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Chapter 9: Q37 E (page 388)

Hexavalent chromium has been identified as an inhalation carcinogen and an air toxin of concern in a number of different locales. The article "Airborne Hexavalent Chromium in Southwestern Ontario"(J . of Air and Waste Mgmnt. Assoc., \(1997: 905 - 910\)) gave the accompanying data on both indoor and outdoor concentration (nanograms\(/{m^3}\)) for a sample of houses selected from a certain

House

\(\begin{array}{*{20}{l}}{\;\;\;\;\;\;\;\;\;\;\;\;1\;\;\;\;\;\;2\;\;\;\;\;3\;\;\;\;\;\;\;4\;\;\;\;\;5\;\;\;\;\;\;\;6\;\;\;\;\;\;\;7\;\;\;\;\;\;\;8\;\;\;\;\;\;\;9}\\{Indoor\;\;\;\;\;\;\;\;\;.07\;\;\;\;.08\;\;\;\;.09\;\;\;\;.12\;\;\;\;.12\;\;\;\;.12\;\;\;\;.13\;\;\;\;.14\;\;\;\;.15}\\{Outdoor\;\;\;\;\;\;.29\;\;\;\;.68\;\;\;\;.47\;\;\;\;.54\;\;\;\;.97\;\;\;\;.35\;\;\;\;.49\;\;\;\;.84\;\;\;\;.86}\end{array}\)

House

\(\begin{array}{*{20}{l}}{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;10\;\;\;\;\;11\;\;\;\;\;12\;\;\;\;\;13\;\;\;\;\;14\;\;\;\;\;15\;\;\;\;\;16\;\;\;\;\;17}\\{Indoor\;\;\;\;\;\;\;\;\;.15\;\;\;\;.17\;\;\;\;.17\;\;\;\;.18\;\;\;\;.18\;\;\;\;.18\;\;\;\;.18\;\;\;\;.19}\\{Outdoor\;\;\;\;\;\;.28\;\;\;\;.32\;\;\;\;.32\;\;\;\;1.55\;\;\;.66\;\;\;\;.29\;\;\;\;.21\;\;\;\;1.02}\end{array}\)

House

\(\begin{array}{*{20}{c}}{}&{18}&{19}&{20}&{21}&{22}&{23}&{24}&{25}\\{Indoor}&{.20}&{.22}&{.22}&{.23}&{.23}&{.25}&{.26}&{.28}\\{Outdoor}&{1.59}&{.90}&{.52}&{.12}&{.54}&{.88}&{.49}&{1.24}\end{array}\)

House

\(\begin{array}{*{20}{c}}{}&{26}&{27}&{28}&{29}&{30}&{31}&{32}&{33}\\{Indoor}&{.28}&{.29}&{.34}&{.39}&{.40}&{.45}&{.54}&{.62}\\{Outdoor}&{.48}&{.27}&{.37}&{1.26}&{.70}&{.76}&{.99}&{.36}\end{array}\)

a. Calculate a confidence interval for the population mean difference between indoor and outdoor concentrations using a confidence level of\(95\% \), and interpret the resulting interval.

b. If a\(34\)th house were to be randomly selected from the population, between what values would you predict the difference in concentrations to lie?region.

Short Answer

Expert verified

(a) \(( - 0.5611, - 0.2867)\)

(b) \(( - 1.2237,0.3759)\)

Step by step solution

01

A)Step 1:

Given:

\(c = 95\% = 0.95\)

Determine the difference in value of each pair

02

Step 2:

Determine the sample mean of the differences. The mean is the sum of all values divided by the number of values.

\(\bar d = \frac{{ - 0.22 - 0.6 - 0.38 + \ldots - 0.31 - 0.45 + 0.26}}{{33}} \approx - 0.4239\)

Determine the sample standard deviation of the differences:

\({s_d} = \sqrt {\frac{{{{( - 0.22 - ( - 0.4239))}^2} + \ldots + {{(0.26 - ( - 0.4239))}^2}}}{{33 - 1}}} \approx 0.3868\)

Determine the \({t_{\alpha /2}}\) using the Student's T distribution table in the appendix with \(df = n - 1 = 33 - 1 = 32\) :

\({t_{0.025}} = 2.037\)

The margin of error is then:

\(E = {t_{\alpha /2}} \cdot \frac{{{s_d}}}{{\sqrt n }} = 2.037 \cdot \frac{{0.3868}}{{\sqrt {33} }} \approx 0.1372\)

The endpoints of the confidence interval for $\mu_{d}$ are: \(\begin{array}{l}\bar d - E = - 0.4239 - 0.1372 = - 0.5611\\\bar d + E = - 0.4239 + 0.1372 = - 0.2867\end{array}\)

03

b)Step 1:

We need to determine a prediction interval for the\(34\)th house.

The margin of error is then:

\(E = {t_{\alpha /2}} \cdot {s_d}\sqrt {1 + \frac{1}{n}} = 2.037 \cdot 0.3868\sqrt {1 + \frac{1}{{33}}} \approx 0.7998\)

The endpoints of the prediction interval for\({\mu _d}\)are:

\(\begin{array}{l}\bar d - E = - 0.4239 - 0.7998 = - 1.2237\\\bar d + E = - 0.4239 + 0.7998 = 0.3759\end{array}\)

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Most popular questions from this chapter

The level of monoamine oxidase (MAO) activity in blood platelets (nm/mg protein/h) was determined for each individual in a sample of \(43\) chronic schizophrenics, resulting in \(\bar x = 2.69\) and \({s_1} = 2.30,\), as well as for \(45\) normal subjects, resulting in \(\bar y = 6.35\) and \({s_2} = 4.03.\). Does this data strongly suggest that true average MAO activity for normal subjects is more than twice the activity level for schizophrenics? Derive a test procedure and carry out the test using \(\alpha = .01\)

. (Hint: \({H_0}\) and \({H_a}\) here have a different form from the three standard cases. Let \({\mu _1}\) and \({\mu _2}\) refer to true average MAO activity for schizophrenics and normal subjects, respectively, and consider the parameter \(\theta = 2{\mu _1} - {\mu _2}\). Write \({H_0}\) and \({H_a}\) in terms of \(\theta \), estimate \(\theta \), and derive \({\hat \sigma _{\tilde \theta }}\) (鈥淩educed Monoamine Oxidase Activity in Blood Platelets from Schizophrenic Patients,鈥� Nature, July 28, 1972: 225鈥�226).) \(\alpha = .01\)

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\(\begin{array}{*{20}{l}}{ Epoxy }&{1.75}&{2.12}&{2.05}&{1.97}\\{ MMA prepolymer }&{1.77}&{1.59}&{1.70}&{1.69}\end{array}\)

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