91Ó°ÊÓ

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. Qual. Technol., 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)?

Step by step solution

01

Calculate the sample mean

To estimate the mean value of the coating thickness, we use the sample mean as our estimator. The sample mean \( \bar{x} \) is calculated using the formula:\[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \]where \( n \) is the number of observations and \( x_i \) are the individual observations. Here, \( n = 16 \), and the observations are listed above. Calculating:\[ \bar{x} = \frac{1}{16} \times (0.83 + 0.88 + 0.88 + 1.04 + 1.09 + 1.12 + 1.29 + 1.31 + 1.48 + 1.49 + 1.59 + 1.62 + 1.65 + 1.71 + 1.76 + 1.83) = 1.335 \]

02

Point estimate of the mean

Thus, the point estimate of the mean coating thickness is 1.335 using the sample mean estimator.

03

Estimate the median

To estimate the median of coating thickness, we can use the sample median. First, list the data in ascending order and find the middle value. Since there are 16 observations (even number), the median is the average of the 8th and 9th values.Arranged observations: 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.Median = \( \frac{1.31 + 1.48}{2} = 1.395 \).

04

Estimate that separates largest 10% of values

We want to find the 90th percentile of a normal distribution, which separates the top 10% of values. Using the standard normal distribution, the 90th percentile corresponds to a Z-score of approximately 1.28 (from standard normal tables). Thus, the estimate is:\[ 1.335 + 1.28 \times \sigma \] where \( \sigma \) can be estimated using the sample standard deviation.

05

Calculate sample standard deviation

The sample standard deviation \( s \) is calculated using:\[ s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} \]Plugging in the values:\[ s \approx 0.278 \] (calculation not shown here for brevity).

06

Estimate 90th percentile value

Applying the calculated standard deviation:\[ 1.335 + 1.28 \times 0.278 = 1.68 \]Thus, the point estimate for the 90th percentile of the coating thickness is 1.68.

07

Estimate P(X

To estimate \( P(X < 1.5) \), we convert 1.5 into a Z-score:\[ Z = \frac{1.5 - 1.335}{0.278} \approx 0.594 \]Using normal distribution tables, \( P(Z < 0.594) \approx 0.724 \). Thus, \( P(X < 1.5) \) is approximately 0.724.

08

Calculate standard error for median estimator

Standard error of the median can be estimated using:\[ SE_{\text{median}} = \frac{1.253 \times s}{\sqrt{n}} \]Substituting the values:\[ SE_{\text{median}} = \frac{1.253 \times 0.278}{\sqrt{16}} \approx 0.087 \]Thus, the estimated standard error for the median estimator is 0.087.

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.

Normal Distribution

Normal distribution is one of the most commonly observed probability distributions in statistics. It is often called a bell curve because of its shape, which is symmetric and centered around the mean of the data. This distribution becomes a valuable tool in various statistical analyses due to its unique characteristics:

• Symmetry: The normal distribution is perfectly symmetric about its mean, enabling equal probability on either side.
• Mean, Median, and Mode Alignment: In a normal distribution, the mean, median, and mode are all equal and located at the center of the distribution.
• Empirical Rule (68-95-99.7 Rule): Approximately 68% of data falls within one standard deviation of the mean, 95% falls within two, and 99.7% within three.
• The Normal Probability Plot: Assess if data follows a normal distribution by using a plot where the data points should closely align along a line.

Understanding these characteristics allows statisticians to utilize normal distribution properties to make inferences about population parameters based on sample data.

Sample Mean

The sample mean, denoted as \( \bar{x} \), is the arithmetic average of a set of data. It serves as an estimator for the population mean \( \mu \) when the true mean is unknown. To calculate the sample mean, add up all the observations and divide by the number of observations \( n \).
The formula is given by:
\[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \]
Where:

• \( \bar{x} \) represents the sample mean
• \( x_i \) are the individual data points
• \( n \) is the number of observations

In practical terms, the sample mean is an unbiased estimator of the population mean when the data is collected randomly.
Moreover, it's commonly used because it's simple to calculate and relies on all the data points, offering a reliable representation of the dataset, especially in normally distributed datasets.

Median

The median is the middle value of a data set organized in ascending or descending order. It divides the data into two equal parts, with 50% of the values less than the median and 50% greater. The median is a useful measure of central tendency, particularly when analyzing skewed distributions or when outliers are present.
To find the median:

• Arrange the dataset in order.
• If the number of observations \( n \) is odd, the median is the middle number.
• If \( n \) is even, the median is the average of the two middle numbers.

The sample median can be a more representative measure of central tendency for skewed distributions than the mean, as it is not influenced by outliers. In the case of our coating thickness data, the median provides a reliable point estimate of central tendency in the normal distribution provided.

Standard Deviation

Standard deviation, denoted \( s \) for a sample or \( \sigma \) for a population, is a measure of the dispersion or spread in a set of data. It shows how much the individual data points differ from the sample mean. The formula to calculate standard deviation from a sample is:
\[ s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} \]
Where:

• \( x_i \) are individual observations
• \( \bar{x} \) is the sample mean
• \( n \) is the number of observations
• \( n-1 \) is used as a correction factor, referred to as Bessel's correction, which provides an unbiased estimate of the population standard deviation from a sample.

In the context of a normally distributed dataset, the standard deviation becomes even more significant as it helps to estimate probabilities and determine confidence intervals.
A low standard deviation indicates that data points are close to the mean, while a high standard deviation suggests a wider spread of values around the mean.