/*! 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 29 Data on oxide thickness of semic... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 29

Data on oxide thickness of semiconductors are as follows: \(425,431,416,419,421,436,418,410,431,433,423,\) 426,410,435,436,428,411,426,409,437,422,428,413,416 (a) Calculate a point estimate of the mean oxide thickness for all wafers in the population. (b) Calculate a point estimate of the standard deviation of oxide thickness for all wafers in the population. (c) Calculate the standard error of the point estimate from part (a). (d) Calculate a point estimate of the median oxide thickness for all wafers in the population. (e) Calculate a point estimate of the proportion of wafers in the population that have oxide thickness greater than 430 angstroms.

Short Answer

Expert verified
a) 424.58, b) 5.78, c) 1.18, d) 422.5, e) 0.2917

Step by step solution

01

Calculate the Mean

To find the mean, sum all oxide thickness values and divide by the number of observations. We have 24 data points. The total sum is \( \sum X = 425 + 431 + 416 + 419 + 421 + 436 + 418 + 410 + 431 + 433 + 423 + 426 + 410 + 435 + 436 + 428 + 411 + 426 + 409 + 437 + 422 + 428 + 413 + 416 = 10190 \). The mean \( \bar{X} \) is given by \( \bar{X} = \frac{10190}{24} = 424.58 \).
02

Calculate the Standard Deviation

First, find the sum of squared differences from the mean. \( \sum (X - \bar{X})^2 = (425 - 424.58)^2 + (431 - 424.58)^2 + \ldots + (416 - 424.58)^2 = 765.67 \). The formula for standard deviation \( s \) is \( s = \sqrt{\frac{\sum (X - \bar{X})^2}{n-1}} \), where \( n \) is the number of observations. So, \( s = \sqrt{\frac{765.67}{23}} \approx 5.78 \).
03

Calculate the Standard Error

The standard error of the mean is \( SE = \frac{s}{\sqrt{n}} \). Using \( n = 24 \) and \( s = 5.78 \), we have \( SE = \frac{5.78}{\sqrt{24}} \approx 1.18 \).
04

Calculate the Median

To find the median, arrange the data in ascending order. The middle value(s) are the median. Since \( n = 24 \), the median is the average of the 12th and 13th values when sorted: \( 422, 423 \). So, the median \( M \) is \( M = \frac{422 + 423}{2} = 422.5 \).
05

Calculate the Proportion Greater Than 430

Count how many wafers have oxide thickness greater than 430. The values are 431, 433, 435, 436, 436, 431, and 437 鈥� 7 values in total. The proportion \( P \) is given by \( P = \frac{7}{24} \approx 0.2917 \).

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 91影视!

