91影视

"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."

(a)

Consider the given,

\({\rm{P}}\left( {\frac{{{\rm{\bar X - \mu }}}}{{{\rm{\sigma /}}\sqrt {\rm{n}} }}{\rm{ < 1}}{\rm{.645}}} \right){\rm{ = 0}}{\rm{.95}}\)

\({\rm{\mu }}\)has a one-sided interval that can be calculated as

\(\begin{array}{l}\frac{{{\rm{\bar x - \mu }}}}{{{\rm{\sigma /}}\sqrt {\rm{n}} }}{\rm{ < 1}}{\rm{.645}}\\{\rm{\bar x - \mu < 1}}{\rm{.645 \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}\\{\rm{\mu > \bar x - 1}}{\rm{.645 \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}\end{array}\)

or equally

\(\left( {{\rm{\bar x - 1}}{\rm{.645 \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{,}} + \infty } \right)\)

From the exercise\({\rm{5,(a)}}\), for \({\rm{\bar x = 4}}{\rm{.85,\sigma = 0}}{\rm{.75,n = 20}}\), the one-sided interval is

\(\begin{array}{c}\left( {\bar x - 1.645 \cdot \frac{\sigma }{{\sqrt n }}, + \infty } \right) = \left( {4.85 - 1.645 \cdot \frac{{0.75}}{{\sqrt {20} }}, + \infty } \right)\\ = (4.5741, + \infty ).\end{array}\)

Thus, Interval for the data \(\left( {\bar x - 1.645 \cdot \frac{\sigma }{{\sqrt n }}, + \infty } \right);(4.5741, + \infty );\)

(b)

Generally, from

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

For \({\rm{\mu }}\), an universal one-sided confidence interval with a confidence level of \({\rm{100(1 - \alpha )\% }}\) percent can be calculated as follows:

\(\begin{array}{c}\frac{{{\rm{\bar x - \mu }}}}{{{\rm{\sigma /}}\sqrt {\rm{n}} }}{\rm{ < }}{{\rm{z}}_{\rm{\alpha }}}\\{\rm{\bar x - \mu < }}{{\rm{z}}_{\rm{\alpha }}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}\\{\rm{\mu > \bar x - }}{{\rm{z}}_{\rm{\alpha }}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}\end{array}\)

or equally

Thus, the result is \(\left( {\bar x - {{\rm{z}}_{\rm{\alpha }}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{,}} + \infty } \right) \cdot \)

(c)

Consider the formula,

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

For\({\rm{\mu }}\), an universal one-sided confidence interval with a confidence level of \({\rm{100(1 - \alpha )\% }}\) percent can be calculated as follows:

\(\begin{array}{l}{{\rm{z}}_{\rm{\alpha }}}{\rm{ < }}\frac{{{\rm{\bar x - \mu }}}}{{{\rm{\sigma /}}\sqrt {\rm{n}} }}\\{{\rm{z}}_{\rm{\alpha }}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{ < \bar x - \mu }}\\{\rm{\mu < \bar x + }}{{\rm{z}}_{\rm{\alpha }}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}\end{array}\)

or equally

\(\left( { - \infty ,\bar x + {z_{\rm{\alpha }}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}} \right)\)

From the exercise\({\rm{4,(a)}}\), for \({\rm{\bar x = 58}}{\rm{.3,\sigma = 3,n = 25}}\), the one-sided interval is

\(\begin{array}{c}\left( {\bar x - 1.645 \cdot \frac{\sigma }{{\sqrt n }}, + \infty } \right) = \left( { - \infty ,58.3 + 2.33 \cdot \frac{3}{{\sqrt {25} }}} \right)\\ = ( - \infty ,59.7)\end{array}\)

Where

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

and

\({{\rm{z}}_{\rm{\alpha }}}{\rm{ = }}{{\rm{z}}_{{\rm{0}}{\rm{.05}}}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{2}}{\rm{.33}}\)

(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, software can also be used to calculate the probability\(\left( { - \infty ,\bar x + {z_\alpha } \cdot \frac{\sigma }{{\sqrt n }}} \right);( - \infty ,59.7)\)