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Chapter 7: Q13E (page 292)

The article 鈥淕as Cooking, Kitchen Ventilation, and Exposure to Combustion Products鈥� 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

Expert verified

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

01

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 \({\rm{(608}}{\rm{.58,699}}{\rm{.74)}}\)

02

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 \({\rm{n - 189}}\)

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Most popular questions from this chapter

Example 1.11 introduced the accompanying observations on bond strength.

11.5 12.1 9.9 9.3 7.8 6.2 6.6 7.0

13.4 17.1 9.3 5.6 5.7 5.4 5.2 5.1

4.9 10.7 15.2 8.5 4.2 4.0 3.9 3.8

3.6 3.4 20.6 25.5 13.8 12.6 13.1 8.9

8.2 10.7 14.2 7.6 5.2 5.5 5.1 5.0

5.2 4.8 4.1 3.8 3.7 3.6 3.6 3.6

a.Estimate true average bond strength in a way that conveys information about precision and reliability.

(Hint: \(\sum {{{\bf{x}}_{\bf{i}}}} {\bf{ = 387}}{\bf{.8}}\) and \(\sum {{{\bf{x}}^{\bf{2}}}_{\bf{i}}} {\bf{ = 4247}}{\bf{.08}}\).)

b. Calculate a 95% CI for the proportion of all such bonds whose strength values would exceed 10.

The Pew Forum on Religion and Public Life reported on \({\rm{Dec}}{\rm{. 9, 2009}}\), that in a survey of \({\rm{2003}}\) American adults, \({\rm{25\% }}\)said they believed in astrology.

a. Calculate and interpret a confidence interval at the \({\rm{99\% }}\)confidence level for the proportion of all adult Americans who believe in astrology.

b. What sample size would be required for the width of a \({\rm{99\% }}\)CI to be at most .05 irrespective of the value of p?

A journal article reports that a sample of size 5 was used as a basis for calculating a 95% CI for the true average natural frequency (Hz) of delaminated beams of a certain type. The resulting interval was (229.764, 233.504). You decide that a confidence level of 99% is more appropriate than the 95% level used. What are the limits of the 99% interval? (Hint: Use the center of the interval and its width to determine \(\overline x \) and s.)

The negative effects of ambient air pollution on children鈥檚 lung function has been well established, but less research is available about the impact of indoor air pollution. The authors of 鈥淚ndoor Air Pollution and Lung Function Growth Among Children in Four Chinese Cities鈥� investigated the relationship between indoor air-pollution metrics and lung function growth among children ages \({\rm{6--13}}\)years living in four Chinese cities. For each subject in the study, the authors measured an important lung-capacity index known as FEV1, the forced volume (in ml) of air that is exhaled in \({\rm{1}}\) second. Higher FEV1 values are associated with greater lung capacity. Among the children in the study, \({\rm{514}}\) came from households that used coal for cooking or heating or both. Their FEV1 mean was \({\rm{1427}}\)with a standard deviation of \({\rm{325}}\). (A complex statistical procedure was used to show that burning coal had a clear negative effect on mean FEV1 levels.)

a. Calculate and interpret a \({\rm{95\% }}\) (two-sided) confidence interval for true average FEV1 level in the population of all children from which the sample was selected. Does it appear that the parameter of interest has been accurately estimated?

b. Suppose the investigators had made a rough guess of \({\rm{320}}\) for the value of s before collecting data. What sample size would be necessary to obtain an interval width of \({\rm{50}}\)ml for a confidence level of \({\rm{95\% }}\)?

A study of the ability of individuals to walk in a straight line reported the accompanying data on cadence (strides per second) for a sample of n =\({\rm{20}}\) randomly selected healthy men.

\({\rm{.95 }}{\rm{.85 }}{\rm{.92 }}{\rm{.95 }}{\rm{.93 }}{\rm{.86 1}}{\rm{.00 }}{\rm{.92 }}{\rm{.85 }}{\rm{.81 }}{\rm{.78 }}{\rm{.93 }}{\rm{.93 1}}{\rm{.05 }}{\rm{.93 1}}{\rm{.06 1}}{\rm{.06 }}{\rm{.96 }}{\rm{.81 }}{\rm{.96}}\)

A normal probability plot gives substantial support to the assumption that the population distribution of cadence is approximately normal. A descriptive summary of the data from Minitab follows:

Variable N Mean Median TrMean StDev SEMean cadence

\({\rm{20 0}}{\rm{.9255 0}}{\rm{.9300 0}}{\rm{.9261 0}}{\rm{.0809 0}}{\rm{.0181}}\)

Variable Min Max Q1 Q3 cadence

\({\rm{0}}{\rm{.7800 1}}{\rm{.0600 0}}{\rm{.8525 0}}{\rm{.9600}}\)

a. Calculate and interpret a \({\rm{95\% }}\) confidence interval for population mean cadence.

b.Calculate and interpret a \({\rm{95\% }}\)prediction interval for the cadence of a single individual randomly selected from this population.

c. Calculate an interval that includes at least \({\rm{99\% }}\)of the cadences in the population distribution using a confidence level of \({\rm{95\% }}\)
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