Question

Suppose indoor air quality monitoring results, for a sample of 70 kitchens with natural gas cooking appliances, show a sample mean CO2 concentration of 553.6 ppm and a sample standard deviation of 173.8 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.)

Answer 1

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 = t_0.025,69 = 1.995

Margin of error = E = t_/2,df * (s /n)

= 1.995 * (173.8 / 70)

= 44.4

The 95% confidence interval estimate of the population mean is,

- 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