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."

If When you give a normal population, you'll get 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.

(a)

The \({\rm{95\% }}\)confidence interval for given values \({\rm{\sigma = 0}}{\rm{.75}}\), n=20\({\rm{,\bar x = 4}}{\rm{.85}}\)is

\(\begin{array}{l}\left( {{\rm{\bar x - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{,\bar x + }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}} \right)\\{\rm{ = }}\left( {{\rm{4}}{\rm{.85 - 1}}{\rm{.96 \times }}\frac{{{\rm{0}}{\rm{.75}}}}{{\sqrt {{\rm{20}}} }}{\rm{,4}}{\rm{.85 + 1}}{\rm{.96 \times }}\frac{{{\rm{0}}{\rm{.75}}}}{{\sqrt {{\rm{20}}} }}} \right)\\{\rm{ = (4}}{\rm{.52,5}}{\rm{.18)}}\end{array}\)

Where,

\(\begin{array}{c}{\rm{100(1 - \alpha ) = 95}}\\{\rm{\alpha = 0}}{\rm{.05,}}\end{array}\)

and

\(\begin{array}{c}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ = }}{{\rm{z}}_{{\rm{0}}{\rm{.05/2}}}}\\{\rm{ = }}{{\rm{z}}_{{\rm{0}}{\rm{.025}}}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{1}}{\rm{.96}}\end{array}\)

(1): this is obtained from\({\rm{P}}\left( {{\rm{Z > }}{{\rm{z}}_{{\rm{0}}{\rm{.025}}}}} \right){\rm{ = 0}}{\rm{.025}}\) and from the normal probability table in the appendix.

As a result, the likelihood can be calculated using software\({\rm{(4}}{\rm{.52,5}}{\rm{.18)}}\).

(b)

The \({\rm{98\% }}\)confidence interval for given values \({\rm{\sigma = 0}}{\rm{.75,n = 16,\bar x = 4}}{\rm{.56}}\)is\(\left( {{\rm{\bar x - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{,\bar x + }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}} \right)\)

\({\rm{ = }}\left( {{\rm{4}}{\rm{.56 - 2}}{\rm{.33 \times }}\frac{{{\rm{0}}{\rm{.75}}}}{{\sqrt {{\rm{16}}} }}{\rm{,4}}{\rm{.56 + 2}}{\rm{.33 \times }}\frac{{{\rm{0}}{\rm{.75}}}}{{\sqrt {{\rm{16}}} }}} \right)\)

\({\rm{ = (4}}{\rm{.12,5)}}\)

Where

\(\begin{aligned}100(1 - \alpha ) &= 98 \\{\rm{\alpha = 0}}{\rm{.02}}\end{aligned}\)

and

\(\begin{array}{c}{z_{\alpha /2}} = {z_{0.02/2}}\\ = {z_{0.01}}\mathop = \limits^{(1)} 2.33\end{array}\)

As a result, the likelihood can be calculated using software\({\rm{(4}}{\rm{.12,5) }}\)

(c)

In order to calculate the 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 \({\rm{w}}\), the necessary sample size \({\rm{n}}\)is

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

The smaller width \({\rm{w}}\), the larger \({\rm{n}}\)must be.

Given \({\rm{w = 0}}{\rm{.4,}}{{\rm{z}}_{{\rm{0}}{\rm{.025}}}}{\rm{ = 1}}{\rm{.96(see}}\left. {{\rm{(a)}}} \right){\rm{,n}}\)has to be

\(\begin{aligned}n &= {\left( {{\rm{2}}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\rm{w}}}} \right)^{\rm{2}}}\\&= {\left( {{\rm{2 \times 1}}{\rm{.96 \times }}\frac{{{\rm{0}}{\rm{.75}}}}{{{\rm{0}}{\rm{.4}}}}} \right)^{\rm{2}}}\\&= 54{\rm{.02}}\end{aligned}\)

Thus, because an integer number is required, round it up to \({\rm{n = 55}}{\rm{.}}\)

(d)

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 \({\rm{w}}\), the necessary sample size \({\rm{n}}\)is

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

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

Given \({\rm{w = 2 \times 0}}{\rm{.2 = 0}}{\rm{.4}}\left( {{\rm{0}}{\rm{.2}}} \right.\)to the left of the mean and \({\rm{0}}{\rm{.2}}\)to the right - within), \({{\rm{z}}_{{\rm{0}}{\rm{.005}}}}{\rm{ = 2}}{\rm{.58,n}}\)has to be

\(\begin{array}{c}{\rm{n = }}{\left( {{\rm{2}}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\rm{w}}}} \right)^{\rm{2}}}{\rm{ = }}{\left( {{\rm{2 \times 2}}{\rm{.58 \times }}\frac{{{\rm{0}}{\rm{.75}}}}{{{\rm{0}}{\rm{.4}}}}} \right)^{\rm{2}}}\\{\rm{ = 93}}{\rm{.61}}\end{array}\)

Because an integer number is required, round it up to the nearest integer

\({\rm{n = 94}}\)The \({{\rm{z}}_{{\rm{\alpha /2}}}}\)is computed from

\(\begin{aligned}100(1 - \alpha ) &= 99\\alpha &= 0{\rm{.01}}\end{aligned}\)

therefore,

\(\begin{gathered}{z_{\alpha /2}} &= {z_{0.01/2}} \\ &= {z_{0.005}}\mathop = \limits^{(1)} 2.58 \\ \end{gathered} \)

(1): this is obtained from

\({\rm{P}}\left( {{\rm{Z > }}{{\rm{z}}_{{\rm{0}}{\rm{.005}}}}} \right){\rm{ = 0}}{\rm{.005}}\)

and from the appendix's normal probability table

Thus, software programme can also be used to calculate the likelihood \({\rm{n = 94;}}\)