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Chapter 7: Q21E (page 276)

In a sample of \({\rm{1000}}\) randomly selected consumers who had opportunities to send in a rebate claim form after purchasing a product, \({\rm{250}}\) of these people said they never did so. Reasons cited for their behaviour included too many steps in the process, amount too small, missed deadline, fear of being placed on a mailing list, lost receipt, and doubts about receiving the money. Calculate an upper confidence bound at the \({\rm{95\% }}\)confidence level for the true proportion of such consumers who never apply for a rebate. Based on this bound, is there compelling evidence that the true proportion of such consumers is smaller than \({\rm{1/3}}\)? Explain your reasoning

Short Answer

Expert verified

Yes, there is compelling evidence.

Step by step solution

01

To explain the reasoning

Denote with

\(\begin{array}{l}{\rm{\bar p - }}\left( {\frac{{{\rm{\hat p + z}}_{{\rm{\alpha /2}}}^{\rm{2}}}}{{{\rm{2n}}}}} \right){\rm{/}}\left( {{\rm{1 + }}\frac{{{\rm{z}}_{{\rm{\alpha /2}}}^{\rm{2}}}}{{\rm{n}}}} \right)\\{\rm{\hat q - 1 - \hat p}}{\rm{.}}\end{array}\)

A confidence interval for a

population proportion

p with confidence level approximately\({\rm{100(1 - \alpha )}}\)is

\(\left( {{\rm{\tilde p - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\sqrt {{\rm{\hat p\hat q/n + z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{/4}}{{\rm{n}}^{\rm{2}}}} }}{{{\rm{1 + z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{/n}}}}{\rm{,\bar p + }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\sqrt {{\rm{\hat p\hat q/n + z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{/4}}{{\rm{n}}^{\rm{2}}}} }}{{{\rm{1 + z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{/n}}}}} \right)\)

This is also called a score\({\rm{Cl for p}}\)

By replacing\({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{with }}{{\rm{z}}_{\rm{\alpha }}}{\rm{ and in \pm }}\) only – or + stand, the large-sample upper confidence bound for\({\rm{\mu }}\)and the large-sample lower confidence bound for\({\rm{\mu }}\)are obtained.

The upper confidence bound at the\(95\% \)confidence level for the true proportion can be computed using

\({\rm{\bar p + }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{,}}\frac{{\sqrt {{\rm{\hat p\hat q/n + z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{/4}}{{\rm{n}}^{\rm{2}}}} }}{{{\rm{1 + z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{/n}}}}{\rm{.}}\)

The values in the formula are

\(\begin{array}{l}{\rm{\bar p - }}\frac{{{\rm{0}}{\rm{.25 + 1}}{\rm{.64}}{{\rm{5}}^{\rm{2}}}{\rm{/2000}}}}{{{\rm{1 + 1}}{\rm{.64}}{{\rm{5}}^{\rm{2}}}{\rm{/1000}}}}{\rm{ - 0}}{\rm{.2507,}}\\{\rm{\hat p - }}\frac{{{\rm{250}}}}{{{\rm{1000}}}}{\rm{ - 0}}{\rm{.25,}}\\{\rm{\hat q - 1 - \hat p - 0}}{\rm{.75,}}\end{array}\)

where

\(\begin{array}{*{20}{r}}{{\rm{100(1 - \alpha ) - 95}}}\\{{\rm{\alpha - 0}}{\rm{.05}}}\end{array}\)

and

\({{\rm{z}}_{\rm{\alpha }}}{\rm{ - }}{{\rm{z}}_{{\rm{0}}{\rm{.05}}}}\mathop {\rm{ - }}\limits^{{\rm{(1)}}} {\rm{1}}{\rm{.645}}\)

(1) : this is obtained from

\({\rm{P}}\left( {{\rm{Z > }}{{\rm{z}}_{{\rm{0}}{\rm{.05}}}}} \right){\rm{ - 0}}{\rm{.05}}\)

and from the normal probability table in the appendix. The probability can also be computed with a software.

