Question

An article reported that for a sample of 56 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 162.

(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 183 for the value of s before collecting data. What sample size would be necessary to obtain an interval width of 49 ppm for a confidence level of 95%? (Round your answer up to the nearest whole number.)

Answer 1

Solution :

Given that,

a)

Point estimate = sample mean = = 654.16

sample standard deviation = s = 162

sample size = n = 56

Degrees of freedom = df = n - 1 = 55

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,55 = 2.004

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

= 2.004 * (162 / 56)

= 43.38

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

- E < < + E

654.16 - 43.38 < < 654.16 + 43.38

610.78 < < 697.54

(610.78 , 697.54)

b)

Given that,

Population standard deviation = s = 183

Margin of error = E = width / 2 = 49/2 = 24.5

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

sample size = n = (Z_/2* s / E) ²

n = (1.96 *183 / 24.5)²

n = 214.33

n = 215

Sample size = 215