An article reported that for a sample of 55 kitchens with gas cooking appliances

Question

An article reported that for a sample of 55 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 163.53.

(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.

(b) 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 59 ppm for a confidence level of 95%? (Round your answer up to the nearest whole number.)

Explanation / Answer

a.
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=654.16
Standard deviation( sd )=163.53
Sample Size(n)=55
Confidence Interval = [ 654.16 ± t a/2 ( 163.53/ Sqrt ( 55) ) ]
= [ 654.16 - 2.005 * (22.05) , 654.16 + 2.005 * (22.05) ]
= [ 609.95,698.37 ]

b.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )
Standard Deviation ( S.D) = 163.53
ME =59
n = ( 1.96*163.53/59) ^2
= (320.519/59 ) ^2
= 29.512 ~ 30  

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