/*! 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 53 The fraction of defective integr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The fraction of defective integrated circuits produced in a photolithography process is being studied. A random sample of 300 circuits is tested, revealing 13 defectives. (a) Calculate a \(95 \%\) two-sided CI on the fraction of defective circuits produced by this particular tool. (b) Calculate a \(95 \%\) upper confidence bound on the fraction of defective circuits.

Short Answer

Expert verified
(a) 95% CI: (0.0200, 0.0666). (b) 95% upper bound: 0.0629.

Step by step solution

01

Determine Sample Proportion

We first calculate the sample proportion of defective circuits. This is done by dividing the number of defective circuits by the total sample size: \[ \hat{p} = \frac{13}{300} = 0.0433. \]
02

Calculate Standard Error

The standard error of the sample proportion is calculated using the formula: \[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}. \] Substituting the known values, we get: \[ SE = \sqrt{\frac{0.0433(1-0.0433)}{300}} \approx 0.0119. \]
03

Find Z-value for 95% Confidence

At a 95% confidence level, the critical z-value (two-tailed) is approximately 1.96.
04

Calculate 95% Confidence Interval (CI)

Using the sample proportion, standard error, and z-value, compute the confidence interval: \[ \hat{p} \pm Z \cdot SE = 0.0433 \pm 1.96 \times 0.0119. \] This gives: \[ (0.0200, 0.0666). \]
05

Calculate Upper Confidence Bound Z-value

For the upper 95% confidence bound, use a z-value of 1.645 (one-tailed).
06

Calculate 95% Upper Confidence Bound

Using the sample proportion, standard error, and z-value for the upper bound, we find: \[ \hat{p} + Z \cdot SE = 0.0433 + 1.645 \times 0.0119 = 0.0629. \]

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 gives us an estimate of what fraction of a population is expected to show a particular characteristic, based on a random sample. In our example, we are interested in knowing the proportion of defective circuits in a lot of integrated circuits.
To calculate the sample proportion:
  • Count the number of targets showing the characteristic of interest (defective circuits).
  • Divide that number by the total number of samples tested.
For instance, in a test of 300 circuits where 13 are found to be defective, the sample proportion, denoted as \( \hat{p}, \)is:\[ \hat{p} = \frac{13}{300} = 0.0433. \]This means about 4.33% of the circuits in this sample are defective.
The sample proportion is crucial in inferential statistics as it helps in estimating the proportion for the whole population with a certain level of confidence.
Standard Error
The standard error (SE) of a sample proportion provides a measure of how much variability there is around that proportion. It helps us understand how accurately our sample proportion can estimate the true population proportion. The lower the standard error, the more precise our estimate. To calculate the standard error of a sample proportion, use the formula:
\[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}. \]Here, \( \hat{p} \)is the sample proportion, and \( n \)is the sample size.
In our case, substituting the sample proportion and sample size,\[ SE = \sqrt{\frac{0.0433(1-0.0433)}{300}} \approx 0.0119. \]This value informs us about how much deviation we might expect from the true proportion if we were to take multiple samples of the same size.
Z-value
A Z-value is a measure used in statistics that relates to the standard normal distribution curve. It indicates how many standard deviations a point is from the mean of the distribution. In confidence intervals, it helps us define the coverage probability. For a given confidence level, like 95%, the corresponding Z-value tells us the range within which the true parameter lies with that confidence.
In our example, we first need a Z-value for a 95% two-sided confidence interval where it is roughly 1.96. This means there's a 95% chance the true proportion is within 1.96 standard errors on either side of the estimated proportion. For a one-sided confidence bound (just the upper limit, in this case), the Z-value is slightly lower, about 1.645.
These Z-values are critical in calculating how confident we are in our statistical estimations and predictions.
Upper Confidence Bound
The upper confidence bound provides an assurance that the true proportion of a population characteristic is less than a specified value with a set level of confidence. Here, the upper bound is specifically about the potential maximum proportion of defectives in the circuits beyond the sample mean.
To compute this bound, we use the sample proportion, the standard error, and a one-tailed Z-value. For a 95% upper bound, the Z-value, as mentioned, is about 1.645.
The formula is:\[ \hat{p} + Z \cdot SE, \]where substituting values, we get:\[ 0.0433 + 1.645 \times 0.0119 = 0.0629. \]This suggests that, with 95% confidence, the true proportion of defective circuits doesn't exceed 6.29%.
This method is useful in quality control and assurance, especially when it is more critical to assess the maximum possible defect rate rather than the full range.

