Question

Reliance on solid biomass fuel for cooking and heating exposes many children from developing countries to high levels of indoor air pollution. The article “Domestic Fuels, Indoor Air Pollution, and Children’s Health” (Annals of the N.Y. Academy of Sciences, \(2008:209 - 217\)) presented information on various pulmonary characteristics in samples of children whose households in India used either biomass fuel or liquefied petroleum gas (\(LPG\)). For the \(755\) children in biomass households, the sample mean peak expiratory flow (a person’s maximum speed of expiration) was \(3.30L/s\), and the sample standard deviation was \(1.20\). For the \(750\) children whose households used liquefied petroleum gas, the sample mean \(PEF\) was \(4.25\) and the sample standard deviation was \(1.75\).

a. Calculate a confidence interval at the \(95\% \) confidence level for the population mean \(PEF\) for children in biomass households and then do likewise for children in \(LPG\) households. What is the simultaneous confidence level for the two intervals?

b. Carry out a test of hypotheses at significance level \(.01\) to decide whether true average \(PEF\) is lower for children in biomass households than it is for children in \(LPG\) households (the cited article included a P-value for this test)

c. \(FE{V_1}\), the forced expiratory volume in \(1\) second, is another measure of pulmonary function. The cited article reported that for the biomass households the sample mean FEV1 was \(2.3L/s\) and the sample standard deviation was \(.5L/s\). If this information is used to compute a \(95\% \) \(CI\) for population mean \(FE{V_1}\), would the simultaneous confidence level for this interval and the first interval calculated in (a) be the same as the simultaneous confidence level determined there? Explain

Short answer

the solution is

a)

\(\begin{array}{l}(3.2144,3.3856);\\(4.1248,4.3752);\\0.0975\end{array}\)

b) There is sufficient data to support the hypothesis that children in biomass households have a lower true average PEF than children in LPG households.

c) no

Step-by-step solution

calculate a confidence interval

a)The sample mean was for the \({n_1} = 755\) children in the biomass family.

\({\bar x_1} = 3.30\)

The standard deviation of the sample was.

\({s_1} = 1.2\)

The sample mean was for \({n_2} = 750\) children whose families used liquefied petroleum gas.

\({\bar x_2} = 4.25\)

The standard deviation of the sample was

\({s_2} = 1.75\)

The standardized random variable, which is huge \(n\).

\(Z = \frac{{\bar X - \mu }}{{S/\sqrt n }}\)

With an expectation of \(0\) and a standard deviation of \(1\), the distribution is roughly normal. as a result,

Confidence interval for \(\mu \) based on a high sample size.

\(\bar x \pm {z_{\alpha /2}} \times \frac{s}{{\sqrt n }}100(1 - \alpha )\% .\)

with a \(100(1 - \alpha )\% \) confidence level Regardless of population distribution, this remains the case.

find confidence level

biomass household sample:

Because \({n_1} = 755\) is large enough \(( \ge 30)\), the \(95\% \)confidence interval for the population mean may be calculated using the procedure above. \(100(1 - \alpha ) = 95,\alpha = 0.05 and {z_{\alpha /2}} = {z_{0.025}} = 1.96\) (from the table in the appendix). As a result, the mean's \(CI\) value is.

\(\left( {3.3 - 1.96 \times \frac{{1.2}}{{\sqrt {755} }},3.3 + 1.96 \times \frac{{1.2}}{{\sqrt {755} }}} \right) = (3.2144,3.3856)\)

LPG sample

Because \({n_2} = 750\) is large enough \(( \ge 30)\), the \(95\% \) confidence interval for the population mean may be calculated using the procedure above. \(100(1 - \alpha ) = 95,\alpha = 0.05 and {z_{\alpha /2}} = {z_{0.025}} = 1.96\) (from the table in the appendix). As a result, the mean's \(CI\) value is.

\(\left( {4.25 - 1.96 \times \frac{{1.75}}{{\sqrt {750} }},4.25 + 1.96 \times \frac{{1.75}}{{\sqrt {750} }}} \right) = (4.1248,4.3752)\)

Simultaneous confidence level:

\(P\)( at least one type \(I\) error committed) \( = 1 - P\)(no type \(I\)errors committed)

\( = 1 - P\)( no type \(I\) error in first ). \(P\)(no type \(I\) error committed)

\(\begin{array}{l} = 1 - 0.95 \times 0.95 = 1 - 0.9025\\ = 0.0975\end{array}\)

determine value of test static

b) given

\(\begin{array}{l}{{\bar x}_1} = 3.30\\{{\bar x}_2} = 4.25\\{s_1} = 1.20\\{s_2} = 1.75\end{array}\)

\(\begin{array}{l}{n_1} = 755\\{n_2} = 750\\\alpha = 0.01\end{array}\)

We can employ the \(z\)-test because the samples are huge \((n \ge 30)\). (instead of a t-test).

Lower, as stated.

Either the null hypothesis or the alternative hypothesis is asserted. The null hypothesis and the alternative hypothesis are diametrically opposed. An equality must be included in the null hypothesis.

\(\begin{array}{l}{H_0}:{\mu _1} = {\mu _2}\\{H_a}:{\mu _1} < {\mu _2}\end{array}\)

Determine the value of the test statistic:

\(z = \frac{{{{\bar x}_1} - {{\bar x}_2}}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }} = \frac{{3.30 - 4.25}}{{\sqrt {\frac{{1.2{0^2}}}{{755}} + \frac{{1.7{5^2}}}{{750}}} }} \approx - 12.27\)

If the null hypothesis is true, the P-value is the probability of getting a result more extreme or equal to the standardised test statistic \(z\). Using the normal probability table, determine the probability.

\(P = P(Z < - 12.27) \approx 0\)

The null hypothesis is rejected if the P-value is less than the alpha significance level.

There is sufficient data to support the conclusion that the genuine average PEF for children in biomass families is lower than for children in non-biomass households.

explain confidence level

There is sufficient data to support the hypothesis that children in biomass households have a lower true average PEF than children in LPG households.

c)

\(\begin{array}{l}{{\bar x}_1} = 3.30\\{{\bar x}_2} = 4.25\\{s_1} = 1.20\\{s_2} = 1.75\\{n_1} = 755\end{array}\)

\(\begin{array}{l}{n_2} = 750\\c = 95\% = 0.95\\\bar x = 2.3\\s = 0.5\end{array}\)

The simultaneous confidence interval will be different from the simultaneous confidence interval in part (a), because this simultaneous confidence interval will only be for biomass households, whereas the simultaneous confidence interval in part (a) will be for both biomass and LPG households.

Because the sample mean and standard deviation differ from those of the other two samples, the values of the confidence intervals cannot be the same.

conclusion

a)

\(\begin{array}{l}(3.2144,3.3856);\\(4.1248,4.3752);\\0.0975\end{array}\)

b)

There is sufficient data to support the hypothesis that children in biomass households have a lower true average PEF than children in LPG households.

c) no