Question

An article reported that for a sample of 60 kitchens with gas cooking appliances monitored during a one-week period, the sample mean CO₂ level (ppm) was 654.16, and the sample standard deviation was 166.42.

(a) Calculate and interpret a 95% (two-sided) confidence interval for true average CO₂ level in the population of all homes from which the sample was selected. (Round your answers to two decimal places.

(          ,           )ppm

(b) Suppose the investigators had made a rough guess of 178 for the value of s before collecting data. What sample size would be necessary to obtain an interval width of 59 ppm for a confidence level of 95%? (Round your answer up to the nearest whole number.)
kitchens

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 654.16

sample standard deviation = s = 166.42

sample size = n = 60

Degrees of freedom = df = n - 1 = 60 - 1 = 59

a) At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t0.025,59 = 2.001

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

= 2.001 * (166.42 / 60 )

Margin of error = E = 42.99

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

  ± E  

= 654.16 ± 42.99

= ( 611.17, 697.15 )

b) sample standard deviation = s = 178

margin of error = E = width / 2 = 59 / 2 = 29.5

sample size = n = [t/2,df* s / E]2

n = [2.001 * 178 / 29.5 ]2

n = 145.77

Sample size = n = 146