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Chapter 6: Problem 25

The shear strength of each of ten test spot welds is determined, yielding the following data (psi): \(\begin{array}{llllllllll}392 & 376 & 401 & 367 & 389 & 362 & 409 & 415 & 358 & 375\end{array}\) a. Assuming that shear strength is normally distributed, estimate the true average shear strength and standard deviation of shear strength using the method of maximum likelihood. b. Again assuming a nomal distribution, estimate the strength value below which \(95 \%\) of all welds will have their strengths. [Hint: What is the 95 th percentile in terms of \(\mu\) and \(\sigma\) ? Now use the invariance principle.]

Short Answer

Expert verified
Estimated mean shear strength: 384.4 psi, standard deviation: 15.09 psi. The 95th percentile exceeds 409.17 psi.

Step by step solution

01

Organize and Understand the Data

We are given ten shear strength values (in psi): 392, 376, 401, 367, 389, 362, 409, 415, 358, and 375. These are considered a sample from a normally distributed population. We are tasked with estimating the population parameters 鈥� the true mean (\(\mu\)) and standard deviation (\(\sigma\)).
02

Calculate the Sample Mean

The sample mean is calculated as \( \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \), where \( n = 10 \) and \( x_i \) are the sample values. Compute: \[ \bar{x} = \frac{392 + 376 + 401 + 367 + 389 + 362 + 409 + 415 + 358 + 375}{10} = 384.4. \] Thus, the sample mean (\(\mu\)) is 384.4 psi.
03

Calculate the Sample Standard Deviation

The sample standard deviation (\( s \)) is calculated using the formula \( s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} \). Compute each term \((x_i - \bar{x})^2\), sum them, and divide by \(n-1\): \[ s = \sqrt{\frac{1}{9} ((392 - 384.4)^2 + (376 - 384.4)^2 + ... + (375 - 384.4)^2)} = \sqrt{227.8222} = 15.09. \] Thus, \(\sigma\) is estimated to be 15.09 psi.
04

Estimate the 95th Percentile for Shear Strength

Since shear strength is normally distributed, the 95th percentile can be calculated as \( \mu + z_{0.95} \cdot \sigma \), where \( z_{0.95} \) is the z-score for the 95th percentile, which is approximately 1.645. Substitute \( \mu = 384.4 \) and \( \sigma = 15.09 \): \[ x_{0.95} = 384.4 + 1.645 \times 15.09 = 409.17. \] The 95th percentile of the shear strength is estimated to be about 409.17 psi.

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Key Concepts

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

Maximum Likelihood Estimation
Maximum Likelihood Estimation (MLE) is a method used in statistics to estimate the parameters of a statistical model. Here, we wish to find the best estimates for the mean (\(\mu\)) and standard deviation (\(\sigma\)) of our normally distributed data. MLE works by maximizing the likelihood function for the given data. This likelihood function represents how likely it is to observe the given sample data for different parameter values.In the context of a normal distribution, the MLE approach tells us that the sample mean (\(\bar{x}\)) is the best estimate for (\(\mu\)). Similarly, the sample standard deviation (\(s\)) is used to estimate (\(\sigma\)), which is equivalent to the Maximum Likelihood Estimate for the standard deviation when the sample size is large.MLE is popular because it provides estimates with nice statistical properties such as consistency, meaning that as the sample size grows, the MLE estimates converge to the true parameter values.
Sample Mean
The sample mean is a fundamental concept in statistics and is denoted by (\(\bar{x}\)). It serves as an estimate of the population mean (\(\mu\)) when the population is normally distributed. Calculating the sample mean involves summing all the observed data values and dividing by the number of observations, denoted by (\(n\)).For our data, the formula is:
  • \[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \]
  • In our exercise, with values 392, 376, ..., 375, the calculation results in (\(\bar{x} = 384.4\) psi).
The sample mean provides a central value for the data and indicates where most data points are located. This makes it an important measure for summarizing data.
Standard Deviation
The standard deviation (\(\sigma\) or \(s\)) expresses the dispersion or spread of data points relative to the mean. It provides insight into how much the data varies. A smaller standard deviation indicates data points are close to the mean, while a larger standard deviation suggests more spread.For calculation, the sample standard deviation is determined by:
  • The formula: \[ s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} \]
  • We compute each squared difference from the mean, sum them, and then divide by (\(n-1\)).
  • This exercise results in (\(s = 15.09\) psi).
The standard deviation is key in understanding variability and for conducting further statistical analysis, like confidence intervals and hypothesis tests.
Percentile Calculation
Percentile calculation determines the value below which a given percentage of observations fall in a dataset. For normally distributed data, this involves using the z-score, which relates to standard deviations away from the mean.To find a particular percentile, like the 95th, perform the following:
  • Identify the corresponding z-score. For the 95th percentile, (\(z_{0.95} = 1.645\)).
  • Use the formula: \[ x_{0.95} = \mu + z_{0.95} \cdot \sigma \]
  • Substituting our estimates (\(\mu = 384.4\)) and (\(\sigma = 15.09\)), the 95th percentile value becomes approximately (\(x_{0.95} = 409.17\) psi).
This calculation is valuable in quality control or risk assessment, helping identify threshold values for future analysis or decision-making.

