An article reported that for a sample of 60 kitchens with gas cooking appliances

Question

An article reported that for a sample of 60 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 162.68. (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.) 612.156 X696.164 x ppnm 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. ( (b) Suppose the investigators had made a rough guess of 179 for the value of s before collecting data. What sample size would be necessary to obtain an interval width of 40 ppm for a confidence level of 95%? (Round your answer up to the nearest whole number.) X kitchens You may need to use the appropriate table in the Appendix of Tables to answer this question.

Explanation / Answer

Solution:- Z(0.025) = 1.96, mean = 654.16,sd = 162.68 n = 60

a) 654.16 ± 1.96*(162.68/sqrt60)
= 654.16 ± 1.96*21.0018
= (654.16 - 41.1635 , 654.16 - 41.1635)
= (612.9965 , 695.3235) ppm
= (613.00 , 695.32)(rounded)

=> we are 95% confident that this interval contains the true population mean.

  
b) by given s = 179

2*Z*s/sqrtn = 40
= 2*1.96*( 179/sqrtn) = 40
= 1.96*( 179/sqrtn) = 20
= sqrtn = (1.96*179)/20
= sqrtn = 17.542
= n = 17.5422 = 307.7217
= n = 308
  

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