Suppose indoor air quality monitoring results, for a sample of 80 kitchens with
ID: 3217120 • 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 558.2 ppm and a sample standard deviation of 178.2 ppm. a.) Compute a 95% confidence interval for the mean CO2 concentration in the population of 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.)
Explanation / Answer
a.
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=558.2
Standard deviation( sd )=178.2
Sample Size(n)=80
Confidence Interval = [ 558.2 ± t a/2 ( 178.2/ Sqrt ( 80) ) ]
= [ 558.2 - 1.99 * (19.923) , 558.2 + 1.99 * (19.923) ]
= [ 518.553,597.847 ]
b.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )
Standard Deviation ( S.D) = 175
ME =50
n = ( 1.96*175/50) ^2
= (343/50 ) ^2
= 47.06 ~ 48