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Chapter 7: Q14E (page 292)

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 \({\rm{95\% }}\)?

Short Answer

Expert verified

a) The boundaries of the confidence interval then become\((1398.9083,1455.0969)\)

b) The sample size is \({\rm{n = 630}}\)

Step by step solution

01

Given by

\(\begin{array}{l}{\rm{n = 514}}\\{\rm{\bar x - 1427}}\\{\rm{s = 325}}\\{\rm{c = 95\% = 0}}{\rm{.95}}\end{array}\)

Central limit theorem: If the sample size is large (more than 30), then the sampling distribution of the sample mean \({\rm{\bar x}}\)is approximately normal.

The data set contains \(514\)data values, thus we can use the central limit theorem and we then know that the sampling distribution of the sample mean is approximately normal.

02

Large-sample confidence interval

\({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ - 1}}{\rm{.96}}\)

The margin of error is then:

\({\rm{E - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}{\rm{ - 1}}{\rm{.96 \times }}\frac{{{\rm{325}}}}{{\sqrt {{\rm{514}}} }}{\rm{\gg 28}}{\rm{.0969}}\)

The boundaries of the confidence interval then become:

\(\begin{array}{l}{\rm{\bar x - E - 1427 - 28}}{\rm{.0969 - 1398}}{\rm{.9031}}\\{\rm{\bar x + E - 1427 + 28}}{\rm{.0969 - 1455}}{\rm{.0969}}\end{array}\)

Hence The boundaries of the confidence interval then become\((1398.9083,1455.0969)\)

03

To find the sample size

(b)

\(\begin{array}{l}{\rm{s = 320 }}\\{\rm{c = 95\% = 0}}{\rm{.95}}\end{array}\)

Width interval \( - 50\)

he margin of error E is half the width of the confidence interval.

\({\rm{E - }}\frac{{{\rm{ Width interval }}}}{{\rm{2}}}{\rm{ - }}\frac{{{\rm{50}}}}{{\rm{2}}}{\rm{ - 25}}\)

When the sample is large, it is appropriate to assume that the population standard deviation is approximately the sample standard deviation:

\({\rm{\sigma \gg s - 320}}\)

formula sample size:

\({\rm{n - }}{\left( {\frac{{{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{\sigma }}}}{{\rm{E}}}} \right)^{\rm{2}}}\)

or confidence level \({\rm{1 - \alpha - 0}}{\rm{.95}}\),determine \({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ - }}{{\rm{z}}_{{\rm{0}}{\rm{.025}}}}\)using table the normal probability table in the appendix (look up $0.025$ in the table, the z-score is then the found z-score with opposite sign):

\({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ - 1}}{\rm{.96}}\)

The sample size is then (round up to the nearest integer!):

\({\rm{n - }}{\left( {\frac{{{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{\sigma }}}}{{\rm{E}}}} \right)^{\rm{2}}}{\rm{ - }}{\left( {\frac{{{\rm{1}}{\rm{.96 \times 320}}}}{{{\rm{25}}}}} \right)^{\rm{2}}}{\rm{\gg 630}}\)

Hence the sample size is \({\rm{n = 630}}\)

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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?

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?

On the basis of extensive tests, the yield point of a particular type of mild steel-reinforcing bar is known to be normally distributed with\({\rm{\sigma = 100,}}\)The composition of bars has been slightly modified, but the modification is not believed to have affected either the normality or the value of\({\rm{\sigma }}\).

a. Assuming this to be the case, if a sample of 25 modified bars resulted in a sample average yield point of 8439 lb, compute a 90% CI for the true average yield point of the modified bar.

b. How would you modify the interval in part (a) to obtain a confidence level of 92%?

A normal probability plot of the n=\({\rm{26}}\) observations on escape time shows a substantial linear pattern; the sample mean and sample standard deviation are \({\rm{370}}{\rm{.69 and 24}}{\rm{.36}}\), respectively.

a. Calculate an upper confidence bound for population mean escape time using a confidence level of \({\rm{95\% }}\)

b. Calculate an upper prediction bound for the escape time of a single additional worker using a prediction level of \({\rm{95\% }}\). How does this bound compare with the confidence bound of part (a)?

c. Suppose that two additional workers will be chosen to participate in the simulated escape exercise. Denote their escape times by \({\rm{X27 and X28}}\), and let X new denote the average of these two values. Modify the formula for a PI for a single x value to obtain a PI for X new, and calculate a 95% two-sided interval based on the given escape data.

The article 鈥淢easuring and Understanding the Aging of Kraft Insulating Paper in Power Transformers鈥� contained the following observations on degree of polymerization for paper specimens for which viscosity tim\({\rm{418 421 421 422 425 427 431 434 437 439 446 447 448 453 454 463 465}}\)es concentration fell in a certain middle range:

a. Construct a boxplot of the data and comment on any interesting features.

b. Is it plausible that the given sample observations were selected from a normal distribution?

c. Calculate a two-sided \({\rm{95\% }}\)confidence interval for true average degree of polymerization (as did the authors of the article). Does the interval suggest that \({\rm{440}}\) is a plausible value for true average degree of polymerization? What about \({\rm{450}}\)?
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