduced to 4,898.

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