91Ó°ÊÓ

"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."

When given a normal population,

A \({\rm{100(1 - \alpha )\% }}\)confidence interval

for the mean is given by

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

When the value of \({{\rm{\sigma }}^{\rm{2}}}\)is known.

This is the mentioned \({\rm{Cl}}\) in the exercise.

In order for confidence interval

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

To have width\({\rm{w}}\), \({\rm{n}}\)is the required sample size

\({\rm{n = }}{\left( {{\rm{2}}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\rm{w}}}} \right)^{\rm{2}}}\)

The lower the width \({\rm{w}}\), the larger the number \({\rm{n}}\).

This suggests that

\({\rm{w = 2}}{{\rm{z}}_{{\rm{\alpha /2}}}}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}\)

As a result, if the new width is \({\rm{w/2}}\), the sample size may be calculated using

\(\frac{{\rm{w}}}{{\rm{2}}}{\rm{ = 2}}{{\rm{z}}_{{\rm{\alpha /2}}}}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}\)

As a result, the sample size must be 4 times \({\rm{n}}\)in order to attain a width of \({\rm{w/2}}\). This statement is correct since the following holds true for sample size\({n^\prime }{\rm{ = 4n}}\).

\(\begin{aligned}{l}{w^\prime } &= {\rm{2}}{{\rm{z}}_{{\rm{\alpha /2}}}}\frac{\sigma }{{\sqrt {{n^\prime }} }}\\ &= 2 {{\rm{z}}_{{\rm{\alpha /2}}}}\frac{{\rm{\sigma }}}{{\sqrt {{\rm{4n}}} }} &= \frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times }}\left( {{\rm{2}}{{\rm{z}}_{{\rm{\alpha /2}}}}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}} \right)\\ &= \frac{{\rm{w}}}{{\rm{2}}}\end{aligned}\)

As a result, the sample size \({\rm{n}}\) should be raised by four times.

Assume that \({n^{\prime \prime }}{\rm{ = 25n}}\). \({w^{\prime \prime }}\)Becomes

\(\begin{aligned}{l}{w^{\prime \prime }} &= {\rm{2}}{{\rm{z}}_{{\rm{a/2}}}}\frac{\sigma }{{\sqrt {{n^{\prime \prime }}} }}\\ &= 2 {{\rm{z}}_{{\rm{\alpha /2}}}}\frac{{\rm{\sigma }}}{{\sqrt {{\rm{25n}}} }} &= \frac{{\rm{1}}}{{\rm{5}}}{\rm{ \times }}\left( {{\rm{2}}{{\rm{z}}_{{\rm{\alpha /2}}}}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}} \right)\\ &= \frac{{\rm{w}}}{{\rm{5}}}\end{aligned}\)

So, the width would decrease by a factor of 5.