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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

Determine the t critical value for a lower or an upper confidence bound for each of the situations

\(\begin{array}{*{20}{l}}{{\rm{a}}{\rm{. Confidence level = 95\% , df = 10}}}\\{{\rm{b}}{\rm{. Confidence level = 95\% , df = 15}}}\\{{\rm{c}}{\rm{. Confidence level = 99\% , df = 15}}}\\{{\rm{d}}{\rm{. Confidence level = 99\% , n = 5}}}\\{{\rm{\;e}}{\rm{. Confidence level = 98\% , df = 24}}}\\{{\rm{f}}{\rm{. Confidence level = 99\% , n = 38}}}\end{array}\)

A sample of 14 joint specimens of a particular type gave a sample mean proportional limit stress of \({\rm{8}}{\rm{.48}}\)MPa and a sample standard deviation of . \({\rm{79}}\)MPa a. Calculate and interpret a \({\rm{95\% }}\)lower confidence bound for the true average proportional limit stress of all such joints. What, if any, assumptions did you make about the distribution of proportional limit stress?

b. Calculate and interpret a \({\rm{95\% }}\)lower prediction bound for the proportional limit stress of a single joint of this type.

The technology underlying hip replacements has changed as these operations have become more popular . Starting in 2003, highly durable ceramic hips were marketed. Unfortunately, for too many patients the increased durability has been counterbalanced by an increased incidence of squeaking. The May 11, 2008, issue of the New York Times reported that in one study of 143 individuals who received ceramic hips between 2003 and 2005, 10 of the hips developed squeaking.

a. Calculate a lower confidence bound at the \({\rm{95\% }}\)confidence level for the true proportion of such hips that develop squeaking.

b. Interpret the \({\rm{95\% }}\)confidence level used in (a).

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?)

Consider the next \({\rm{1000}}\) 95% CIs for m that a statistical consultant will obtain for various clients. Suppose the data sets on which the intervals are based are selected independently of one another. How many of these\({\rm{1000}}\)intervals do you expect to capture the corresponding value of m? What is the probability that between \({\rm{940 and 960}}\)Do these intervals contain the corresponding value of m? (Hint: Let Y = the number among the \({\rm{1000}}\) intervals that contain m. What kind of random variable is Y?)
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