Question

An article reported that for a sample of 50 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 163.02.

(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

Interpret the resulting interval.

We are 95% confident that this interval contains the true population mean.

We are 95% confident that the true population mean lies above this interval.   

We are 95% confident that the true population mean lies below this interval.

We are 95% confident that this interval does not contain the true population mean.

(b) Suppose the investigators had made a rough guess of 185 for the value of s before collecting data. What sample size would be necessary to obtain an interval width of 48 ppm for a confidence level of 95%? (Round your answer up to the nearest whole number.)

( ) kitchens

Answer 1

solution:-

given that mean = 654.16 , standard deviation = 163.02

and n = 50

(a) the value of 95% confidence from z table is z = 1.96

confidence interval formula

=> mean +/- z * standard deviation/sqrt(n)

=> 654.16 +/- 1.96 * 163.02/sqrt(50)

=> (608.97 , 699.35)

Interpret the resulting interval.

option : We are 95% confident that the true population mean lies above this interval.   

(b) here given that width = 48 , standard deviation s = 185

the value of 95% confidence from z table is z = 1.96

formula for width

w = 2 *z*(s/sqrt(n))

sqrt(n) = 2 * z * (s/w)

n = (2 * z * (s/w))^2

= (2 * 1.96 * (185/48))^2

= 228