Suppose indoor air quality monitoring results, for a sample of 80 kitchens with
ID: 3063160 • Letter: S
Question
Suppose indoor air quality monitoring results, for a sample of 80 kitchens with natural gas cooking appliances, show a sample mean CO2 concentration of 555.2 ppm and a sample standard deviation of 171 ppm. a) Compute a 95% confidence interval for the mean CO2 concentration in the populationof all homes from which the sample was selected. (Give decimal answer to one place past decimal.)Lower bound: - Upper bound: | b) What sample size would be necessary to obtain a 95% CI interval width of 50 ppm, if the population standard deviation is estimated to be 175 ppm, before collecting the data? (Give answer as a whole number.) Tries 0/5 - Tries 0/5Explanation / Answer
x=555.2
=171
n=80
Z/2 for 95% confidence is 1.96
a)
CI = x ± Z/2 × (/n)
=555.2 ± 1.96 x 171/80
=555.2 ± 37.472
Lower limit =555.2 - 37.472 = 517.7
Upper limit =555.2 + 37.472 = 592.7
b)
E=50
=175
Z/2 for 95% confidence is 1.96
E =Z/2 × (/n)
50 =2*1.96 x (175)/n
50 =2*343/n
n=2*343/50
n=13.72
n=(13.72)^2
n=189