Question

The article “Gas Cooking, Kitchen Ventilation, and Exposure to Combustion Products” reported that for a sample of 50 kitchens with gas cooking appliances monitored during a one week period, the sample mean CO2 level (ppm) was \({\rm{654}}{\rm{.16}}\), and the sample standard deviation was \({\rm{164}}{\rm{.43}}{\rm{.}}\)

a. Calculate and interpret a \({\rm{95\% }}\) (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 \({\rm{175}}\) for the value of s before collecting data. What sample size would be necessary to obtain an interval width of ppm for a confidence level of \({\rm{95\% }}\)

Short answer

a) The \(95\% \)confidence interval for true average is \({\rm{(608}}{\rm{.58,699}}{\rm{.74)}}\)

b) The necessary sample size n is \({\rm{n - 189}}\)

Step-by-step solution

To calculate and interpret a 95% (two-sided) confidence interval

(a):

For large n, the standardized random variable

\({\rm{Z = }}\frac{{{\rm{\bar X - \mu }}}}{{{\rm{S/}}\sqrt {\rm{n}} }}\)

has approximately a normal distribution with expectation \(0\) and standard deviation\(1\). Therefore, a

large-sample confidence interval for mu is

\({\rm{\bar x \pm }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}\)

with confidence level of approximately\({\rm{100(1 - \alpha )\% }}\). This stands regardless of the population distribution.

The \(95\% \)confidence interval for true average is

\(\begin{array}{l}\left( {{\rm{\bar x - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}{\rm{,\bar x + }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}} \right){\rm{ = }}\left( {{\rm{654}}{\rm{.16 - 1}}{\rm{.96 \times }}\frac{{{\rm{165}}{\rm{.43}}}}{{\sqrt {{\rm{50}}} }}{\rm{,654}}{\rm{.16 + 1}}{\rm{.96 \times }}\frac{{{\rm{165}}{\rm{.43}}}}{{\sqrt {{\rm{50}}} }}} \right)\\{\rm{ = (608}}{\rm{.58,699}}{\rm{.74)}}\end{array}\)

Where

\({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ - }}{{\rm{z}}_{{\rm{0}}{\rm{.05/2}}}}{\rm{ - }}{{\rm{z}}_{{\rm{0}}{\rm{.025}}}}{\rm{ - 1}}{\rm{.96}}\)

this is obtained from

\({\rm{P}}\left( {{\rm{Z > }}{{\rm{z}}_{{\rm{0}}{\rm{.025}}}}} \right){\rm{ - 0}}{\rm{.025}}\)

and from the normal probability table in the appendix. The probability can also be computed with software.

Hence The \(95\% \)confidence interval for true average is \({\rm{(608}}{\rm{.58,699}}{\rm{.74)}}\)

To find the sample size

(b)

In order for confidence intervals.

\(\left( {{\rm{\bar x - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}{\rm{,\bar x + }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{,}}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}} \right)\)

to have width w, the

necessary sample size n

is

\({\rm{n - }}{\left( {{\rm{2}}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{s}}}{{\rm{w}}}} \right)^{\rm{2}}}\)

The smaller width w, the larger n must be. The necessary sample size to obtain an width of\(\;50\)ppm with \(95\% \)confidence level is

\({\rm{n - }}{\left( {{\rm{2}}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{s}}}{{\rm{w}}}} \right)^{\rm{2}}}{\rm{ - }}{\left( {{\rm{2 \times 1}}{\rm{.96 \times }}\frac{{{\rm{175}}}}{{{\rm{50}}}}} \right)^{\rm{2}}}{\rm{ - 13}}{\rm{.7}}{{\rm{2}}^{\rm{2}}}{\rm{ - 188}}{\rm{.24}}\)

Because we need an integer, the first bigger integer is \(189\) and

Therefore the necessary sample size n is \({\rm{n - 189}}\)