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 165.4.
Required:
a. Calculate and interpret a 9596 (two-sided) confidence interval for true average C02 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 sample size would be necessary to obtain an interval width of 50 ppm for a confidence level of 95%?

Answer 1

Answer:

a) $654.16-2.01\frac{165.4}{\sqrt{52}}=608.06$    

$654.16+2.01\frac{165.4}{\sqrt{52}}=700.26$    

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

So the answer for this case would be n=189 rounded up to the nearest integer

Step-by-step explanation:

Part a

$\bar X=654.16$ represent the sample mean

$\mu$ population mean (variable of interest)

s=165.4 represent the sample standard deviation

n =52represent the sample size  

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

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

The degrees of freedom aregiven by:

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

Since the Confidence is 0.95 or 95%, the significance $\alpha=0.05$ and $\alpha/2 =0.025$, and the critical value would be $t_{\alpha/2}=2.01$

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

$654.16-2.01\frac{165.4}{\sqrt{52}}=608.06$    

$654.16+2.01\frac{165.4}{\sqrt{52}}=700.26$    

Part b

The margin of error is given by this formula:

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

And on this case we have that ME =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 this case wuld be $z_{\alpha/2}=1.960$, replacing into formula (b) we got:

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

So the answer for this case would be n=189 rounded up to the nearest integer