02

To find the upper confidence bound

Therefore, the upper confidence bound at the\(95\% \)confidence level for the true proportion of such consumers is

\(\begin{array}{l}{\rm{\bar p + }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\sqrt {{\rm{\hat p\hat q/n + z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{/4}}{{\rm{n}}^{\rm{2}}}} }}{{{\rm{1 + z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{/n}}}}{\rm{ - \& 0}}{\rm{.2507 + }}\frac{{{\rm{1}}{\rm{.645 \times }}\sqrt {{\rm{0}}{\rm{.25 \times 0}}{\rm{.75/1000 + 1}}{\rm{.64}}{{\rm{5}}^{\rm{2}}}{\rm{/}}\left( {{\rm{4 \times 100}}{{\rm{0}}^{\rm{2}}}} \right)} }}{{{\rm{1 + 1}}{\rm{.64}}{{\rm{5}}^{\rm{2}}}{\rm{/1000}}}}\\{\rm{ - 0}}{\rm{.2507 + 0}}{\rm{.0225 - 0}}{\rm{.2732}}{\rm{.}}\end{array}\)

There is\(95\% \)confidence that the true proportion is less than\(0.2732\),which indicated that there is compelling evidence that the true proportion is smaller than\({\rm{1 / 3}}\)The answer is\(0.2732\)

Yes, there is compelling evidence.

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Most popular questions from this chapter

A CI is desired for the true average stray-load loss \({\rm{\mu }}\) (watts) for a certain type of induction motor when the line current is held at \({\rm{10 amps}}\) for a speed of \({\rm{1500 rpm}}\). Assume that stray-load loss is normally distributed with \({\rm{\sigma = 3}}{\rm{.0}}\). a. Compute a \({\rm{95\% }}\) CI for \({\rm{\mu }}\) when \({\rm{n = 25}}\) and \({\rm{\bar x = 58}}{\rm{.3}}\). b. Compute a \({\rm{95\% }}\) CI for \({\rm{\mu }}\) when \({\rm{n = 100}}\) and \({\rm{\bar x = 58}}{\rm{.3}}\). c. Compute a \({\rm{99\% }}\) CI for \({\rm{\mu }}\) when \({\rm{n = 100}}\) and \({\rm{\bar x = 58}}{\rm{.3}}\). d. Compute an \({\rm{82\% }}\) CI for \({\rm{\mu }}\) when \({\rm{n = 100}}\) and \({\rm{\bar x = 58}}{\rm{.3}}\). e. How large must n be if the width of the \({\rm{99\% }}\) interval for \({\rm{\mu }}\) is to be \({\rm{1}}{\rm{.0}}\)?

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 \({\rm{75\% }}\).

Let X1, X2,…, Xn be a random sample from a continuous probability distribution having median \(\widetilde {\bf{\mu }}\) (so that \({\bf{P(}}{{\bf{X}}_{\bf{i}}} \le \widetilde {\bf{\mu }}{\bf{) = P(}}{{\bf{X}}_{\bf{i}}} \le \widetilde {\bf{\mu }}{\bf{) = }}{\bf{.5}}\)).

a. Show that

\({\bf{P(min(}}{{\bf{X}}_{\bf{i}}}{\bf{) < }}\widetilde {\bf{\mu }}{\bf{ < max(}}{{\bf{X}}_{\bf{i}}}{\bf{)) = 1 - }}{\left( {\frac{{\bf{1}}}{{\bf{2}}}} \right)^{{\bf{n - 1}}}}\)

