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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)?

Step by step solution

01

Calculate the Sample Mean

The sample mean is a point estimate of the mean value of the coating thickness. To find the sample mean, sum up all the observed values and divide by the number of observations:\[\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} = \frac{0.83 + 0.88 + 0.88 + 1.04 + 1.09 + 1.12 + 1.29 + 1.31 + 1.48 + 1.49 + 1.59 + 1.62 + 1.65 + 1.71 + 1.76 + 1.83}{16}\]Calculating, we get \(\bar{x} \approx 1.3381\). We use the sample mean (\(\bar{x}\)) as the estimator.

02

Calculate the Sample Median

To estimate the median, we find the middle value(s) of the ordered data set:1. Order the 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, 1.83.2. Find the median position, which is the \((n+1)/2\)th observation. For \(n=16\), the median is the average of the 8th and 9th values.The 8th and 9th values are 1.31 and 1.48, respectively:\[\text{Median} = \frac{1.31 + 1.48}{2} = 1.395\]The sample median is used as the estimator.

03

Calculate the 90th Percentile

To find the 90th percentile of a normal distribution with known mean \(\mu\) and standard deviation \(\sigma\), we use the Z-table. Express the percentile position as \(Z\), and find that \(P(Z < z_{0.9}) = 0.9\).Since \(Z_{0.9} \approx 1.2816\) for \(P(Z < Z_{0.9}) = 0.9\), the 90th percentile (denoted as \(x_{0.9}\)) can be expressed as:\[x_{0.9} = \mu + z_{0.9} \cdot \sigma\]

04

Calculate P(X < 1.5)

Estimate \(P(X < 1.5)\) by finding the Z-score of \(1.5\):\[Z = \frac{1.5 - \bar{x}}{s} = \frac{1.5 - 1.3381}{s}\]Using the sample standard deviation \(s\), compute the Z-score, look up the cumulative probability associated with \(Z\) in the standard normal distribution table.

05

Calculate Standard Error of the Median

The standard error of the median requires computing based on the binomial distribution since the median involves order statistics. It can be approximated for large samples by \(\sigma/\sqrt{n}\), where \(\sigma\) is the sample standard deviation, and \(n\) is the sample size, as median variance approaches that of the mean in large samples.Using the sample obtained \(\sigma\approx 0.30\), Standard Error is:\[SE \approx \frac{0.30}{\sqrt{16}} \approx 0.075\].

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Key Concepts

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

Point Estimation

Point estimation is a method in statistics used to provide a single value guess for an unknown parameter within a population. It's designed to estimate population parameters like the mean or median, using sample data.
The key here is choosing an estimator, a function of the sample data, which provides the best estimation of the population parameter. For instance, if you're aiming to find the mean coating thickness from a sample, the sample mean is normally used as the point estimate.
The estimator is chosen for its favorable properties such as being unbiased, consistent, and efficient. This means it hopefully hits the true population parameter, provides close approximations consistently, and does so with minimal spread (variance) from the true value. Point estimators are critical for statistical inference as they serve the foundational building block for a variety of analyses.

Sample Mean

The sample mean, denoted \(\bar{x}\), is the average of all the measurements or observations in a sample. It serves as a point estimate for the population mean \(\mu\). To calculate it, add up all sample values and divide the result by the number of observations.

• Mathematically, \(\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}\).
• This method is linear and simple, making it widely applicable and reliable.
• It helps summarize the data set with one central value, making it a straightforward estimator.

However, the sample mean is only an estimator because we don't have data from the entire population. Yet, for normally distributed data, as the sample size increases, \(\bar{x}\) becomes more accurate at estimating \(\mu\) due to the central limit theorem.

Probability Distribution

A probability distribution describes how the values of a random variable are distributed. For coating thickness, imagine each outcome's likelihood represented on a graph.
In real-world scenarios, we often assume normal distribution due to the central limit theorem. It's represented by a bell-shaped curve on a graph. The mean of the distribution is the peak's center.
This pattern is significant because it allows statisticians to apply specific methods and formulas like the Z-score. For instance, to find where most coating thickness values lie, or to identify outliers. By modeling data this way, we can also predict the likelihood of various outcomes. For example, knowing the probability distribution can help estimate the probability of coating thickness being less than a specific value, like 1.5.

Standard Error

The standard error (SE) measures the accuracy with which a sample represents a population. It's the standard deviation of the sample mean's distribution. In simpler terms, it tells us how far the sample mean is likely to be from the true population mean.
It's calculated by dividing the sample's standard deviation by the square root of the sample size, \(SE = \frac{s}{\sqrt{n}}\).

• SE decreases as the sample size increases, indicating more precise mean estimates.
• It's vital for constructing confidence intervals and hypothesis tests, as it gauges estimate reliability.

Understanding SE is crucial for statistical analysis as it tells you how "noisy" a sample estimate might be. Smaller standard errors suggest the sample mean is a reliable estimate of the population mean, whereas larger ones indicate less confidence.