An article reported that for a sample of 50 kitchens with ga

An article reported that for a sample of 50 kitchens with gas cooking appliances monitored during a one-week period, the sample mean CO_2 level (ppm) was 654.16, and the sample standard deviation was 165.4. Calculate and interpret a 95% (two-sided) confidence interval for true average CO_2 level in the population of all homes from which the sample was selected. (Round your answers to two decimal places.) Interpret the resulting interval. 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. We are 95% confident that this interval contains the true population mean. Suppose the investigators had made a rough guess of 180 for the value of s before collecting data. What sample size would be necessary to obtain an interval width of 44 ppm for a confidence level of 95%? (Round your answer up to the nearest whole number.)

Solution

5 . a)

n =51

Mean = 654.16

Standard Deviation SD = 166.59

z for 95% confidence interval = 1.96

Margin of error ME = (z*SD)/sqrt(n)   

= (1.96*166.59)/sqrt(51)

= 45.7214

Confidence Interval : (Mean - ME , Mean +ME)

= (654.16- 45.7214 , 654.16+ 45.7214 )

=( 608.44 , 699.88 ) Answer

Answer D. >> We ae 95% confident that interval contains the true population men.

5.b)

SD =168

Interval width = 49

Interval Width = 2 * Margin of error

=> 49 = 2 * ( z*SD)/sqrt(n)

=> sqrt(n) = (2*( 1.96*168))/49

=> n = 180.63 ~ 181 kitchens Answer