Question

Let\({{\rm{\alpha }}_{\rm{1}}}{\rm{ > 0,}}{{\rm{\alpha }}_{\rm{2}}}{\rm{ > 0,}}\)with\({{\rm{\alpha }}_{\rm{1}}}{\rm{ + }}{{\rm{\alpha }}_{\rm{2}}}{\rm{ = 0}}\)

\({\rm{P}}\left( {{\rm{ - }}{{\rm{z}}_{{\alpha _{\rm{1}}}}}{\rm{ < }}\frac{{{\rm{\bar X - m}}}}{{{\rm{s/}}\sqrt {\rm{n}} }}{\rm{ < }}{{\rm{z}}_{{\alpha _{\rm{2}}}}}} \right){\rm{ = 1 - }}\alpha \)

a. Use this equation to derive a more general expression for a \({\rm{100(1 - \alpha )\% }}\)CI for \({\rm{\mu }}\)of which the interval (7.5) is a special case.

b. Let \({\rm{\alpha = 0}}{\rm{.5}}\)and \({{\rm{\alpha }}_{\rm{1}}}{\rm{ = \alpha /4,}}\)\({{\rm{\alpha }}_{\rm{2}}}{\rm{ = 3\alpha /4}}{\rm{.}}\)Does this result in a narrower or wider interval than the interval (7.5)?

Short answer

(a) When the value of \({{\rm{\sigma }}^{\rm{2}}}\)is known \(\left( {{\rm{\bar x - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{,\bar x + }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}} \right){\rm{;}}\)

(b) The required result is wider interval.

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."

Finding part (a)

When a normal population is given,

(a)

From

\({\rm{P}}\left( {{\rm{ - }}{{\rm{z}}_{{{\rm{a}}_{\rm{1}}}}}{\rm{ < }}\frac{{{\rm{\bar X - m}}}}{{{\rm{s/}}\sqrt {\rm{n}} }}{\rm{ < }}{{\rm{z}}_{{{\rm{a}}_{\rm{2}}}}}} \right){\rm{ = 1 - a}}\)

By isolating \({\rm{\mu }}\)as follows, the more general expression for \({\rm{100(1 - \alpha )\% }}\)confidence interval for \({\rm{\mu }}\)can be produced.

\(\begin{array}{c} - {z_{{\alpha _1}}} < \frac{{\bar X - \mu }}{{\sigma /\sqrt n }} < {z_{{\alpha _2}}}\\ - {z_{{\alpha _1}}}\frac{\sigma }{{\sqrt n }} < \bar X - \mu < {z_{{\alpha _2}}}\frac{\sigma }{{\sqrt n }}\\\bar X - {z_{{\alpha _2}}}\frac{\sigma }{{\sqrt n }} < \mu < \bar X + {z_{{\alpha _1}}}\frac{\sigma }{{\sqrt n }}\end{array}\)

As a result, when a normal population is provided,

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

for the mean is given by:

\(\left( {{\rm{\bar x - }}{{\rm{z}}_{{{\rm{\alpha }}_{\rm{2}}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{,\bar x + }}{{\rm{z}}_{{{\rm{\alpha }}_{\rm{1}}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}} \right)\)

Thus , when the value of \({{\rm{\sigma }}^{\rm{2}}}\)is known.

  Does this result in a narrower or wider interval

(b)

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.

The \({\rm{100(1 - \alpha )\% = 95\% }}\)confidence interval is

\(\begin{array}{l}\left( {{\rm{\bar x - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{,\bar x + }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}} \right)\\{\rm{ = }}\left( {{\rm{\bar x - 1}}{\rm{.96 \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{,\bar x + 1}}{\rm{.96 \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}} \right)\end{array}\)

Where,

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

(1): this is obtained from

\({\rm{P}}\left( {{\rm{Z > }}{{\rm{z}}_{{\rm{0}}{\rm{.025}}}}} \right){\rm{ = 0}}{\rm{.025}}\)

And from the appendix's normal probability table A software can also be used to calculate the probability.

The length of such an interval is its width.

\({\rm{L = \bar x + 1}}{\rm{.96 \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{ - \bar x + 1}}{\rm{.96 \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{ = 3}}{\rm{.92 \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}\)

The more general \({\rm{100(1 - \alpha )\% = 95\% }}\)confidence interval for \({\rm{\mu }}\)is,

\(\begin{array}{c}\left( {{\rm{\bar x - }}{{\rm{z}}_{{{\rm{\alpha }}_{\rm{2}}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{,\bar x + }}{{\rm{z}}_{{{\rm{\alpha }}_{\rm{1}}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}} \right)\\{\rm{ = }}\left( {{\rm{\bar x - }}{{\rm{z}}_{\scriptstyle{\rm{3}}\atop\scriptstyle{\rm{ \times 0}}{\rm{.05/4}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{,\bar x + }}{{\rm{z}}_{{\rm{0}}{\rm{.05/4}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}} \right)\\{\rm{ = }}\left( {{\rm{\bar x - }}{{\rm{z}}_{{\rm{0}}{\rm{.0375}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{,\bar x + }}{{\rm{z}}_{{\rm{0}}{\rm{.0125}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}} \right)\\{\rm{ = }}\left( {{\rm{\bar x - 1}}{\rm{.78 \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{,\bar x + 2}}{\rm{.24 \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}} \right)\end{array}\)

Where,

\(\begin{array}{l}{{\rm{z}}_{{\rm{0}}{\rm{.0375}}}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{1}}{\rm{.78}}\\{{\rm{z}}_{{\rm{0}}{\rm{.0125}}}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{2}}{\rm{.24}}\end{array}\)

(1) : this is obtained from

\(\begin{array}{l}{\rm{P}}\left( {{\rm{Z > }}{{\rm{z}}_{{\rm{0}}{\rm{.0375}}}}} \right){\rm{ = 0}}{\rm{.0375;}}\\{\rm{P}}\left( {{\rm{Z > }}{{\rm{z}}_{{\rm{0}}{\rm{.0125}}}}} \right){\rm{ = 0}}{\rm{.0125;}}\end{array}\)

And from the appendix's normal probability table A software can also be used to calculate the probability.

The length of such an interval is its width

\({L^\prime }{\rm{ = \bar x + 2}}{\rm{.24 \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{ - \bar x + 1}}{\rm{.78 \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{ = (2}}{\rm{.24 + 1}}{\rm{.78) \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{ = 4}}{\rm{.02 \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}\)

Obviously,

\({L^\prime }{\rm{ = 4}}{\rm{.02 \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{ > 3}}{\rm{.92 \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{ = L}}\)

Conclusion: A broader interval is obtained using the more generic equation for a \({\rm{95\% }}\) confidence interval.

The two graphs below show the visual difference.

The 95% confidence interval is as follows:

The 95 percent general confidence interval is as follows:

Thus, the required result is wider interval.