# 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.)

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 /$\sqrt$n)

= 1.995 * (173.8 / $\sqrt$70)

= 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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