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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

As the population ages, there is increasing concern about accident-related injuries to the elderly. The article "Age and Gender Differences in Single-Step Recovery from a Forward Fall" (J. of Gerontology, \(1999: M44 - M50\)) reported on an experiment in which the maximum lean angle-the farthest a subject is able to lean and still recover in one step-was determined for both a sample of younger females (\(21 - 29\)years) and a sample of older females (\(67 - 81\)years). The following observations are consistent with summary data given in the article:

YF:\(29,34,33,27,28,32,31,34,32,27\)

OF:\(18,15,23,13,12\)

Does the data suggest that true average maximum lean angle for older females is more than\(10\)degrees smaller than it is for younger females? State and test the relevant hypotheses at significance level . \(10\).

Cushing's disease is characterized by muscular weakness due to adrenal or pituitary dysfunction. To provide effective treatment, it is important to detect childhood Cushing's disease as early as possible. Age at onset of symptoms and age at diagnosis (months) for 15 children suffering from the disease were given in the article "Treatment of Cushing's Disease in Childhood and Adolescence by Transphenoidal Microadenomectomy" (New Engl. J. of Med., 1984: 889). Here are the values of the differences between age at onset of symptoms and age at diagnosis:

\(\begin{array}{*{20}{l}}{ - 24}&{ - 12}&{ - 55}&{ - 15}&{ - 30}&{ - 60}&{ - 14}&{ - 21}\\{ - 48}&{ - 12}&{ - 25}&{ - 53}&{ - 61}&{ - 69}&{ - 80}&{}\end{array}\)

a. Does the accompanying normal probability plot cast strong doubt on the approximate normality of the population distribution of differences?

b. Calculate a lower\(95\% \)confidence bound for the population mean difference, and interpret the resulting bound.

c. Suppose the (age at diagnosis) - (age at onset) differences had been calculated. What would be a\(95\backslash \% \)upper confidence bound for the corresponding population mean difference?

Use the data of Exercise 19 to calculate a \(95\% CI\) for the difference between true average stopping distance for cars equipped with system \(1\) and cars equipped with system \(2\). Does the interval suggest that precise information about the value of this difference is available?

In an experiment to compare bearing strengths of pegs inserted in two different types of mounts, a sample of 14 observations on stress limit for red oak mounts resulted in a sample mean and sample standard deviation of \(8.48MPa\) and .79 MPa, respectively, whereas a sample of 12 observations when Douglas fir mounts were used gave a mean of \(9.36\) and a standard deviation of \(1.52\) ('Bearing Strength of White Oak Pegs in Red Oak and Douglas Fir Timbers," J. of Testing and Evaluation, 1998, 109-114). Consider testing whether or not true average stress limits are identical for the two types of mounts. Compare df's and P-values for the unpooled and pooled t tests.

Which way of dispensing champagne, the traditional vertical method or a tilted beer-like pour,preserves more of the tiny gas bubbles that improve flavor and aroma? The following data was reported in the article 鈥淥n the Losses of Dissolved \(C{O_2}\) during Champagne Serving鈥� (J. Agr. Food Chem., 2010: 8768鈥�8775)

\(\begin{array}{*{20}{c}}{ Temp \left( {^^\circ C} \right)}&{ Type of Pour }&n&{ Mean (g/L)}&{ SD }\\{18}&{ Traditional }&4&{4.0}&{.5}\\{18}&{ Slanted }&4&{3.7}&{.3}\\{12}&{ Traditional }&4&{3.3}&{.2}\\{12}&{ Slanted }&4&{2.0}&{.3}\\{}&{}&{}&{}&{}\end{array}\)

Assume that the sampled distributions are normal.

a. Carry out a test at significance level \(.01\) to decide whether true average\(C{O_2}\)loss at \(1{8^o}C\) for the traditional pour differs from that for the slanted pour.

b. Repeat the test of hypotheses suggested in (a) for the \(1{2^o}\) temperature. Is the conclusion different from that for the \(1{8^o}\) temperature? Note: The \(1{2^o}\) result was reported in the popular media
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