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 1423 with a standard deviation of 329. (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.)

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Part a

[tex]\bar X=1423[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

[tex]s=329[/tex] represent the sample standard deviation

n=512 represent the sample size

95% or 0.95 confidence interval

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)

We can assume that the sample deviation is a good estimate of the population deviation since the sample size is large enough.

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that [tex]z_{\alpha/2}=1.96[/tex]

Now we have everything in order to replace into formula (1):

[tex]1423-1.96\frac{329}{\sqrt{512}}=1394.502[/tex]

[tex]1423+1.96\frac{329}{\sqrt{512}}=1451.498[/tex]

So on this case the 95% confidence interval would be given by (1394.502;1451.498)

Part b

We can use to solve this part of the question the formula for the margin of error given by:

[tex]Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]

And on this case we have that ME =50 and we are interested in order to find the value of n, if we solve n from equation for Me we got:

[tex]n=(\frac{z_{\alpha/2} s}{ME})^2[/tex]

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got [tex]z_{\alpha/2}=1.960[/tex], replacing into formula for n we got:

[tex]n=(\frac{1.960(330)}{50})^2 =167.34 \approx 168[/tex]