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. 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 Sample Mean for Part (a)

To estimate the mean value of the coating thickness, use the sample mean formula, \( \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \), where \( n \) is the number of observations and \( x_i \) represents each observation of coating thickness. Here, \( n = 16 \) and the observations are: 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. The mean is calculated as follows: \( \bar{x} = \frac{0.83 + 0.88 + 0.88 + 1.04 + 1.09 + 1.12 + 1.29 + 1.31 + 1.48 + 1.49 + 1.59 + 1.62 + 1.65 + 1.71 + 1.76 + 1.83}{16} = 1.389 \).

02

Median Calculation for Part (b)

To find the median, order the observations and find the middle value(s). Since there are 16 observations, the median is the average of the 8th and 9th values. First, reorder the values if necessary: 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. The 8th and 9th values are 1.31 and 1.48, so the median is \( \text{Median} = \frac{1.31 + 1.48}{2} = 1.395 \).

03

Determine 90th Percentile for Part (c)

To find the 90th percentile, or the point where the largest 10% of values begin, express it in terms of the mean \( \mu \) and standard deviation \( \sigma \). Thus, calculate the z-score corresponding to the 90th percentile, \( z = 1.2816 \). The formula for a percentile in a normal distribution is \( x = \mu + z \sigma \). Use the sample mean (\( 1.389 \)) and estimate the sample standard deviation \( s \) with \( s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n}(x_i-\bar{x})^2} = 0.301 \). The 90th percentile: \( x = 1.389 + 1.2816 \times 0.301 = 1.775 \).

04

Estimate Probability for Part (d)

To estimate \( P(X<1.5) \), calculate the z-score for 1.5 using \( z = \frac{1.5 - \bar{x}}{s} = \frac{1.5 - 1.389}{0.301} \approx 0.369 \). Using a standard normal distribution table or calculator, find that \( P(Z < 0.369) \approx 0.644 \).

05

Standard Error for Part (e)

The standard error of the median for a normal distribution can be approximated using \( SE = \frac{\sigma}{\sqrt{n}} \sqrt{\frac{\pi}{2}} \approx \frac{0.301}{\sqrt{16}} \sqrt{\frac{\pi}{2}} \approx 0.073 \).

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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 technique used to calculate a single value from sample data to serve as a best guess for an unknown population parameter. For example, in the given exercise, we perform point estimation to find both the mean and median of the coating thickness distribution. In part (a) of the exercise, the sample mean serves as a point estimate for the population mean. The sample mean is calculated using the formula:

• \( \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \)

where \(n\) is the total number of observations, and \(x_i\) represents each observation. This method provides an efficient point estimate for the average coating thickness in this sample set.
In general, point estimation allows us to make informed guesses about population parameters using sample data, making it a crucial part of statistical analysis.

Standard Deviation

Standard deviation is a measure of the spread or dispersion of a set of data points. It shows us how much individual data points deviate from the mean of the dataset. In the exercise, the sample standard deviation \(s\) is calculated to understand the variability in coating thickness. The formula used for finding sample standard deviation is as follows:

• \( s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n}(x_i-\bar{x})^2} \)

Here, \(x_i\) is each data point, \(\bar{x}\) is the sample mean, and \(n\) is the number of observations.
Standard deviation is crucial in determining how "spread out" the data are from the mean and assists in calculating other values, such as the z-scores and percentiles, which are explored in this exercise.

Z-score

Understanding z-scores is essential for interpreting where a particular observation stands within a distribution. A z-score measures how many standard deviations an element is from the mean. It helps us standardize scores from different datasets, making comparisons possible. In the exercise, z-scores are used to find percentiles and probabilities.
The formula for calculating a z-score is:

• \( z = \frac{x - \mu}{\sigma} \)

where \(x\) is the value of interest, \(\mu\) is the mean of the distribution, and \(\sigma\) is the standard deviation. Given that \(\mu\) and \(\sigma\) are often unknown in practice, they can be estimated using the sample mean \(\bar{x}\) and sample standard deviation \(s\).
Z-scores enable us to work with the standard normal distribution to find probabilities and percentiles, as demonstrated in parts (c) and (d) of the exercise.

Percentile

A percentile indicates the relative standing of a value within a dataset. The nth percentile is the value below which n percent of the observations fall. In the exercise, the 90th percentile is calculated to separate the highest 10% of the coating thickness values.
We utilize the standard normal distribution and z-scores to find this percentile. The formula used is:

• \( x = \mu + z \sigma \)

where \(z\) is the z-score corresponding to the desired percentile (in this case, the 90th percentile), \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
Percentiles provide a way to understand the distribution of data points among a population, and are particularly useful in determining thresholds or cut-off values in a dataset.

Standard Error

Standard error measures the variability of a sample statistic, such as the mean, from the actual population parameter. It tells us how much the sample mean (or other statistic) is expected to vary from the true population mean. In the context of the exercise, the standard error of the median is approximated using the given data.
The formula to estimate the standard error of the median in a normal distribution is:

• \( SE = \frac{\sigma}{\sqrt{n}} \sqrt{\frac{\pi}{2}} \)

where \(\sigma\) is the standard deviation, \(n\) is the sample size, and \(\pi\) is a constant that enhances the approximation for the median.
Understanding the standard error enables us to assess how reliable our sample estimations are, and it plays a significant role in inferential statistics and hypothesis testing.