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Chapter 7: Q1E (page 284)

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

Expert verified

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

01

Define interval

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

02

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,

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

03

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,

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

04

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 \({\rm{2}}{\rm{.96}}\).

05

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 \({\rm{1}}{\rm{.15}}\).

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Most popular questions from this chapter

A manufacturer of college textbooks is interested in estimating the strength of the bindings produced by a particular binding machine. Strength can be measured by recording the force required to pull the pages from the binding. If this force is measured in pounds, how many books should be tested to estimate the average force required to break the binding to within .1 lb with 95% confidence? Assume that s is known to be .8.

Aphid infestation of fruit trees can be controlled either by spraying with pesticide or by inundation with ladybugs. In a particular area, four different groves of fruit trees are selected for experimentation. The first three groves are sprayed with pesticides 1, 2, and 3, respectively, and the fourth is treated with ladybugs, with the following results on yield:

Treatment ni = Number of Trees (Bushels/Tree) si

1 100 10.5 1.5

2 90 10.0 1.3

3 100 10.1 1.8

4 120 10.7 1.6

Let 碌i= the true average yield (bushels/tree) after receiving the ith treatment. Then

\({\bf{\theta = }}\frac{{\bf{1}}}{{\bf{3}}}{\bf{(}}{{\bf{\mu }}_{\bf{1}}}{\bf{ + }}{{\bf{\mu }}_{\bf{2}}}{\bf{ + }}{{\bf{\mu }}_{\bf{3}}}{\bf{) - }}{{\bf{\mu }}_{\bf{4}}}\)

measures the difference in true average yields between treatment with pesticides and treatment with ladybugs. When n1, n2, n3, and n4 are all large, the estimator \(\widehat {\bf{\theta }}\) obtained by replacing each \({{\bf{\mu }}_{\bf{i}}}\)by Xiis approximately normal. Use this to derive a large-sample 100(1 -伪)% CI for \({\bf{\theta }}\), and compute the 95% interval for the given

data.

a. Under the same conditions as those leading to the interval\({\rm{(7}}{\rm{.5),p((}}\overline {\rm{X}} {\rm{ - \mu )/(\sigma /}}\sqrt {\rm{n}} {\rm{) < 1}}{\rm{.645 = }}{\rm{.95}}{\rm{.}}\)Use this to derive a one-sided interval for\({\rm{\mu }}\)that has infinite width and provides a lower confidence bound on m. What is this interval for the data in Exercise 5(a)?

b. Generalize the result of part (a) to obtain a lower bound with confidence level\({\rm{100(1 - \alpha )\% }}\)

c. What is an analogous interval to that of part (b) that provides an upper bound on\({\rm{\mu }}\)? Compute this 99% interval for the data of Exercise 4(a).

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)?

The Pew Forum on Religion and Public Life reported on \({\rm{Dec}}{\rm{. 9, 2009}}\), that in a survey of \({\rm{2003}}\) American adults, \({\rm{25\% }}\)said they believed in astrology.

a. Calculate and interpret a confidence interval at the \({\rm{99\% }}\)confidence level for the proportion of all adult Americans who believe in astrology.

b. What sample size would be required for the width of a \({\rm{99\% }}\)CI to be at most .05 irrespective of the value of p?
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