Question

Determine the confidence level for each of the following large-sample one-sided confidence bounds:

\(\begin{array}{l}{\rm{a}}{\rm{.Upperbound:\bar x + }}{\rm{.84s/}}\sqrt {\rm{n}} \\{\rm{b}}{\rm{.Lowerbound:\bar x - 2}}{\rm{.05s/}}\sqrt {\rm{n}} \\{\rm{c}}{\rm{.Upperbound:\bar x + }}{\rm{.67\;s/}}\sqrt {\rm{n}} \end{array}\)

Short answer

The confidence level are:

a) \({\rm{79}}{\rm{.95\% }}\)

b) \({\rm{97}}{\rm{.98\% }}{\rm{.}}\)

c) \({\rm{74}}{\rm{.86\% }}\)

Step-by-step solution

Given by

For large n, the standardized random variable

\({\rm{Z - }}\frac{{{\rm{\bar X - \mu }}}}{{{\rm{S/}}\sqrt {\rm{n}} }}\)

has approximately a normal distribution with expectation \(0\) and standard deviation \(1\). Therefore, a

large-sample confidence interval for\({\rm{\mu }}\)

\({\rm{\bar x \pm }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}\)

with confidence level of approximately \({\rm{100(1 - \alpha )\% }}\)This stands regardless the population distribution.

By replacing \({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ with }}{{\rm{z}}_{\rm{\alpha }}}\)

the large-sample upper confidence bound for \({\rm{\mu }}\)and the large-sample lower confidence bound for \({\rm{\mu }}\)are obtained.

To Determine the confidence level for \({\rm{Upperbound:\bar x + }}{\rm{.84s/}}\sqrt {\rm{n}} \)

(a) Given upper bound:

\({\rm{\bar x + 0}}{\rm{.84}}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}\)

from

\({{\rm{z}}_{\rm{\alpha }}}{\rm{ - 0}}{\rm{.84,}}\)

where, in this case, the following holds

\(P\left({Z£{z_\alpha}}\right)- 1-\alpha n\& P\left({Z£{z_\alpha}}\right)- P(Z£0.84)\mathop-\limits^{(1)} \Phi (0.94) - 0.7995\)

therefore,

\({\rm{\alpha - 1 - 0}}{\rm{.7995 - 0}}{\rm{.2005}}\)

and the confidence level is

\({\rm{0}}{\rm{.7995 = 79}}{\rm{.95\% }}\)

(1) : from the normal probability table in the appendix. The probability can also be computed with a software.

On the graph you can see the area of \({\rm{1 - \alpha }}\)and the confidence level.

Hence the confidence level is \({\rm{79}}{\rm{.95\% }}\)

To Determine the confidence level for \({\rm{Lowerbound:\bar x - 2}}{\rm{.05s/}}\sqrt {\rm{n}} \)

(b) Given lower bound:

\({\rm{\bar x - 2}}{\rm{.05}}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}\)

from

\({{\rm{z}}_{\rm{a}}}{\rm{ - 2}}{\rm{.05}}\)

where, in this case, the following holds

\(P\left( {Z£{z_\alpha }} \right) - 1 - \alpha n\& P\left( {Z£{z_\alpha }} \right) - P(Z£2.05)\mathop- \limits^{(1)} \Phi (2.05) - 0.9798\)

therefore,

\({\rm{\alpha - 1 - 0}}{\rm{.9798 - 0}}{\rm{.0202}}\)

and the confidence level is

\({\rm{0}}{\rm{.9798 = 97}}{\rm{.98\% }}{\rm{.}}\)

(1) : from the normal probability table in the appendix. The probability can also be computed with a software.

On the graph you can see the area of \({\rm{1 - \alpha }}\)and the confidence level.

Hence the confidence level is \({\rm{97}}{\rm{.98\% }}{\rm{.}}\)

To Determine the confidence level for \({\rm{Upperbound:\bar x + }}{\rm{.67\;s/}}\sqrt {\rm{n}} \)

(c) Given upper bound:

\({\rm{\bar x + 0}}{\rm{.67}}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}\)

from

\({{\rm{z}}_{\rm{\alpha }}}{\rm{ - 0}}{\rm{.67,}}\)

where, in this case, the following holds

\(P\left( {Z£{z_\alpha }} \right) - 1 - \alpha n\& P\left( {Z£{z_\alpha }} \right) - P(Z£0.67)\mathop - \limits^{(1)} \Phi (0.67) - 0.7486\)

therefore,

\({\rm{\alpha - 1 - 0}}{\rm{.7486 - 0}}{\rm{.2514}}\)

and the confidence level is

\({\rm{0}}{\rm{.7486 = 74}}{\rm{.86\% }}\)

(1) : from the normal probability table in the appendix. The probability can also be computed with a software.

On the graph you can see the area of \({\rm{1 - \alpha }}\)and the confidence level.

Hence the confidence level is \({\rm{74}}{\rm{.86\% }}\)