Question

The article “Gas Cooking, Kitchen Ventilation, and Exposure to Combustion Products” reported that for a sample of 50 kitchens with gas cooking appliances monitored during a one-week period, the average CO2 level(ppm) was 654.16, and suppose that the population of all homes kitchen CO2 level was normally distributed and the standard deviation was 164.43. (a) Calculate 95% confidence interval for the true average CO2 level in the population of all homes from which the sample was selected.

Answer 1

Mean, = 654.16 ppm

Standard deviation, = 164.43 ppm

n = 50

The sampling distribution of the sample mean of size n = 50 is Normally distributed with mean = 654.16 and standard deviation, S.e = 164.43/√50 = 23.254

Corresponding to 95% confidence interval, the critical z score = 1.96

Margin of error = z*S.e = 1.96*23.254 = 45.58

Thus, the 95% confidence interval for the true average CO2 level in the population of all homes from which the sample was selected is (654.16 - 45.58, 654.16 + 45.58)

= (608.58, 699.74) ppm