Question

Consider a normal population distribution with the value of \({\rm{\sigma }}\) known. a. What is the confidence level for the interval \({\rm{\bar x \pm 2}}{\rm{.81\sigma /}}\sqrt {\rm{n}} \)? b. What is the confidence level for the interval \({\rm{\bar x \pm 1}}{\rm{.44\sigma /}}\sqrt {\rm{n}} \)? c. What value of \({{\rm{z}}_{{\rm{\alpha /2}}}}\) in the CI formula (\({\rm{7}}{\rm{.5}}\)) results in a confidence level of \({\rm{99}}{\rm{.7\% }}\)? d. Answer the question posed in part (c) for a confidence level of \({\rm{75\% }}\).

Short answer

a.The confidence level is \({\rm{99}}{\rm{.5\% }}\).

b.The confidence level is \({\rm{85\% }}\).

c.The value is \({\rm{2}}{\rm{.96}}\).

d.The value is \({\rm{1}}{\rm{.15}}\).

Step-by-step solution

Define interval

An interval is a set of numbers that includes all the real numbers between the two endpoints of the interval.

Explanation

a.A \({\rm{100(1 - \alpha )\% }}\) confidence interval for the mean is supplied when a normal population is specified.

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

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

The following known connection can be used to get the confidence level\({\rm{100(1 - \alpha )\% }}\).

\({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ = 2}}{\rm{.81}}\)

There are a few different approaches to compute the confidence level, as shown in the diagram below. The area of\({\rm{1 - \alpha }}\)is denoted by the red\({\rm{X}}\). The area to the right of\({{\rm{z}}_{{\rm{\alpha /2}}}}\)(the probability to the right) is\({\rm{\alpha /2}}\), as is the region to the left of\({\rm{ - }}{{\rm{z}}_{{\rm{\alpha /2}}}}\).

The first step is to locate\({\rm{\alpha }}\). Because the probability to the right of\({{\rm{z}}_{{\rm{\alpha /2}}}}\)is\({\rm{\alpha /2}}\)and the population is normally distributed,\({\rm{\alpha /2}}\)can be calculated as the probability previously mentioned.

\(\begin{array}{c}\frac{{\rm{\alpha }}}{{\rm{2}}}{\rm{ = P}}\left( {{\rm{Z > }}{{\rm{z}}_{{\rm{\alpha /2}}}}} \right)\\{\rm{ = 1 - P}}\left( {{\rm{Z}} \le {{\rm{z}}_{{\rm{\alpha /2}}}}} \right)\\{\rm{ = 1 - \Phi (2}}{\rm{.81)}}\end{array}\)

\({\rm{(1)1 - 0}}{\rm{.9975 = 0}}{\rm{.0025}}\)

1. : From the appendix's normal probability table. A software can also be used to calculate the probability.

Therefore,

\(\begin{array}{c}{\rm{\alpha = 2 \times 0}}{\rm{.0025}}\\{\rm{ = 0}}{\rm{.005}}\end{array}\)

and the level of confidence is,

\(\begin{array}{c}{\rm{100(1 - \alpha )\% = 100(1 - 0}}{\rm{.005)}}\\{\rm{ = 99}}{\rm{.5\% }}\end{array}\)

Explanation

a.The following known connection can be used to get the confidence level \({\rm{100(1 - \alpha )\% }}\).

\({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ = 1}}{\rm{.44}}\)

There are a few different approaches to compute the confidence level, as shown in the diagram below. The area of\({\rm{1 - \alpha }}\)is denoted by the blue shade. The area to the right of\({{\rm{z}}_{{\rm{\alpha /2}}}}\)(the probability to the right) is\({\rm{\alpha /2}}\), as is the region to the left of\({\rm{ - }}{{\rm{z}}_{{\rm{\alpha /2}}}}\).

The first step is to locate\({\rm{\alpha }}\). Because the probability to the right of\({{\rm{z}}_{{\rm{\alpha /2}}}}\)is\({\rm{\alpha /2}}\)and the population is normally distributed,\({\rm{\alpha /2}}\)can be calculated as the probability previously mentioned.

\(\begin{array}{c}\frac{{\rm{\alpha }}}{{\rm{2}}}{\rm{ = P}}\left( {{\rm{Z > }}{{\rm{z}}_{{\rm{\alpha /2}}}}} \right)\\{\rm{ = 1 - P}}\left( {{\rm{Z}} \le {{\rm{z}}_{{\rm{\alpha /2}}}}} \right)\\{\rm{ = 1 - \Phi (1}}{\rm{.44)}}\end{array}\)

\({\rm{(1)1 - 0}}{\rm{.925 = 0}}{\rm{.075}}\)

1. : From the appendix's normal probability table. A software can also be used to calculate the probability.

