Question

An article reported that for a sample of 41 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 165.74. (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.)

Answer 1

- x ± zά/2x s/âšn Here zά/2= z0.05/2= 1.96 Substitute the values we have 654.16 ± 1.96 x 164.43/âš41 =(603.83, 704.49)

Answer 2

Given:

Mean of 654.16

Sample size is 41

Standard dev is 165.74

Alpha is 1 – 0.95, Z(0.025) = 1.96

So the computation would be:

Mean ± Z* s/sqrt (n)

= 654.16 ± 1.96*165.74/sqrt(41)

= 654.16 ± 1.96*25.88

= 654.16 ± 50.7248

= 603.4352, 704.8848 is the confidence interval

Interpretation: We are 95% confident that this confidence interval contains the true pop. mean