Question

An article reported that for a sample of 43 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 162.68.

(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 50 ppm for a confidence level of 95%? (Round your answer up to the nearest whole number.).....................
kitchens

Answer 1

Solution :

A ) Given that,

= 654.16

s = 162.68

n = 43

Degrees of freedom = df = n - 1 = 43 - 1 = 42

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,42 = 2.018

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

= 2.018* ( 162.68 / 43)

= 50.06

Margin of error = 50.06

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

- E < < + E

654.16 - 50.06 < < 654.16 + 50.06

604.10 < < 704.22

(604.10, 704.22 )

B ) Given that,

standard deviation = s = 183

margin of error = E = 50 / 2 = 25

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.960

Sample size = n = ((Z_/2 * s ) / E)²

= ((1.960 * 183) / 25 )²

= 205.88

= 206

Sample size = 206