Therefore,

\(\begin{array}{c}{\rm{\alpha = 2 \times 0}}{\rm{.075}}\\{\rm{ = 0}}{\rm{.15}}\end{array}\)

and the level of confidence is,

\(\begin{array}{c}{\rm{100(1 - \alpha )\% = 100(1 - 0}}{\rm{.15)}}\\{\rm{ = 85\% }}\end{array}\)

Explanation

c. Because the confidence level is \({\rm{99}}{\rm{.7\% }}\), the \ ({\rm{\alpha }}\) may be found from the following relationship.

\({\rm{100(1 - \alpha ) = 99}}{\rm{.7}}\)

This suggests that

\({\rm{\alpha = 0}}{\rm{.003}}{\rm{.}}\)

The point on the diagram below is the value of\({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ = }}{{\rm{z}}_{{{\rm{\alpha }}_{{\rm{0}}{\rm{.0015}}}}}}\). The region (probability) to the right of the value is\({\rm{\alpha /2 = 0}}{\rm{.0015}}\).

Therefore,

\({\rm{P}}\left( {{\rm{Z > }}{{\rm{z}}_{{\rm{0}}{\rm{.0015}}}}} \right){\rm{ = 0}}{\rm{.0015}}\)

and

\(\begin{array}{c}{\rm{P}}\left( {{\rm{Z > }}{{\rm{z}}_{{\rm{0}}{\rm{.0015}}}}} \right){\rm{ = 1 - P}}\left( {{\rm{Z}} \le {{\rm{z}}_{{\rm{0}}{\rm{.0015}}}}} \right)\\{\rm{ = 1 - \Phi }}\left( {{{\rm{z}}_{{\rm{0}}{\rm{.0015}}}}} \right)\end{array}\)

suggest that

\(\begin{array}{c}{\rm{1 - \Phi }}\left( {{{\rm{z}}_{{\rm{0}}{\rm{.0015}}}}} \right){\rm{ = 0}}{\rm{.0015}}\\{\rm{\Phi }}\left( {{{\rm{z}}_{{\rm{0}}{\rm{.0015}}}}} \right){\rm{ = 0}}{\rm{.9985}}\end{array}\)

So, because darkened area (blue) is nearly\({\rm{1}}\), almost everything is blue. It can be seen from the normal probability table in the appendix that

\({{\rm{z}}_{{\rm{0}}{\rm{.0015}}}}{\rm{ = 2}}{\rm{.96}}\)

calculates the probability,

\({\rm{\Phi }}\left( {{{\rm{z}}_{{\rm{0}}{\rm{.0015}}}}} \right){\rm{ = 0}}{\rm{.9985}}\)

Therefore, the value is \({\rm{2}}{\rm{.96}}\).

Explanation

d. Because the confidence level is \({\rm{75\% }}\), the\({\rm{\alpha }}\) may be found from the following relationship.

\({\rm{100(1 - \alpha ) = 75}}\)

This suggests that

\({\rm{\alpha = 0}}{\rm{.25}}\)

The point on the diagram below is the value of\({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ = }}{{\rm{z}}_{{{\rm{\alpha }}_{{\rm{0}}{\rm{.125}}}}}}\). The region (probability) to the right of the value is\({\rm{\alpha /2 = 0}}{\rm{.125}}\).

Therefore,

\({\rm{P}}\left( {{\rm{Z > }}{{\rm{z}}_{{\rm{0}}{\rm{.125}}}}} \right){\rm{ = 0}}{\rm{.125}}\)

and

\(\begin{array}{c}{\rm{P}}\left( {{\rm{Z > }}{{\rm{z}}_{{\rm{0}}{\rm{.125}}}}} \right){\rm{ = 1 - P}}\left( {{\rm{Z}} \le {{\rm{z}}_{{\rm{0}}{\rm{.125}}}}} \right)\\{\rm{ = 1 - \Phi }}\left( {{{\rm{z}}_{{\rm{0}}{\rm{.125}}}}} \right)\end{array}\)

suggest that

\(\begin{array}{c}{\rm{1 - \Phi }}\left( {{{\rm{z}}_{{\rm{0}}{\rm{.125}}}}} \right){\rm{ = 0}}{\rm{.125}}\\{\rm{\Phi }}\left( {{{\rm{z}}_{{\rm{0}}{\rm{.125}}}}} \right){\rm{ = 0}}{\rm{.875}}\end{array}\)

So, because darkened area (blue) is\({\rm{0}}{\rm{.875}}\), almost everything is blue. It can be seen from the normal probability table in the appendix that

\({{\rm{z}}_{{\rm{0}}{\rm{.125}}}}{\rm{ = 1}}{\rm{.15}}\)

calculates the probability,

\({\rm{\Phi }}\left( {{{\rm{z}}_{{{\rm{\alpha }}_{{\rm{0}}{\rm{.125}}}}}}} \right){\rm{ = 0}}{\rm{.875}}\)

Therefore, the value is \({\rm{1}}{\rm{.15}}\).