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 an article investigated the relationship between indoor air-pollution metrics and lung function growth among children ages 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 1 second. Higher FEV1 values are associated with greater lung capacity. Among the children in the study, 512 came from households that used coal for cooking or heating or both. Their FEV1 mean was 1429 with a standard deviation of 327. (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 95% (two-sided) confidence interval for true average FEV1 level in the population of all children from which the sample was selected. (Round your answers to one decimal place.)
b. Suppose the investigators had made a rough guess of 330 for the value of s before collecting data. What sample size would be necessary to obtain an interval width of 50 ml for a confidence level of 95%? (Round your answer up to the nearest whole number.)

Answer 1

Answer:

Step-by-step explanation:

Hello!

The objective is to estimate the population mean of the lung-capacity index (FEV1) the forced volume (in ml) of air that is exhaled in 1 second of children that came from households where coal is used for coal or heating or both. The researcher claims that the use of coal has a negative impact (reduces) on the mean FEV1.

a.

A sample of 512 children that came from households where coal is uses was taken, they obtained a sample mean of X[bar]= 1429ml and a standard deviation of S= 327ml

The study variable is X: the forced volume (in ml) of air that is exhaled in 1 second of a child that lives in a household where coal is used.

There is no information about the distribution of the variable, but the sample size is large enough to be valid to apply the Central Limit Theorem and approximate the distribution of the sample mean to normal:

X[bar]≈N(μ;δ²/n)

Using this approximation you can estimate the population mean using an approximate Z and 1 - α: 0.95:

X[bar] ± $Z_{1-\alpha /2}$ * (S/√n)

$Z_{1-\alpha /2}= Z_{0.975}= 1.965$

[1426 ± 1.965 * (327/√512)]

[1397.60; 1454.40]ml

Using a confidence level of 95% you'd expect that the interval [1397.60; 1454.40] contains the true average of the lung-capacity index of Chinese children that live in households where coal is used for cooking, hearing or both.

b.

Now using a value of S= 330 and a CI of 95% you need to calculate the sample size required to get an interval amplitude of 50ml

To calculate the sample size given a determined confidence level and amplitude the best is to use the semiamplitude or margin of error (d) of the interval.

The semiamplitude is half of the amplitude. The general structure of a confidence level for the mean is "point estimate" ± "margin fo error"

Then the formula for the margin of error is:

d=  $Z_{1-\alpha /2}$ * (S/√n)

If a= 50ml then d= 50/2=25ml

$Z_{1-\alpha /2}= Z_{0.975}= 1.965$

(d* $Z_{1-\alpha /2}$)= (S/√n)

√n*(d* $Z_{1-\alpha /2}$)= S

n= [S/(d* $Z_{1-\alpha /2}$)]²

n= [330/(25*1.965)]²

n= 37.29 ≅ 38

You need to take a sample of 38 childre.

I hope it helps!