Question

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

(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

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."

 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 \({\rm{\alpha = 0}}{\rm{.08,}}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ = 1}}{\rm{.75}}\)

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\(\left( {{\rm{8406}}{\rm{.1,8471}}{\rm{.9}}} \right){\rm{;}}\)