Suppose indoor air quality mon
Suppose indoor air quality monitoring results, for a sample of70 kitchens with natural gas cooking appliances, show a sample meanCO2 concentration of 553.6 ppm and a sample standard deviation of173.8 ppm.
a.) Compute a 95% confidence interval for the mean CO2concentration in the population of all homes from which the samplewas selected. (Give decimal answer to one place past decimal.)Lowerbound: Upper bound:
b.) What sample size would be necessary to obtain a 95% CIinterval width of 50 ppm, if the population standard deviation isestimated to be 175 ppm, before collecting the data ? (Give answeras a whole number.)
Answer:
Solution :
Given that,
(a)
Point estimate = sample mean = 
= 553.6
sample standard deviation = s = 173.8
sample size = n = 70
Degrees of freedom = df = n – 1 = 70 – 1 = 69
At 95% confidence level the t is ,
= 1 – 95% = 1 – 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
t/2,df = t0.025,69 = 1.995
Margin of error = E = t/2,df* (s /
= 1.995 * (173.8 /
= 44.4
The 95% confidence interval estimate of the population meanis,
– E < 
< + E
553.6 – 44.4 < < 553.6 + 44.4
509.2 < < 598.0
Lower bound = 509.2
Upper bound = 598.0
(b)
Population standard deviation = 
= 175
width = 50
Margin of error = E = 25
sample size = n = (Z
/2*/ E) 2
n = (1.96 * 175 / 25)2
n = 188.4
n = 189
Sample size = 189
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