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Most popular questions from this chapter

Two different computer systems are monitored for a total of \(n\) weeks. Let \(X_{i}\) denote the number of breakdowns of the first system during the \(i\) th week, and suppose the \(X_{i}\) s are independent and drawn from a Poisson distribution with parameter \(\lambda_{1}\). Similarly, let \(Y_{i}\) denote the number of breakdowns of the second system during the \(i\) th week, and assume independence with each \(Y_{i}\) Poisson with parameter \(\lambda_{2}\). Derive the mle's of \(\lambda_{1}, \lambda_{2}\), and \(\lambda_{1} \quad \lambda_{2}\). [IIint: Using independence, write the joint pmf (likelihood) of the \(X_{i} \mathrm{~s}\) and \(Y_{i}\) s together.]

As an example of a situation in which several different statistics could reasonably be used to calculate a point estimate, consider a population of \(N\) invoices. Associated with each invoice is its "book value," the recorded amount of that invoice. Let \(T\) denote the total book value, a known amount. Some of these book values are erroneous. An audit will be carried out by randomly selecting \(n\) invoices and determining the audited (correct) value for each one. Suppose that the sample gives the following results (in dollars). Let $$ \begin{aligned} &\bar{Y}=\text { sample mean book value } \\ &\bar{X}=\text { sample mean audited value } \\ &\bar{D}=\text { sample mean error } \end{aligned} $$ Propose three different statistics for estimating the total audited (i.e., correct) value-one involving just \(N\) and \(\bar{X}\), \(\frac{\text { another involving } T, N \text {, and } \bar{D} \text {, and the last involving } T \text { and }}{}\) \(\bar{X} / \bar{Y}\). If \(N=5000\) and \(T=1,761,300\), calculate the three corresponding point estimates. (The article "Statistical Models and Analysis in Auditing," Statistical Science, 1989: 2-33). discusses properties of these estimators.)

The article from which the data of Exercise 1 was extracted also gave the accompanying strength observations for cylinders: \(\begin{array}{rrrrrrrrrr}6.1 & 5.8 & 7.8 & 7.1 & 7.2 & 9.2 & 6.6 & 8.3 & 7.0 & 8.3 \\ 7.8 & 8.1 & 7.4 & 8.5 & 8.9 & 9.8 & 9.7 & 14.1 & 12.6 & 11.2\end{array}\) Prior to obtaining data, denote the beam strengths by \(X_{1}, \ldots, X_{m}\) and the cylinder strengths by \(Y_{1}, \ldots, Y_{n^{*}}\). Suppose that the \(X_{i}\) s constitute a random sample from a distribution with mean \(\mu_{1}\) and standard deviation \(\sigma_{1}\) and that the \(Y_{i} \mathrm{~s}\) form a random sample (independent of the \(X_{i} \mathrm{~s}\) ) from another distribution with mean \(\mu_{2}\) and standard deviation \(\sigma_{2^{\circ}}\) a. Use rules of expected value to show that \(\bar{X}^{2}-\bar{Y}\) is an unbiased estimator of \(\mu_{1}-\mu_{2}\). Calculate the estimate for the given data. b. Use rules of variance from Chapter 5 to obtain an expression for the variance and standard deviation (standard error) of the estimator in part (a), and then compute the estimated standard error. c. Calculate a point estimate of the ratio \(\sigma_{1} / \sigma_{2}\) of the two standard deviations. d. Suppose a single beam and a single cylinder are randomly selected. Calculate a point estimate of the variance of the difference \(X-Y\) between beam strength and cylinder strength.

Suppose a certain type of fertilizer has an expected yield per acre of \(\mu_{1}\) with variance \(\sigma^{2}\), whereas the expected yield for a second type of fertilizer is \(\mu_{2}\) with the same variance \(\sigma^{2}\). Let \(S_{1}^{2}\) and \(S_{2}^{2}\) denote the sample variances of yields based on sample sizes \(n_{1}\) and \(n_{2}\), respectively, of the two fertilizers. Show that the pooled (combined) estimator $$ \hat{\sigma}^{2}=\frac{\left(n_{1}-1\right) S_{1}^{2}+\left(n_{2}-1\right) S_{2}^{2}}{n_{1}+n_{2}-2} $$ is an unbiased estimator of \(\sigma^{2}\).

Consider a random sample \(X_{1}, X_{2}, \ldots, X_{n}\) from the shifted exponential pdf $$ f(x ; \lambda, \theta)=\left\\{\begin{array}{cl} \lambda e^{-\lambda(x-\theta)} & x \geq \theta \\ 0 & \text { otherwise } \end{array}\right. $$ Taking \(\theta=0\) gives the pdf of the exponential distribution considered previously (with positive density to the right of zero). An example of the shifted exponential distribution appeared in Example \(4.5\), in which the variable of interest was time headway in traffic flow and \(\theta=.5\) was the minimum possible time headway. a. Obtain the maximum likelihood estimators of \(\theta\) and \(\lambda\). b. If \(n=10\) time headway observations are made, resulting in the values \(3.11, .64,2.55,2.20,5.44,3.42,10.39,8.93\), \(17.82\), and \(1.30\), calculate the estimates of \(\theta\) and \(\lambda\).
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