An article reported that for a sample of 50 kitchens with gas cooking appliances

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 162.17.

(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 does not contain the true population mean.

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

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

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

(b) Suppose the investigators had made a rough guess of 178 for the value of s before collecting data. What sample size would be necessary to obtain an interval width of 52 ppm for a confidence level of 95%? (Round your answer up to the nearest whole number.)

Explanation / Answer

The 95% c.i=Xbar+Z(s/sqrt N-1), where Xbar is sample mean, s is sample standard deviation, N is sample size and Z corresponds to Z score at 955 c.i.

95% c.i=654.16+-1.96(162.17/sqrt 50-1)

=608.75 to 699.57

Option 4.

b) Sample size, n=(2*Zalpha/2*sigma/w)^2, where w refers to interval width

n=(2*1.96*178/52)^2

=180

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