An article reported that for a sample of 46 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 163.7.

Required:
a. Calculate and interpret a 9596 (two-sided) confidence interval for true average CO2 level in the population of all homes from which the sample was selected.
b. Suppose the investigators had made a rough guess of 175 for the value of s before collecting data. What sample size would be necessary to obtain an interval width of 50 ppm for a confidence level of 95%?

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jtellezd jtellezd · Answer 1 · 2020-07-29T13:47:26+02:00

Answer:

a) CI = ( 148,69 ; 243,31 )

b) n = 189

Step-by-step explanation:

a) If the Confidence Interval is 95 %

α = 5 % or α = 0,05 and α/2 = 0,025

citical value for α/2 = 0,025 is z(c) = 1,96

the MOE ( margin of error is )

1,96* s/√n

1,96* 163,7/ √46

MOE = 47,31

Then CI = 196 ± 47,31

CI = ( 148,69 ; 243,31 )

CI look very wide ( it sems that if sample size was too low )

b) Now if s (sample standard deviation) is 175, and we would like to have only 50 ppm width with Confidence level 95 %, we need to make

MOE = 25 = z(c) * s/√n

25*√n = z(c)* 175

√n = 1,96*175/25

√n = 13,72

n = 188,23

as n is an integer number we make n = 189