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Chapter 7: Q25E (page 294)

A state legislator wishes to survey residents of her district to see what proportion of the electorate is aware of her position on using state funds to pay for abortions.

a. What sample size is necessary if the \({\rm{95\% }}\)CI for p is to have a width of at most . \({\rm{10}}\)irrespective of p?

b. If the legislator has strong reason to believe that at least \({\rm{2/3}}\)of the electorate know of her position, how large a sample size would you recommend?

Short Answer

Expert verified

a) The sample size is \({\rm{n = 381}}\)

b) I recommend the sample size of \({\rm{n = 339}}\)

Step by step solution

01

To find the sample size

(a):

The expression for the necessary sample size for a confidence interval to have width w is

\({\rm{n - }}\frac{{{\rm{2z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{\hat p\hat q - z}}_{{\rm{\alpha /2}}}^{\rm{2}}{{\rm{w}}^{\rm{2}}}{\rm{ \pm }}\sqrt {{\rm{4z}}_{{\rm{\alpha /2}}}^{\rm{4}}{\rm{\hat p\hat q}}\left( {{\rm{\hat p\hat q - }}{{\rm{w}}^{\rm{2}}}} \right){\rm{ + }}{{\rm{w}}^{\rm{2}}}{\rm{z}}_{{\rm{\alpha /2}}}^{\rm{4}}} }}{{{{\rm{w}}^{\rm{2}}}}}\)

Therefore, for \({\rm{width w = 0}}{\rm{.1}}\)and\({\rm{\hat p - \hat q - 0}}{\rm{.5}}\) (irrespective of p ) the necessary sample size becomes

\(\begin{array}{l}{\rm{n = }}\frac{{{\rm{2z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{\hat p\hat q - z}}_{{\rm{\alpha /2}}}^{\rm{2}}{{\rm{w}}^{\rm{2}}}{\rm{ \pm }}\sqrt {{\rm{4z}}_{{\rm{\alpha /2}}}^{\rm{4}}{\rm{\hat p\hat q}}\left( {{\rm{\hat p\hat q - }}{{\rm{w}}^{\rm{2}}}} \right){\rm{ + }}{{\rm{w}}^{\rm{2}}}{\rm{z}}_{{\rm{\alpha /2}}}^{\rm{4}}} }}{{{{\rm{w}}^{\rm{2}}}}}\\{\rm{ - }}\frac{{{\rm{2 \times 1}}{\rm{.9}}{{\rm{6}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.5 \times 0}}{\rm{.5 - 1}}{\rm{.9}}{{\rm{6}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.}}{{\rm{1}}^{\rm{2}}}{\rm{ \pm }}\sqrt {{\rm{4 \times 1}}{\rm{.9}}{{\rm{6}}^{\rm{4}}}{\rm{ \times 0}}{\rm{.5 \times 0}}{\rm{.5 \times }}\left( {{\rm{0}}{\rm{.5 \times 0}}{\rm{.5 - 0}}{\rm{.}}{{\rm{1}}^{\rm{2}}}} \right){\rm{ + 0}}{\rm{.}}{{\rm{1}}^{\rm{2}}}{\rm{ \times 1}}{\rm{.9}}{{\rm{6}}^{\rm{4}}}} }}{{{\rm{0}}{\rm{.}}{{\rm{1}}^{\rm{2}}}}}\\{\rm{ = 381}}\end{array}\)

Hence the sample size is \({\rm{n = 381}}\)

02

To find how large a sample size would you recommend

(b):

For \({\rm{width w = 0}}{\rm{.1}}\),and \({\rm{\hat p - 1/3,\hat q - 2/3}}\)the necessary sample size becomes

\(\begin{array}{l}{\rm{n = }}\frac{{{\rm{2z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{\hat p\hat q - z}}_{{\rm{\alpha /2}}}^{\rm{2}}{{\rm{w}}^{\rm{2}}}{\rm{ \pm }}\sqrt {{\rm{4z}}_{{\rm{\alpha /2}}}^{\rm{4}}{\rm{\hat p\hat q}}\left( {{\rm{\hat p\hat q - }}{{\rm{w}}^{\rm{2}}}} \right){\rm{ + }}{{\rm{w}}^{\rm{2}}}{\rm{z}}_{{\rm{\alpha /2}}}^{\rm{4}}} }}{{{{\rm{w}}^{\rm{2}}}}}\\{\rm{ = }}\frac{{{\rm{2 \times 1}}{\rm{.9}}{{\rm{6}}^{\rm{2}}}{\rm{ \times 1/3 \times 2/3 - 1}}{\rm{.9}}{{\rm{6}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.}}{{\rm{1}}^{\rm{2}}}{\rm{ \pm }}\sqrt {{\rm{4 \times 1}}{\rm{.9}}{{\rm{6}}^{\rm{4}}}{\rm{ \times 1/3 \times 2/3 \times }}\left( {{\rm{1/3 \times 2/3 - 0}}{\rm{.}}{{\rm{1}}^{\rm{2}}}} \right){\rm{ + 0}}{\rm{.}}{{\rm{1}}^{\rm{2}}}{\rm{ \times 1}}{\rm{.9}}{{\rm{6}}^{\rm{4}}}} }}{{{\rm{0}}{\rm{.}}{{\rm{1}}^{\rm{2}}}}}\\{\rm{ = 339}}{\rm{.}}\end{array}\)

