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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

Each of the following is a confidence interval for \({\rm{\mu = }}\) true average (i.e., population mean) resonance frequency (Hz) for all tennis rackets of a certain type: \({\rm{(114}}{\rm{.4,115}}{\rm{.6)(114}}{\rm{.1,115}}{\rm{.9)}}\) a. What is the value of the sample mean resonance frequency? b. Both intervals were calculated from the same sample data. The confidence level for one of these intervals is \({\rm{90\% }}\) and for the other is \({\rm{99\% }}\). Which of the intervals has the \({\rm{90\% }}\) confidence level, and why?

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.

A journal article reports that a sample of size 5 was used as a basis for calculating a 95% CI for the true average natural frequency (Hz) of delaminated beams of a certain type. The resulting interval was (229.764, 233.504). You decide that a confidence level of 99% is more appropriate than the 95% level used. What are the limits of the 99% interval? (Hint: Use the center of the interval and its width to determine \(\overline x \) and s.)

a.Use the results of Example 7.5 to obtain a 95% lower confidence bound for the parameter 位 of an exponential distribution, and calculate the bound based on the data given in the example.

b.If lifetime X has an exponential distribution, the probability that lifetime exceeds t is P(X>t) = e-位迟. Use the result of part (a) to obtain a 95% lower confidence bound for the probability that breakdown time exceeds 100 min.

A more extensive tabulation of t critical values than what appears in this book shows that for the t distribution with

\({\rm{20}}\)df, the areas to the right of the values \({\rm{.687, }}{\rm{.860, and 1}}{\rm{.064 are }}{\rm{.25, }}{\rm{.20, and }}{\rm{.15,}}\)respectively. What is the confidence level for each of the following three confidence intervals for the mean m of a normal population distribution? Which of the three intervals would you recommend be used, and why?

\(\begin{array}{l}{\rm{a}}{\rm{.(\bar x - }}{\rm{.687s/}}\sqrt {{\rm{21}}} {\rm{,\bar x + 1}}{\rm{.725s/}}\sqrt {{\rm{21}}} {\rm{)}}\\{\rm{b}}{\rm{.(\bar x - }}{\rm{.860\;s/}}\sqrt {{\rm{21}}} {\rm{,\bar x + 1}}{\rm{.325s/}}\sqrt {{\rm{21}}} {\rm{)}}\\{\rm{c}}{\rm{.(\bar x - 1}}{\rm{.064s/}}\sqrt {{\rm{21}}} {\rm{,\bar x + 1}}{\rm{.064s/}}\sqrt {{\rm{21}}} {\rm{)}}\end{array}\)
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