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 166.59.

(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

(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 50 ppm for a confidence level of 95%?

Explanation / Answer

PART A.

TRADITIONAL METHOD

given that,

sample mean, x =654.16

standard deviation, s =166.59

sample size, n =60

I.

stanadard error = sd/ sqrt(n)

where,

sd = standard deviation

n = sample size

standard error = ( 166.59/ sqrt ( 60) )

= 21.507

II.

margin of error = t /2 * (stanadard error)

where,

ta/2 = t-table value

level of significance, = 0.05

from standard normal table, two tailed value of |t /2| with n-1 = 59 d.f is 2.001

margin of error = 2.001 * 21.507

= 43.035

III.

CI = x ± margin of error

confidence interval = [ 654.16 ± 43.035 ]

= [ 611.125 , 697.195 ]

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DIRECT METHOD

given that,

sample mean, x =654.16

standard deviation, s =166.59

sample size, n =60

level of significance, = 0.05

from standard normal table, two tailed value of |t /2| with n-1 = 59 d.f is 2.001

we use 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

confidence interval = [ 654.16 ± t a/2 ( 166.59/ Sqrt ( 60) ]

= [ 654.16-(2.001 * 21.507) , 654.16+(2.001 * 21.507) ]

= [ 611.125 , 697.195 ]

-----------------------------------------------------------------------------------------------

interpretations:

1) we are 95% sure that the interval [ 611.125 , 697.195 ] contains the true population mean

2) If a large number of samples are collected, and a confidence interval is created

for each sample, 95% of these intervals will contains the true population mean

PART 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) = 166.59

ME =50

n = ( 1.96*166.59/50) ^2

= (326.52/50 ) ^2

= 42.65 ~ 43

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