Key Concepts

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

Mean Calculation
The mean, often referred to as the average, is a point estimate representing the central tendency of data. To calculate the mean, start by totaling all values in the dataset and then divide by the number of values.
For example, with our semiconductor wafer data:
  • Sum of oxide thickness: \( \sum X = 10190 \)
  • Number of data points: \( n = 24 \)
The formula for the mean \( \bar{X} \) is:\[\bar{X} = \frac{10190}{24} = 424.58\]This calculation gives us a mean oxide thickness of \( 424.58 \) angstroms, indicating the typical thickness for wafers in this dataset.
Standard Deviation
The standard deviation is a measure indicating how spread out the data is from the mean. It shows the amount of variation or dispersion.
To calculate the standard deviation, follow these steps:
  • Compute each data point's deviation from the mean and square it
  • Sum these squared deviations
  • Divide by \( n-1 \), where \( n \) is the number of data points
  • Take the square root of the result
In our example:
  • Sum of squared deviations: \( \sum (X - \bar{X})^2 = 765.67 \)
  • Calculate standard deviation: \( s = \sqrt{\frac{765.67}{23}} \approx 5.78 \)
Thus, the standard deviation of \( 5.78 \) indicates moderate variability in oxide thickness across the wafer samples.
Standard Error
The standard error provides an estimate of how far the sample mean is likely to be from the population mean. A smaller standard error suggests a more precise estimate.
You can calculate the standard error of the mean using:
  • The formula \( SE = \frac{s}{\sqrt{n}} \)
  • Here, \( s \) is the standard deviation, and \( n \) is the sample size
For our dataset:
  • Standard deviation \( s = 5.78 \)
  • Sample size \( n = 24 \)
  • Calculate the standard error: \( SE = \frac{5.78}{\sqrt{24}} \approx 1.18 \)
This result of \( 1.18 \) suggests that if we were to sample repeatedly, the sample mean would likely be within about \( 1.18 \) angstroms of the true population mean.
Median Calculation
The median is the middle value that separates the higher half from the lower half of a data sample. It is a useful measure because it is not affected by extremely high or low values.
To find the median, first arrange the data in ascending order:
  • When \( n \) is odd, the median is the middle number
  • When \( n \) is even, like in our example, the median will be the average of the two middle numbers
For our sorted data, which has 24 points, the median is calculated from the 12th and 13th numbers:
  • The two middle values are \( 422 \) and \( 423 \)
Thus, the median \( M \) is:\[M = \frac{422 + 423}{2} = 422.5\]This provides a central value of \( 422.5 \) angstroms for the wafer thickness, providing another perspective on the dataset's average.

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 size \(n=16\) is taken from a normal population with \(\mu=40\) and \(\sigma^{2}=5 .\) Find the probability that the sample mean is less than or equal to \(37 .\)

A random variable \(x\) has probability density function $$f(x)=\frac{1}{2 \theta^{3}} x^{2} e^{-x / \theta}, \quad 0

Two different plasma etchers in a semiconductor factory have the same mean etch rate \(\mu .\) However, machine 1 is newer than machine 2 and consequently has smaller variability in etch rate. We know that the variance of etch rate for machine 1 is \(\sigma_{1}^{2}\) and for machine 2 it is \(\sigma_{2}^{2}=a \sigma_{1}^{2}\). Suppose that we have \(n_{1}\) independent observations on etch rate from machine 1 and \(n_{2}\) independent observations on etch rate from machine 2 (a) Show that \(\hat{\mu}=\alpha \bar{X}_{1}+(1-\alpha) \bar{X}_{2}\) is an unbiased estimator of \(\mu\) for any value of \(\alpha\) between 0 and 1 . (b) Find the standard error of the point estimate of \(\mu\) in part (a). (c) What value of \(\alpha\) would minimize the standard error of the point estimate of \(\mu ?\) (d) Suppose that \(a=4\) and \(n_{1}=2 n_{2}\). What value of \(\alpha\) would you select to minimize the standard error of the point estimate of \(\mu\) ? How "bad" would it be to arbitrarily choose \(\alpha=0.5\) in this case?

Consider the shifted exponential distribution $$f(x)=\lambda e^{-\lambda(x-\theta)}, \quad x \geq \theta$$ When \(\theta=0,\) this density reduces to the usual exponential distribution. When \(\theta>0,\) there is only positive probability to the right of \(\theta\) (a) Find the maximum likelihood estimator of \(\lambda\) and \(\theta,\) based on a random sample of size \(n\). (b) Describe a practical situation in which one would suspect that the shifted exponential distribution is a plausible model.

Suppose we have a random sample of size \(2 n\) from a population denoted by \(X,\) and \(E(X)=\mu\) and \(V(X)=\sigma^{2} .\) Let $$\bar{X}_{1}=\frac{1}{2 n} \sum_{i=1}^{2 n} X_{i} \quad \text { and } \quad \bar{X}_{2}=\frac{1}{n} \sum_{i=1}^{n} X_{i} $$be two estimators of \(\mu\). Which is the better estimator of \(\mu\) ? Explain your choice.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation
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.