Question

An article reported that for a sample of 52 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 164.55.

a) calculate and interpret a 95% 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 sanple size would be necessary to obtain an interval width of 50 ppm for confidence level of 95% ?

Answer 1

a)
`654.16-2.01(164.55)/(√(52))=608.29`

`654.16+2.01(164.55)/(√(52))=700.03`

And we can conclude that we are 95% confident that the true mean of Co2 level is between 608.29 and 700.03 ppm

b)
`n=((1.960(175))/(25))^2 =188.23 \approx 189`

Explanation:

Part a

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=52-1=51`

Since the Confidence is 0.95 or 95%, the value of
`\alpha=0.05` and
`\alpha/2 =0.025`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,51)".And we see that
`t_(\alpha/2)=2.01`

Replacing we got:

`654.16-2.01(164.55)/(√(52))=608.29`

`654.16+2.01(164.55)/(√(52))=700.03`

And we can conclude that we are 95% confident that the true mean of Co2 level is between 608.29 and 700.03 ppm

Part b

The margin of error is given by :

`ME=z_(\alpha/2)(\sigma)/(√(n))` (a)

The desired margin of error is ME =50/2=25 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

`n=((z_(\alpha/2) s)/(ME))^2` (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got
`z_(\alpha/2)=1.960`, and we use an estimator of the population variance the value of 175 replacing into formula (b) we got:

`n=((1.960(175))/(25))^2 =188.23 \approx 189`