Question

The negative effects of ambient air pollution on children’s lung function has been well established, but less research is available about the impact of indoor air pollution. The authors of “Indoor Air Pollution and Lung Function Growth Among Children in Four Chinese Cities” 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

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

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.

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\((1398.9083,1455.0969)\)

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 \({\rm{n = 630}}\)