/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 The article "Limited Yield Estim... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The article "Limited Yield Estimation for Visual Defect Sources" (IEEE Trans. on Semiconductor 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 of dies that pass is (0.5130, 0.6162).

Step by step solution

01

Identify the sample proportion

First, we need to find the sample proportion \(p\) of dies that passed the probe. Given in the problem, 201 out of 356 dies passed the probe. So, we calculate \(\hat{p}\) as follows:\[ \hat{p} = \frac{201}{356} \approx 0.5646. \]
02

Find Standard Error of the Proportion

The standard error (SE) for the sample proportion \(\hat{p}\) is given by the formula:\[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \]where \(n\) is the total number of dies sampled (356 in this case). So,\[ SE = \sqrt{\frac{0.5646 \times (1 - 0.5646)}{356}} \approx 0.0263. \]
03

Determine the Z-score for a 95% Confidence Interval

For a 95% confidence level, the corresponding Z-score from the standard normal distribution is approximately 1.96.
04

Calculate the Confidence Interval

The formula for a 95% confidence interval (CI) for the population proportion is:\[ \hat{p} \pm Z \times SE, \]where \(Z\) is the Z-score for a 95% confidence interval. Plug the values in:\[ 0.5646 \pm 1.96 \times 0.0263. \]This gives:\[ 0.5646 \pm 0.0516, \]which results in the interval:\[ (0.5130, 0.6162). \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Sample Proportion
The sample proportion helps us understand the fraction of a particular outcome within a sample. In this exercise, we calculate the sample proportion of dies that passed the inspection probe.
To find the sample proportion, divide the number of successful outcomes by the total number of samples. Here, 201 out of 356 dies passed, leading to:
  • \[ \hat{p} = \frac{201}{356} \approx 0.5646 \; (56.46\% \text{ pass rate}). \]
The sample proportion is a way of estimating the probability of success (in this case, passing the probe) in the population. It serves as a foundation for further statistics, like confidence intervals and hypothesis testing.
Standard Error
The standard error measures the amount of variability or dispersion of the sample proportion from the true population proportion. It reflects how confident you can be about the accuracy of the sample proportion.In our calculation, the standard error for the sample proportion is computed as follows:

  • \[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.5646 \times (1 - 0.5646)}{356}} \approx 0.0263. \]
A small standard error indicates that the sample proportion is a close approximation of the population proportion. As the sample size increases, the standard error decreases, which provides more certainty in your estimate.
Z-score
A Z-score represents how many standard deviations an element is from the mean of a distribution. In confidence intervals, the Z-score helps determine the range of values that are likely to contain the true population parameter. For a 95% confidence interval, we use a Z-score of 1.96. This value comes from the standard normal distribution and reflects the coverage of the interval, meaning it accounts for 95% of the area under the normal curve.
Knowing the correct Z-score is crucial, as it adjusts the width of the confidence interval to accurately reflect the level of confidence desired by the researcher.
Population Proportion
The population proportion is an estimate of the fraction of elements in the whole population that exhibit a particular characteristic, in this case, passing the probe.Using the sample data and the calculated confidence interval, we can make an educated guess about the true population proportion. Given the bounds of:
  • \[ (0.5130, 0.6162), \]
we can state with 95% confidence that the true proportion of all dies passing the probe lies between approximately 51.30% to 61.62%.Reflecting on the margin of error and the sample size gives you a deeper insight into how close your sample reflects the entire population. Understanding the population proportion is essential for making informed decisions based on statistical data.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose that a random sample of 50 bottles of a particular brand of cough syrup is selected and the alcohol content of each bottle is determined. Let \(\mu\) denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting \(95 \%\) confidence interval is \((7.8,9.4)\). a. Would a \(90 \%\) confidence interval calculated from this same sample have been narrower or wider than the given interval? Explain your reasoning. b. Consider the following statement: There is a \(95 \%\) chance that \(\mu\) is between \(7.8\) and \(9.4\). Is this statement correct? Why or why not? c. Consider the following statement: We can be highly confident that \(95 \%\) of all bottles of this type of cough syrup have an alcohol content that is between \(7.8\) and 9.4. Is this statement correct? Why or why not? d. Consider the following statement: If the process of selecting a sample of size 50 and then computing the corresponding \(95 \%\) interval is repeated 100 times, 95 of the resulting intervals will include \(\mu\). Is this statement correct? Why or why not?

The article "Gas Cooking, Kitchen Ventilation, and Exposure to Combustion Products" (Indoor Air, 2006: 65-73) reported that for a sample of 50 kitchens with gas cooking appliances monitored during a oneweek period, the sample mean \(\mathrm{CO}_{2}\) level (ppm) was \(654.16\), and the sample standard deviation was 164.43. a. Calculate and interpret a \(95 \%\) (two-sided) confidence interval for true average \(\mathrm{CO}_{2}\) level in the population of all homes from which the sample was selected. b. Suppose the investigators had made a rough guess of 175 for the value of \(s\) before collecting data. What sample size would be necessary to obtain an interval width of \(50 \mathrm{ppm}\) for a confidence level of \(95 \%\) ?

A study of the ability of individuals to walk in a straight line ("Can We Really Walk Straight?" Amer. J. of Physical Anthro., 1992: 19-27) reported the accompanying data on cadence (strides per second) for a sample of \(n=20\) randomly selected healthy men. \(\begin{array}{rrrrrrrrr}.95 & .85 & .92 & .95 & .93 & .86 & 1.00 & .92 & .85\end{array}\) A normal probability plot gives substantial support to the assumption that the population distribution of cadence is approximately normal. A descriptive summary of the data from Minitab follows: \(\begin{array}{lcccc}\text { Variable N } & \text { Mean } & \text { Median } & \text { TrMean } & \text { StDev SEMean } \\ \text { cadence } 20 & 0.9255 & 0.9300 & 0.9261 & 0.08090 .0181 \\ \text { Variable } & \text { Min } & \text { Max } & \text { Q1 } & \text { Q3 } \\ \text { cadence } & 0.7800 & 1.0600 & 0.8525 & 0.9600\end{array}\) a. Calculate and interpret a \(95 \%\) confidence interval for population mean cadence. b. Calculate and interpret a \(95 \%\) prediction interval for the cadence of a single individual randomly selected from this population. c. Calculate an interval that includes at least \(99 \%\) of the cadences in the population distribution using a confidence level of \(95 \%\).

Each of the following is a confidence interval for \(\mu=\) true average (i.e., population mean) resonance frequency (Hz) for all tennis rackets of a certain type: \((114.4,115.6)(114.1,115.9)\) a. What is the value of the sample mean resonance frequency? b. Both intervals were calculated from the same sample data. The confidence level for one of these intervals is \(90 \%\) and for the other is \(99 \%\). Which of the intervals has the \(90 \%\) confidence level, and why?

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

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.