Manufacturing and service companies attempt to deliver the best products or services to their customers. Manufacturing companies expect that 100% of the goods they produce are non-defective. Also, service companies such as courier services expect that 100% of the packages they deliver arrive on time at the destinations. In general, zero defect is the target of every business. However, in reality the companies always produce nonconforming goods or services, i.e. the ones that fail to meet one or more of the specifications. It is important for the businesses to know the proportion of nonconforming goods they produce or nonconforming services they deliver.

Suppose that we are to estimate the proportion of nonconforming deliveries that occur. To do this, we first collect n sample data on the deliveries and count the number of nonconforming deliveries. Suppose that x deliveries are nonconforming. Then we can compute the sample proportion p=\frac{x}{n}. The sample proportion provides a point estimate for the proportion of nonconforming deliveries. Based on the point estimate, a 100γ% confidence interval for the actual proportion can be constructed, where γ denotes the confidence coefficient or degree of confidence. The confidence interval is an interval estimate of the true proportion. The confidence interval is determined as follows.

p - z_{\alpha /2} \sqrt{\frac{p(1-p)}{n}} \textless \pi \textless p + z_{\alpha /2} \sqrt{\frac{p(1-p)}{n}} …………………………………………………………………………………………………………………………….. (1)

where

π = the true/actual proportion

p = the point estimate

α = 1 – γ

zα/2 = the z-value leaving an area of α/2 to the right

With such interval, we are 100γ% confident that the true proportion is between (p - z_{\alpha /2} \sqrt{\frac{p(1-p)}{n}}) and (p + z_{\alpha /2} \sqrt{\frac{p(1-p)}{n}}). Statistically speaking, given that (1) is the 100γ% confidence interval, it means that in the long run 100γ out of 100 cases intervals like (1) will contain the true proportion.

 

Example 1

A quality manager at a courier service is interested in calculating the proportion of nonconforming deliveries. For this purpose, he collected a random sample of 40 deliveries and it turned out that 8 of them did not meet the delivery specifications. Find the 95% confidence interval for the proportion of nonconforming deliveries. What does the result mean?

Answer

In this case, n = 40 and x = 8. The point estimate for the proportion is p = \frac{8}{40} = .2.

Since γ = .95, α = 1 – γ = 1 – .95 = .05 and α/2 = .025.

From the standard normal distribution table, the z-value that leaves an area of .025 to the right is 1.96. Thus, z.025 = 1.96.

Substituting the values just obtained into (1), we get:

.2 - 1.96 \sqrt{\frac{.2(1-.2)}{40}} \textless \pi \textless .2 + 1.96 \sqrt{\frac{.2(1-.2)}{40}}

.076 < π < .324

So, the 95% confidence interval for the proportion of nonconforming deliveries is .076 < π < .324.

It means that we are 95% confident that between 7.6% and 32.4% of the deliveries are nonconforming.

 

Example 2

In a manufacturing process 500 units have been randomly selected to estimate the percent defective of the process. It turns out that 23 of them are defective. Find the 90% confidence interval for the percent defective.

Answer

In this case, n = 500 and x = 23. The point estimate for the proportion is p = \frac{23}{500} = .046.

Since γ = .90, α = 1 – γ = 1 – .90 = .10 and α/2 = .05.

From the standard normal distribution table, the z-value that leaves an area of .05 to the right is 1.64. Thus, z.05 = 1.64.

Substituting the values just obtained into (1), we get:

.046 - 1.64 \sqrt{\frac{.046(1-.046)}{500}} \textless \pi \textless .046 + 1.64 \sqrt{\frac{.046(1-.046)}{500}}

.031 < π < .061

So, the 90% confidence interval for the percent defective is 3.1% < π < 6.1%.

 

The Sample Size

What is the minimum sample size? If p is used as an estimate of π, we can be 100γ% confident that the error will be less than a specified amount e when the sample size is approximately

n=(\frac{z_{\alpha / 2}}{e})^2 \cdot p(1-p)  ……………………………………………………………………………………………………………………………………………………………………. (2)

In (2), e is the error, which is the difference between the true proportion π and the point estimate p. In practice, p is determined based on some preliminary sample of size n ≥ 30. But if we have a crude estimate of π, as suggested in Walpole (1993), we could use this value for p and then compute n.

 

Example 3

Referring to Example 1, suppose that the quality manager want to be 95% confident that his estimate of π is within 0.10. Assuming that he treats the 40 sample as a preliminary sample, find the minimum sample size.

Answer

In this case, γ = .95 and consequently α = 1 – .95 = .05. and α/2 = .025. From the standard normal distribution table, we get z.025 = 1.96. Based on the preliminary sample p = \frac{8}{40} =.20. Thus, the minimum sample size is n = (\frac{1.96}{0.10})^2 \cdot 0.20 \cdot (1-0,20) \approx 61.47. In conclusion, the minimum sample size is 62.

 

Reference

Walpole, R. E., and R. H. Myers. Probability and Statistics for Engineers and Scientists. New York: Macmillan Publishing Company, 1993.

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