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

A random sample of 50 suspension helmets used by motorcycle riders and automobile race-car drivers was subjected to an impact test, and on 18 of these helmets some damage was observed. (a) Find a \(95 \%\) two-sided confidence interval on the true proportion of helmets of this type that would show damage from this test. (b) Using the point estimate of \(p\) obtained from the preliminary sample of 50 helmets, how many helmets must be tested to be \(95 \%\) confident that the error in estimating the true value of \(p\) is less than \(0.02 ?\) (c) How large must the sample be if we wish to be at least \(95 \%\) confident that the error in estimating \(p\) is less than \(0.02,\) regardless of the true value of \(p ?\)

The maker of a shampoo knows that customers like this product to have a lot of foam. Ten sample bottles of the product are selected at random and the foam heights observed are as follows (in millimeters): 210,215,194,195,211,201 \(198,204,208,\) and \(196 .\) (a) Is there evidence to support the assumption that foam height is normally distributed? (b) Find a \(95 \% \mathrm{Cl}\) on the mean foam height. (c) Find a \(95 \%\) prediction interval on the next bottle of shampoo that will be tested. (d) Find an interval that contains \(95 \%\) of the shampoo foam heights with \(99 \%\) confidence. (e) Explain the difference in the intervals computed in parts (b), (c), and (d).

A particular brand of diet margarine was analyzed to determine the level of polyunsaturated fatty acid (in percentages). A sample of six packages resulted in the following data: 16.8,17.2,17.4,16.9,16.5,17.1 (a) Check the assumption that the level of polyunsaturated fatty acid is normally distributed. (b) Calculate a \(99 \%\) confidence interval on the mean \(\mu\). Provide a practical interpretation of this interval. (c) Calculate a \(99 \%\) lower confidence bound on the mean. Compare this bound with the lower bound of the two-sided confidence interval and discuss why they are different.

A manufacturer produces piston rings for an automobile engine. It is known that ring diameter is normally distributed with \(\sigma=0.001\) millimeters. A random sample of 15 rings has a mean diameter of \(\bar{x}=74.036\) millimeters. (a) Construct a \(99 \%\) two-sided confidence interval on the mean piston ring diameter. (b) Construct a \(99 \%\) lower-confidence bound on the mean piston ring diameter. Compare the lower bound of this confidence interval with the one in part (a)

An article in the Journal of Agricultural Science [ "The Use of Residual Maximum Likelihood to Model Grain Quality Characteristics of Wheat with Variety, Climatic and Nitrogen Fertilizer Effects" (1997, Vol. 128, pp. \(135-142\) ) ] investigated means of wheat grain crude protein content (CP) and Hagberg falling number (HFN) surveyed in the UK. The analysis used a variety of nitrogen fertilizer applications (kg N/ha), temperature \(\left({ }^{\circ} \mathrm{C}\right),\) and total monthly rainfall \((\mathrm{mm})\). The data shown below describe temperatures for wheat grown at Harper Adams Agricultural College between 1982 and \(1993 .\) The temperatures measured in June were obtained as follows: $$ \begin{array}{llllll} 15.2 & 14.2 & 14.0 & 12.2 & 14.4 & 12.5 \\ 14.3 & 14.2 & 13.5 & 11.8 & 15.2 & \end{array} $$ Assume that the standard deviation is known to be \(\sigma=0.5\). (a) Construct a \(99 \%\) two-sided confidence interval on the mean temperature. (b) Construct a \(95 \%\) lower-confidence bound on the mean temperature. (c) Suppose that we wanted to be \(95 \%\) confident that the error in estimating the mean temperature is less than 2 degrees Celsius. What sample size should be used? (d) Suppose that we wanted the total width of the two-sided confidence interval on mean temperature to be 1.5 degrees Celsius at \(95 \%\) confidence. What sample size should be used?

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.