/*! 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 3 Consider the following sample of... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 3

Consider the following sample of observations on coating thickness for low- viscosity paint ("Achieving a Target Value for a Manufacturing Process: A Case Study," \(J\). of Quality Technology, 1992: 22-26): \(\begin{array}{rrrrrrrr}.83 & .88 & .88 & 1.04 & 1.09 & 1.12 & 1.29 & 1.31 \\\ 1.48 & 1.49 & 1.59 & 1.62 & 1.65 & 1.71 & 1.76 & 1.83\end{array}\) Assume that the distribution of coating thickness is normal (a normal probability plot strongly supports this assumption). a. Calculate a point estimate of the mean value of coating thickness, and state which estimator you used. b. Calculate a point estimate of the median of the coating thickness distribution, and state which estimator you used. c. Calculate a point estimate of the value that separates the largest \(10 \%\) of all values in the thickness distribution from the remaining \(90 \%\), and state which estimator you used. [Hint: Express what you are trying to estimate in terms of \(\mu\) and \(\sigma\).] d. Estimate \(P(X<1.5)\), i.e., the proportion of all thickness values less than 1.5. [Hint: If you knew the values of \(\mu\) and \(\sigma\), you could calculate this probability. These values are not available, but they can be estimated.] e. What is the estimated standard error of the estimator that you used in part (b)?

Short Answer

Expert verified
a. Mean: 1.473 b. Median: 1.395 c. 90th Percentile: 1.874 d. P(X < 1.5): 0.534 e. SE of Median: 0.098

Step by step solution

01

Calculate the Sample Mean

To estimate the mean value of the coating thickness, use the sample mean as the estimator. The formula is \( \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \), where \(x_i\) are the observations and \(n\) is the number of observations.Given data: \[ x = \{0.83, 0.88, 0.88, 1.04, 1.09, 1.12, 1.29, 1.31, 1.48, 1.49, 1.59, 1.62, 1.65, 1.71, 1.76, 1.83\} \]Calculate:\[ \bar{x} = \frac{0.83 + 0.88 + 0.88 + 1.04 + 1.09 + 1.12 + 1.29 + 1.31 + 1.48 + 1.49 + 1.59 + 1.62 + 1.65 + 1.71 + 1.76 + 1.83}{16} = \frac{23.57}{16} \approx 1.473 \] The point estimate for the mean using the sample mean is 1.473.
02

Calculate the Sample Median

To estimate the median value of the coating thickness, use the sample median as the estimator. For an ordered data set, if \(n\) is even, the median is the average of the \( (n/2) \)-th and \( (n/2 + 1) \)-th terms. Ordered data: \[ \{0.83, 0.88, 0.88, 1.04, 1.09, 1.12, 1.29, 1.31, 1.48, 1.49, 1.59, 1.62, 1.65, 1.71, 1.76, 1.83\} \]Calculate:Median = \( \frac{1.31 + 1.48}{2} = 1.395 \)The point estimate for the median using the sample median is 1.395.
03

Calculate the 90th Percentile

To estimate the value separating the largest 10% from the rest, we need the 90th percentile of a normal distribution, often denoted as \(x_{0.9}\). This value is given by \( x_{0.9} = \mu + z_{0.9} \cdot \sigma \). Use the sample mean and standard deviation as estimates for \(\mu\) and \(\sigma\).Sample mean \(\bar{x} = 1.473\) (from Step 1) and calculate the sample standard deviation \(s\):\[ s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} \approx 0.313 \]From the z-table, \( z_{0.9} \) for 90th percentile is approximately 1.28.Calculate:\( x_{0.9} = 1.473 + 1.28 \times 0.313 \approx 1.874 \)The point estimate for the 90th percentile is 1.874.
04

Estimate Probability P(X < 1.5)

To estimate \(P(X<1.5)\), determine the standardized z-score using \( z = \frac{1.5 - \bar{x}}{s} \).Using \(\bar{x} = 1.473\) and \(s = 0.313\) from previous steps:\[ z = \frac{1.5 - 1.473}{0.313} \approx 0.086 \]Consulting the z-table for \(z = 0.086\), we find \(P(Z < 0.086)\) approximately 0.534. Thus, the estimated probability \(P(X < 1.5)\) is 0.534.
05

Calculate the Standard Error of the Median

The standard error of the median (SE_m) can be estimated as \( SE_m \approx \frac{1.2533 \times s}{\sqrt{n}} \).Using \(s = 0.313\) and \(n = 16\):\[ SE_m = \frac{1.2533 \times 0.313}{\sqrt{16}} = \frac{0.391}{4} \approx 0.098 \]The estimated standard error of the sample median is approximately 0.098.

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Key Concepts

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

Point Estimate
A point estimate provides a single value as an estimate of an unknown population parameter. It's like summarizing lots of data into one representative number. In our case, when estimating the mean coating thickness of the paint, the sample mean serves as our point estimate. This is because the sample mean, denoted as \( \bar{x} \), is calculated by averaging all the measurements. For our data, we calculated the sample mean to be approximately 1.473. This value gives us a tangible figure to use when discussing the average thickness of the paint layer.
Normal Distribution
The normal distribution is fundamental in statistics and vital in data interpretation. It describes a bell-shaped frequency distribution that is symmetrical about the mean. In practical terms, most of the data points cluster around the central peak (mean) and taper off equally towards both ends. This is crucial when making predictions or estimates since many natural phenomena follow a normal distribution.
When we assume our sample is normally distributed, it allows us to apply various statistical tools and techniques confidently. For instance, knowing the data fits a normal distribution lets us estimate probabilities and use the z-table effectively, as seen when estimating the probability \( P(X < 1.5) \). In our original exercise, this normality idea is supported by the normal probability plot.
Sample Median
The sample median is a measure of central tendency, reflecting the middle point of a data set when arranged in ascending order. If the number of observations \( n \) is even, the median is the average of the two central numbers. For our paint thickness data, with 16 observations, the median is found by averaging the 8th and 9th values in the ordered list: \( \frac{1.31 + 1.48}{2} = 1.395 \).
The median is particularly useful because it provides a robust measure of central tendency that is not heavily affected by extreme values or outliers, unlike the mean. When analyzing non-symmetrical data distributions, the median gives a more accurate picture of the dataset's central location.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It quantifies the average distance of each data point from the mean. When the data points are close to the mean, the standard deviation is small; if they are spread out over a wider range, it is larger.
Understanding standard deviation is crucial because it not only tells us about the variability in our dataset but also plays a key role in predicting probabilities and making estimates. For instance, in our paint thickness problem, the sample standard deviation was calculated to be approximately 0.313. This value helps estimate the 90th percentile and assess probabilities like \( P(X<1.5) \). A smaller standard deviation indicates more consistent data, which is often desirable in a manufacturing context.

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Most popular questions from this chapter

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