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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

Torsion during hip external rotation (ER) and extension may be responsible for certain kinds of injuries in golfers and other athletes. The article "Hip Rotational Velocities During the Full Golf Swing" (J. of Sports Science and Medicine, 2009: 296-299) reported on a study in which peak ER velocity and peak IR (internal rotation) velocity (both in deg.sec\(^{ - 1}\)) were determined for a sample of 15 female collegiate golfers during their swings. The following data was supplied by the article's authors.

\(\begin{aligned}{*{20}{r}}{ Golfer }&{ \backslash multicolumn1c ER }&{ IR }&{ \backslash multicolumn1c diff }&{z perc }\\1&{ - 130.6}&{ - 98.9}&{ - 31.7}&{ - 1.28}\\2&{ - 125.1}&{ - 115.9}&{ - 9.2}&{ - 0.97}\\3&{ - 51.7}&{ - 161.6}&{109.9}&{0.34}\\4&{ - 179.7}&{ - 196.9}&{17.2}&{ - 0.73}\\5&{ - 130.5}&{ - 170.7}&{40.2}&{ - 0.34}\\6&{ - 101.0}&{ - 274.9}&{173.9}&{0.97}\\7&{ - 24.4}&{ - 275.0}&{250.6}&{1.83}\\8&{ - 231.1}&{ - 275.7}&{44.6}&{ - 0.17}\\9&{ - 186.8}&{ - 214.6}&{27.8}&{ - 0.52}\\{10}&{ - 58.5}&{ - 117.8}&{59.3}&{0.00}\\{11}&{ - 219.3}&{ - 326.7}&{107.4}&{0.17}\\{12}&{ - 113.1}&{ - 272.9}&{159.8}&{0.73}\\{13}&{ - 244.3}&{ - 429.1}&{184.8}&{1.28}\\{14}&{ - 184.4}&{ - 140.6}&{ - 43.8}&{ - 1.83}\\{15}&{ - 199.2}&{ - 345.6}&{146.4}&{0.52}\\{}&{}&{}&{}&{}\end{aligned}\)

a. Is it plausible that the differences came from a normally distributed population?

b. The article reported that mean\(( \pm SD) = - 145.3(68.0)\)for ER velocity and\( = - 227.8(96.6)\)for IR velocity. Based just on this information, could a test of hypotheses about the difference between true average IR velocity and true average ER velocity be carried out? Explain.

c. The article stated that " The lead hip peak IR velocity was significantly greater than the trail hip ER velocity\(\;\left( {p = 0.003, t value = 3.65} \right)\). "$ (The phrasing suggests that an upper-tailed test was used.) Is that in fact the case? (Note: "\(p = .033 \)in Table 2 of the article is erroneous.)

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.

An experimenter wishes to obtain a CI for the difference between true average breaking strength for cables manufactured by company I and by company II. Suppose breaking strength is normally distributed for both types of cable with s1 5 30 psi and s2 5 20 psi.

  1. If costs dictate that the sample size for the type I cable should be three times the sample size for the type II cable, how many observations are required if the 99% CI is to be no wider than 20 psi?
  2. Suppose a total of 400 observations is to be made. How many of the observations should be made on type I cable samples if the width of the resulting interval is to be a minimum?

Sometimes experiments involving success or failure responses are run in a paired or before/after manner. Suppose that before a major policy speech by a political candidate, n individuals are selected and asked whether \((S)\)or not (F) they favor the candidate. Then after the speech the same n people are asked the same question. The responses can be entered in a table as follows:

Before

After

S

F

S

\({{\bf{X}}_{\bf{1}}}\)

\({{\bf{X}}_{\bf{2}}}\)

F

\({{\bf{X}}_{\bf{3}}}\)

\({{\bf{X}}_{\bf{4}}}\)

Where\({{\bf{x}}_{\bf{1}}}{\bf{ + }}{{\bf{x}}_{\bf{2}}}{\bf{ + }}{{\bf{x}}_{\bf{3}}}{\bf{ + }}{{\bf{x}}_{\bf{4}}}{\bf{ = n}}\). Let\({{\bf{p}}_{\bf{1}}}{\bf{,}}{{\bf{p}}_{\bf{2}}}{\bf{,}}{{\bf{p}}_{\bf{3}}}\), and \({p_4}\)denote the four cell probabilities, so that \({p_1} = P(S\) before and S after), and so on. We wish to test the hypothesis that the true proportion of supporters (S) after the speech has not increased against the alternative that it has increased.

a. State the two hypotheses of interest in terms of\({p_1},{p_2}\),\({p_3}\), and \({p_4}\).

b. Construct an estimator for the after/before difference in success probabilities

c. When n is large, it can be shown that the rv \(\left( {{{\bf{X}}_{\bf{i}}}{\bf{ - }}{{\bf{X}}_{\bf{j}}}} \right){\bf{/n}}\) has approximately a normal distribution with variance given by\(\left[ {{{\bf{p}}_{\bf{i}}}{\bf{ + }}{{\bf{p}}_{\bf{j}}}{\bf{ - }}{{\left( {{{\bf{p}}_{\bf{i}}}{\bf{ - }}{{\bf{p}}_{\bf{j}}}} \right)}^{\bf{2}}}} \right]{\bf{/n}}\). Use this to construct a test statistic with approximately a standard normal distribution when \({H_0}\)is true (the result is called McNemar's test).

d. If\({{\bf{x}}_{\bf{1}}}{\bf{ = 350,}}\;\;\;{{\bf{x}}_{\bf{2}}}{\bf{ = 150,}}\;\;\;{{\bf{x}}_{\bf{3}}}{\bf{ = 200}}\), and\({x_4} = 300\), what do you conclude?

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 “On the Losses of Dissolved \(C{O_2}\) during Champagne Serving” (J. Agr. Food Chem., 2010: 8768–8775)

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

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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