Hence I recommend the sample size of \({\rm{n = 339}}\)

Where in both a and b

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

and

\({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ - }}{{\rm{z}}_{{\rm{0}}{\rm{.05/2}}}}{\rm{ - }}{{\rm{z}}_{{\rm{0}}{\rm{.025}}}}\mathop {\rm{ - }}\limits^{{\rm{(1)}}} {\rm{1}}{\rm{.96}}\)

(1) : this is obtained from

\({\rm{P}}\left( {{\rm{Z > }}{{\rm{z}}_{{\rm{0}}{\rm{.025}}}}} \right){\rm{ - 0}}{\rm{.025}}\)

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

Hence There were no assumptions made about the distribution.

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Most popular questions from this chapter

TV advertising agencies face increasing challenges in reaching audience members because viewing TV programs via digital streaming is gaining in popularity. The Harris poll reported on November 13, 2012, that 53% of 2343 American adults surveyed said they have watched digitally streamed TV programming on some type of device.

a. Calculate and interpret a confidence interval at the 99% confidence level for the proportion of all adult Americans who watched streamed programming up to that point in time.

b. What sample size would be required for the width of a 99% CI to be at most .05 irrespective of the value of p 藛?

A random sample of n=\({\rm{15}}\)heat pumps of a certain type yielded the following observations on lifetime (in years):

\(\begin{array}{*{20}{l}}{{\rm{2}}{\rm{.0 1}}{\rm{.3 6}}{\rm{.0 1}}{\rm{.9 5}}{\rm{.1 }}{\rm{.4 1}}{\rm{.0 5}}{\rm{.3}}}\\{{\rm{15}}{\rm{.7 }}{\rm{.7 4}}{\rm{.8 }}{\rm{.9 12}}{\rm{.2 5}}{\rm{.3 }}{\rm{.6}}}\end{array}\)

a. Assume that the lifetime distribution is exponential and use an argument parallel to that to obtain a \({\rm{95\% }}\) CI for expected (true average) lifetime.

b. How should the interval of part (a) be altered to achieve a confidence level of \({\rm{99\% }}\)?

c. What is a \({\rm{95\% }}\)CI for the standard deviation of the lifetime distribution? (Hint: What is the standard deviation of an exponential random variable?)

Exercise 72 of Chapter 1 gave the following observations on a receptor binding measure (adjusted distribution volume) for a sample of 13 healthy individuals: 23, 39, 40, 41, 43, 47, 51, 58, 63, 66, 67, 69, 72.

a. Is it plausible that the population distribution from which this sample was selected is normal?

b. Calculate an interval for which you can be 95% confident that at least 95% of all healthy individuals in the population have adjusted distribution volumes lying between the limits of the interval.

c. Predict the adjusted distribution volume of a single healthy individual by calculating a 95% prediction interval. How does this interval鈥檚 width compare to the width of the interval calculated in part (b)?

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).

The negative effects of ambient air pollution on children鈥檚 lung function has been well established, but less research is available about the impact of indoor air pollution. The authors of 鈥淚ndoor Air Pollution and Lung Function Growth Among Children in Four Chinese Cities鈥� investigated the relationship between indoor air-pollution metrics and lung function growth among children ages \({\rm{6--13}}\)years living in four Chinese cities. For each subject in the study, the authors measured an important lung-capacity index known as FEV1, the forced volume (in ml) of air that is exhaled in \({\rm{1}}\) second. Higher FEV1 values are associated with greater lung capacity. Among the children in the study, \({\rm{514}}\) came from households that used coal for cooking or heating or both. Their FEV1 mean was \({\rm{1427}}\)with a standard deviation of \({\rm{325}}\). (A complex statistical procedure was used to show that burning coal had a clear negative effect on mean FEV1 levels.)

a. Calculate and interpret a \({\rm{95\% }}\) (two-sided) confidence interval for true average FEV1 level in the population of all children from which the sample was selected. Does it appear that the parameter of interest has been accurately estimated?

b. Suppose the investigators had made a rough guess of \({\rm{320}}\) for the value of s before collecting data. What sample size would be necessary to obtain an interval width of \({\rm{50}}\)ml for a confidence level of \({\rm{95\% }}\)?

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