So that \({\bf{(min(}}{{\bf{x}}_{\bf{i}}}{\bf{),max(}}{{\bf{x}}_{\bf{i}}}{\bf{))}}\)is a \({\bf{100(1 - \alpha )\% }}\) confidence interval for \(\widetilde {\bf{\mu }}\) with (\({\bf{\alpha = 1 - }}{\left( {\frac{{\bf{1}}}{{\bf{2}}}} \right)^{{\bf{n - 1}}}}\).Hint :The complement of the event \(\left\{ {{\bf{min(}}{{\bf{X}}_{\bf{i}}}{\bf{) < \mu < max(}}{{\bf{X}}_{\bf{i}}}{\bf{)}}} \right\}\)is \({\bf{\{ max(}}{{\bf{X}}_{\bf{i}}}{\bf{)}} \le \widetilde {\bf{\mu }}{\bf{\} }} \cup {\bf{\{ min(}}{{\bf{X}}_{\bf{i}}}{\bf{)}} \ge \widetilde {\bf{\mu }}{\bf{\} }}\). But \({\bf{max(}}{{\bf{X}}_{\bf{i}}}{\bf{)}} \le \widetilde {\bf{\mu }}\) iff \({\bf{(}}{{\bf{X}}_{\bf{i}}}{\bf{)}} \le \widetilde {\bf{\mu }}\)for all i.

b.For each of six normal male infants, the amount of the amino acid alanine (mg/100 mL) was determined while the infants were on an isoleucine-free diet,

resulting in the following data:

2.84 3.54 2.80 1.44 2.94 2.70

Compute a 97% CI for the true median amount of alanine for infants on such a diet(“The Essential Amino Acid Requirements of Infants,” Amer. J. Of Nutrition, 1964: 322–330).

c.Let x(2)denote the second smallest of the xi’s and x(n-1) denote the second largest of the xi’s. What is the confidence level of the interval(x(2), x(n-1)) for \(\widetilde \mu \)?

Let X1, X2,…, Xn be a random sample from a uniform distribution on the interval (0,θ), so that

\({\rm{f(x) = }}\left\{ \begin{aligned}{l}\frac{{\rm{1}}}{{\rm{\theta }}}{\rm{ 0}} \le {\rm{x}} \le {\rm{\theta }}\\{\rm{0 otherwise}}\end{aligned} \right.\)

Then if Y= max(Xi ), it can be shown that the rv U=Y/θ has density function

\({{\rm{f}}_U}{\rm{(u) = }}\left\{ \begin{aligned}{l}{\rm{n}}{{\rm{u}}^{n - 1}}{\rm{ 0}} \le u \le 1\\{\rm{0 otherwise}}\end{aligned} \right.\)

a. Use\({{\rm{f}}_U}{\rm{(u)}}\) to verify that

\({\rm{P}}\left( {{{(\alpha /2)}^{1/n}} < \frac{Y}{\theta } \le {{(1 - \alpha /2)}^{1/n}}} \right) = 1 - \alpha \)

and use this derive a \(100(1 - \alpha )\% \) CI for θ.

b. Verify that \({\rm{P}}\left( {{\alpha ^{1/n}} < Y/\theta \le 1} \right) = 1 - \alpha \), and derive a 100(1 - α)% CI for θ based on this probability statement.

c.Which of the two intervals derived previously is shorter? If my waiting time for a morning bus is uniformly distributed and observed waiting times are x1 = 4.2, x2 = 3.5, x­3 = 1.7, x4 = 1.2, and x5 = 2.4, derive a 95% CI for θ by using the shorter of the two intervals.

In Example 6.8, we introduced the concept of a censored experiment in which n components are put on test and the experiment terminates as soon as r of the components have failed. Suppose component lifetimes are independent, each having an exponential distribution with parameter λ. Let Y1 denote the time at which the first failure occurs, Y2 the time at which the second failure occurs, and so on, so that Tr= Y1 + … + Yr+ (n- r)Yr is the total accumulated lifetime at termination. Then it can be shown that 2λTr has a chi-squared distribution with 2r df. Use this fact to develop a 100(1 - α)% CI formula for true average lifetime \(\frac{{\bf{1}}}{{\bf{\lambda }}}\) . Compute a 95% CI from the data in Example 6.8.
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