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Chapter 8: Problem 8

The article "Limited Yield Estimation for Visual Defect Sources" (IEEE Trans. Semicon. Manuf., 1997: 17-23) reported that, in a study of a particular wafer inspection process, 356 dies were examined by an inspection probe and 201 of these passed the probe. Assuming a stable process, calculate a \(95 \%\) (two- sided) confidence interval for the proportion of all dies that pass the probe.

Short Answer

Expert verified
The 95% confidence interval for the proportion is approximately (0.5129, 0.6163).

Step by step solution

01

Identify Sample Proportion

First, let's identify the sample proportion \( \hat{p} \). This is the proportion of examined dies that passed the probe. We have 201 dies passing out of 356 dies examined. Thus, \( \hat{p} = \frac{201}{356} \approx 0.56461 \).
02

Determine Standard Error

Next, we calculate the standard error (SE) of the sample proportion using the formula: \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \( n \) is the sample size. Plugging in the values, \[ SE = \sqrt{\frac{0.56461 \times (1-0.56461)}{356}} \approx 0.0264 \].
03

Find the Z-score for 95% Confidence Interval

A 95% confidence interval corresponds to a Z-score of approximately 1.96 for a two-sided test. This means you'll need to use 1.96 to create the confidence interval.
04

Calculate Margin of Error

Now, calculate the margin of error (ME) using the formula: \( ME = Z \times SE \), where \( Z = 1.96 \). So, \[ ME = 1.96 \times 0.0264 \approx 0.0517 \].
05

Compute Confidence Interval

Finally, calculate the 95% confidence interval for the proportion by adding and subtracting the margin of error from the sample proportion: \( \hat{p} - ME \) and \( \hat{p} + ME \). This gives: \( 0.56461 - 0.0517 \) and \( 0.56461 + 0.0517 \), which results in the interval \( (0.51291, 0.61631) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Proportion
The sample proportion, often symbolized as \( \hat{p} \), represents the ratio of a particular outcome observed in a sample to the total sample size. In the context of our exercise, it is the proportion of dies that passed the inspection probe out of the total dies examined. To calculate it, you divide the number of successes (passes) by the total number of trials (examinations).
  • For example, if 201 dies pass out of 356, the calculation looks like this: \( \hat{p} = \frac{201}{356} \approx 0.56461 \).
The sample proportion is crucial because it serves as an estimate of the true proportion in the entire population. It becomes the foundation of subsequent calculations, such as finding the confidence interval.
Standard Error
The standard error (SE) is a statistical measure that gives us an idea about how much our sample proportion \( \hat{p} \) might differ from the true population proportion. It provides a sense of the variation and accuracy across multiple samples.
  • The formula for SE when dealing with proportions is: \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \).
  • Here, \( n \) is the total number of observations or sample size.
In our example, the calculation is: \( SE = \sqrt{\frac{0.56461 \times (1-0.56461)}{356}} \approx 0.0264 \).
This tells us that the estimated proportion can vary by about 0.0264 due to random sampling.
Z-score
The Z-score is a measurement of how many standard deviations an element is from the mean. In the context of confidence intervals, it is used to determine the margin within which we can be confident the true proportion lies.
  • For a 95% confidence interval, a Z-score of approximately 1.96 is typically used.
This value comes from the standard normal distribution and reflects the probability associated with that level of confidence.
The choice of Z-score ensures that the area in the tails of the distribution covers 5%, allowing us to be 95% confident.
Margin of Error
The margin of error (ME) represents the range of values below and above the sample statistic in a confidence interval. It quantifies the uncertainty or "wiggle room" around our sample proportion.
  • You calculate it using the formula: \( ME = Z \times SE \), where \( Z \) is the Z-score chosen based on the desired confidence level.
  • For a 95% confidence interval, we use \( Z = 1.96 \).
  • In our calculation, \( ME = 1.96 \times 0.0264 \approx 0.0517 \).
This ME shows that our sample proportion can vary by \( \pm 0.0517 \) from the calculated \( \hat{p} \), indicating the range wherein the population proportion likely falls.

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Most popular questions from this chapter

Nine Australian soldiers were subjected to extreme conditions, which involved a 100 -min walk with a 25 -lb pack when the temperature was \(40^{\circ} \mathrm{C}\left(104^{\circ} \mathrm{F}\right)\). One of them overheated (above \(39^{\circ} \mathrm{C}\) ) and was removed from the study. Here are the rectal Celsius temperatures of the other eight at the end of the walk ("Neural Network Training on Human Body Core Temperature Data," Combatant Protection and Nutrition Branch, Aeronautical and Maritime Research Laboratory of Australia, DSTO TN-0241, 1999): \(\begin{array}{llllllll}38.4 & 38.7 & 39.0 & 38.5 & 38.5 & 39.0 & 38.5 & 38.6\end{array}\) We would like to get a \(95 \%\) confidence interval for the population mean. a. Compute the \(t\)-based confidence interval of Section 8.3. b. Check for the validity of part (a). c. Generate a bootstrap sample of 999 means. d. Use the standard deviation for part (c) to get a \(95 \%\) confidence interval for the population mean. e. Investigate the distribution of the bootstrap means to see if part (d) is valid. f. Use part (c) to form the \(95 \%\) confidence interval using the percentile method. g. Compare the intervals and explain your preference. h. Based on your knowledge of normal body temperature, would you say that body temperature can be influenced by environment?

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Even as traditional markets for sweetgum lumber have declined, large section solid timbers traditionally used for construction bridges and mats have become increasingly scarce. The article "Development of Novel Industrial Laminated Planks from Sweetgum Lumber" (J. of Bridge Engr., 2008: 64-66) described the manufacturing and testing of composite beams designed to add value to low- grade sweetgum lumber. Here is data on the modulus of rupture (psi; the article contained summary data expressed in MPa): \(\begin{array}{llllll}6807.99 & 7637.06 & 6663.28 & 6165.03 & 6991.41 & 6992.23 \\ 6981.46 & 7569.75 & 7437.88 & 6872.39 & 7663.18 & 6032.28 \\\ 6906.04 & 6617.17 & 6984.12 & 7093.71 & 7659.50 & 7378.61 \\ 7295.54 & 6702.76 & 7440.17 & 8053.26 & 8284.75 & 7347.95 \\ 7422.69 & 7886.87 & 6316.67 & 7713.65 & 7503.33 & 7674.99\end{array}\) a. Verify the plausibility of assuming a normal population distribution. b. Estimate the true average modulus of rupture in a way that conveys information about precision and reliability. c. Predict the modulus for a single beam in a way that conveys information about precision and reliability. How does the resulting prediction compare to the estimate in (b).

Determine the \(t\) critical value that will capture the desired \(t\) curve area in each of the following cases: a. Central area \(=.95\), df \(=10\) b. Central area \(=.95, \mathrm{df}=20\) c. Central area \(=99\), df \(=20\) d. Central area \(=.99, \mathrm{df}=50\) e. Upper-tail area \(=.01, \mathrm{df}=25\) f. Lower-tail area \(=.025\), df \(=5\)

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