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

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)(\sigma)/(√(n))`

`=654.16\pm(1.96)(165.23)/(√(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=((z_(\alpha/2)s)/(E))^2\\\\\Rightarrow\ n=(((1.96)(167))/(57))^2=32.9758025239\approx33`

Hence, the minimum required sample size would be 33.