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Chapter 7: Q6E (page 283)

On the basis of extensive tests, the yield point of a particular type of mild steel-reinforcing bar is known to be normally distributed with\({\rm{\sigma = 100,}}\)The composition of bars has been slightly modified, but the modification is not believed to have affected either the normality or the value of\({\rm{\sigma }}\).

a. Assuming this to be the case, if a sample of 25 modified bars resulted in a sample average yield point of 8439 lb, compute a 90% CI for the true average yield point of the modified bar.

b. How would you modify the interval in part (a) to obtain a confidence level of 92%?

Short Answer

Expert verified

(a) The probability can also be computed with a software\({\rm{\alpha = 0}}{\rm{.08,}}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ = 1}}{\rm{.75}}\).

(b) Software can also be used to calculate the probability\(\left( {{\rm{8406}}{\rm{.1,8471}}{\rm{.9}}} \right)\).

Step by step solution

01

Concept Introduction

"A (p, 1) tolerance interval (TI) based on a sample is designed in such a way that it includes at least a proportion p of the sampled population with confidence 1; such a TI is sometimes referred to as p-content (1) coverage TI."

02

 Compute a 90% CI for the true average yield point of the modified bar.

When a normal population is given,

A \({\rm{100(1 - \alpha )\% }}\)confidence interval for the mean is given by:

\(\left( {{\rm{\bar x - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{,\bar x + }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}} \right)\)

(a) When you know the value of \({{\rm{\sigma }}^{\rm{2}}}\)

For given values, the \({\rm{90\% }}\) confidence interval \({\rm{\sigma = 100,n = 25,\bar x = 8349}}\) is

\(\begin{aligned}\left( {{\rm{\bar x - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{,\bar x + }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}} \right)\\ &= \left( {{\rm{8349 - 1}}{\rm{.645 \times }}\frac{{{\rm{100}}}}{{\sqrt {{\rm{100}}} }}{\rm{,8349 + 1}}{\rm{.645 \times }}\frac{{{\rm{100}}}}{{\sqrt {{\rm{100}}} }}} \right)\\ &= 8406{\rm{.1,8471}}{\rm{.9)}}\end{aligned}\)

Where

\(\begin{array}{c}{\rm{100(1 - \alpha ) = 90}}\\{\rm{\alpha = 0}}{\rm{.1}}\end{array}\)

and

\({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ = }}{{\rm{z}}_{{\rm{0}}{\rm{.1/2}}}}{\rm{ = }}{{\rm{z}}_{{\rm{0}}{\rm{.05}}}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{1}}{\rm{.645}}\)

(1): this is obtained from

\({\rm{P}}\left( {{\rm{Z > }}{{\rm{z}}_{{\rm{0}}{\rm{.05}}}}} \right){\rm{ = 0}}{\rm{.05}}\)and from the appendix's normal probability table.

Thus, the probability can also be computed with software \({\rm{\alpha = 0}}{\rm{.08,}}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ = 1}}{\rm{.75}}\)

03

How would you modify the interval in part (a)

(b)

Choose a different \({\rm{\alpha }}\)and a different value of \({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{.}}\)

Thus,

\(\begin{aligned}{\rm{100(1 - \alpha ) = 92}}\\{\rm{\alpha = 0}}{\rm{.08}}\end{aligned}\)

and

\(\begin{aligned}{{\rm{z}}_{{\rm{\alpha /2}}}} = {{\rm{z}}_{{\rm{0}}{\rm{.08/2}}}}\\ = {{\rm{z}}_{{\rm{0}}{\rm{.04}}}}\mathop = \limits^{{\rm{(1)}}} {\rm{1}}{\rm{.75}}\end{aligned}\)

(1): from the appendix's normal probability table.

Thus, software can also be used to calculate the probability\(\left( {{\rm{8406}}{\rm{.1,8471}}{\rm{.9}}} \right){\rm{;}}\)

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Most popular questions from this chapter

A random sample of n=\({\rm{15}}\)heat pumps of a certain type yielded the following observations on lifetime (in years):

\(\begin{array}{*{20}{l}}{{\rm{2}}{\rm{.0 1}}{\rm{.3 6}}{\rm{.0 1}}{\rm{.9 5}}{\rm{.1 }}{\rm{.4 1}}{\rm{.0 5}}{\rm{.3}}}\\{{\rm{15}}{\rm{.7 }}{\rm{.7 4}}{\rm{.8 }}{\rm{.9 12}}{\rm{.2 5}}{\rm{.3 }}{\rm{.6}}}\end{array}\)

a. Assume that the lifetime distribution is exponential and use an argument parallel to that to obtain a \({\rm{95\% }}\) CI for expected (true average) lifetime.

b. How should the interval of part (a) be altered to achieve a confidence level of \({\rm{99\% }}\)?

c. What is a \({\rm{95\% }}\)CI for the standard deviation of the lifetime distribution? (Hint: What is the standard deviation of an exponential random variable?)

A normal probability plot of the n=\({\rm{26}}\) observations on escape time shows a substantial linear pattern; the sample mean and sample standard deviation are \({\rm{370}}{\rm{.69 and 24}}{\rm{.36}}\), respectively.

a. Calculate an upper confidence bound for population mean escape time using a confidence level of \({\rm{95\% }}\)

b. Calculate an upper prediction bound for the escape time of a single additional worker using a prediction level of \({\rm{95\% }}\). How does this bound compare with the confidence bound of part (a)?

c. Suppose that two additional workers will be chosen to participate in the simulated escape exercise. Denote their escape times by \({\rm{X27 and X28}}\), and let X new denote the average of these two values. Modify the formula for a PI for a single x value to obtain a PI for X new, and calculate a 95% two-sided interval based on the given escape data.

In a sample of \({\rm{1000}}\) randomly selected consumers who had opportunities to send in a rebate claim form after purchasing a product, \({\rm{250}}\) of these people said they never did so. Reasons cited for their behaviour included too many steps in the process, amount too small, missed deadline, fear of being placed on a mailing list, lost receipt, and doubts about receiving the money. Calculate an upper confidence bound at the \({\rm{95\% }}\)confidence level for the true proportion of such consumers who never apply for a rebate. Based on this bound, is there compelling evidence that the true proportion of such consumers is smaller than \({\rm{1/3}}\)? Explain your reasoning

a.Use the results of Example 7.5 to obtain a 95% lower confidence bound for the parameter 位 of an exponential distribution, and calculate the bound based on the data given in the example.

b.If lifetime X has an exponential distribution, the probability that lifetime exceeds t is P(X>t) = e-位迟. Use the result of part (a) to obtain a 95% lower confidence bound for the probability that breakdown time exceeds 100 min.

a. Under the same conditions as those leading to the interval\({\rm{(7}}{\rm{.5),p((}}\overline {\rm{X}} {\rm{ - \mu )/(\sigma /}}\sqrt {\rm{n}} {\rm{) < 1}}{\rm{.645 = }}{\rm{.95}}{\rm{.}}\)Use this to derive a one-sided interval for\({\rm{\mu }}\)that has infinite width and provides a lower confidence bound on m. What is this interval for the data in Exercise 5(a)?

b. Generalize the result of part (a) to obtain a lower bound with confidence level\({\rm{100(1 - \alpha )\% }}\)

c. What is an analogous interval to that of part (b) that provides an upper bound on\({\rm{\mu }}\)? Compute this 99% interval for the data of Exercise 4(a).
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