Question

An article reported that for a sample of 52 kitchens with gas cooking appliances monitored during a one-week period, the sample mean CO2 level (ppm) was 654.16, and the sample standard deviation was 164.55.


a) calculate and interpret a 95% confidence interval for true average CO2 level in the population of all homes from which the sample was selected .

b) Suppose the investigators had made a rough guess of 175 for the value of s before collecting data .What sanple size would be necessary to obtain an interval width of 50 ppm for confidence level of 95% ?

Answer 1

Answer:

a) $654.16-2.01\frac{164.55}{\sqrt{52}}=608.29$    

$654.16+2.01\frac{164.55}{\sqrt{52}}=700.03$    

And we can conclude that we are 95% confident that the true mean of Co2 level is between 608.29 and 700.03 ppm

b) $n=(\frac{1.960(175)}{25})^2 =188.23 \approx 189$

Step-by-step explanation:

Part a

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

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=52-1=51$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,51)".And we see that $t_{\alpha/2}=2.01$

Replacing we got:

$654.16-2.01\frac{164.55}{\sqrt{52}}=608.29$    

$654.16+2.01\frac{164.55}{\sqrt{52}}=700.03$    

And we can conclude that we are 95% confident that the true mean of Co2 level is between 608.29 and 700.03 ppm

Part b

The margin of error is given by :

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$    (a)

The desired margin of error is ME =50/2=25 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$   (b)

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 $z_{\alpha/2}=1.960$, and we use an estimator of the population variance the value of 175 replacing into formula (b) we got:

$n=(\frac{1.960(175)}{25})^2 =188.23 \approx 189$