Question

An article reported that for a sample of 40 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 165.23.

Calculate and interpret a 95% (two-sided) confidence interval for true average CO2 level in the population of all homes from which the sample was selected. (Round your answers to two decimal places.)

=(_________,__________) ppm

(b) Suppose the investigators had made a rough guess of 167 for the value of s before collecting data. What sample size would be necessary to obtain an interval width of 57ppm for a confidence level of 95%? (Round your answer up to the nearest whole number.)

=__________ kitchens

Answer 1

Answer:  (a)  (602.95,705.37)

(b) 33

Step-by-step explanation:

(a) Given : Sample size : $n=40$

Sample mean : $\overline{x}=654.16$

Standard deviation : $\sigma= 165.23$

Significance level :$\alpha=1-0.95=0.05$

Critical value : $z_{\alpha/2}=1.96$

The confidence interval for population mean is given by :-

$\mu\ \pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

$=654.16\pm(1.96)\dfrac{165.23}{\sqrt{40}}\\\\\approx654.16\pm51.21\\\\=(654.16-51.21,\ 654.16+51.21)=(602.95,705.37)$

Hence, the 95% (two-sided) confidence interval for true average $CO_2$ level in the population of all homes from which the sample was selected.

(b) Given : Standard deviation : $s= 167\text{ ppm}$

Margin of error : $E=\pm57\text{ ppm}$

Significance level :$\alpha=1-0.95=0.05$

Critical value : $z_{\alpha/2}=1.96$

The formula to calculate the sample size is given by :-

$n=(\dfrac{z_{\alpha/2}s}{E})^2\\\\\Rightarrow\ n=(\dfrac{(1.96)(167)}{57})^2=32.9758025239\approx33$

Hence, the minimum required sample size would be 33.