/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q37 E Hexavalent chromium has been ide... [FREE SOLUTION] | 魅影直播

魅影直播

Log out

Learning Materials

Features

Discover

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:

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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 魅影直播!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Many freeways have service (or logo) signs that give information on attractions, camping, lodging, food, and gas services prior to off-ramps. These signs typically do not provide information on distances. The article "Evaluation of Adding Distance Information to Freeway-Specific Service (Logo) Signs" \((J\). of Transp. Engr., 2011: 782-788) reported that in one investigation, six sites along Virginia interstate highways where service signs are posted were selected. For each site, crash data was obtained for a three-year period before distance information was added to the service signs and for a one-year period afterward. The number of crashes per year before and after the sign changes were as follows:

\(\begin{array}{*{20}{l}}{Before:\;\;15\;26\;66\;115\;62\;64}\\{After:\;\;\;\;\;16\;24\;42\;80\;\;78\;73}\end{array}\)

a. The cited article included the statement "A paired\(t\)test was performed to determine whether there was any change in the mean number of crashes before and after the addition of distance information on the signs." Carry out such a test. (Note: The relevant normal probability plot shows a substantial linear pattern.)

b. If a seventh site were to be randomly selected among locations bearing service signs, between what values would you predict the difference in number of crashes to lie?

Reliance on solid biomass fuel for cooking and heating exposes many children from developing countries to high levels of indoor air pollution. The article 鈥淒omestic Fuels, Indoor Air Pollution, and Children鈥檚 Health鈥� (Annals of the N.Y. Academy of Sciences, \(2008:209 - 217\)) presented information on various pulmonary characteristics in samples of children whose households in India used either biomass fuel or liquefied petroleum gas (\(LPG\)). For the \(755\) children in biomass households, the sample mean peak expiratory flow (a person鈥檚 maximum speed of expiration) was \(3.30L/s\), and the sample standard deviation was \(1.20\). For the \(750\) children whose households used liquefied petroleum gas, the sample mean \(PEF\) was \(4.25\) and the sample standard deviation was \(1.75\).

a. Calculate a confidence interval at the \(95\% \) confidence level for the population mean \(PEF\) for children in biomass households and then do likewise for children in \(LPG\) households. What is the simultaneous confidence level for the two intervals?

b. Carry out a test of hypotheses at significance level \(.01\) to decide whether true average \(PEF\) is lower for children in biomass households than it is for children in \(LPG\) households (the cited article included a P-value for this test)

c. \(FE{V_1}\), the forced expiratory volume in \(1\) second, is another measure of pulmonary function. The cited article reported that for the biomass households the sample mean FEV1 was \(2.3L/s\) and the sample standard deviation was \(.5L/s\). If this information is used to compute a \(95\% \) \(CI\) for population mean \(FE{V_1}\), would the simultaneous confidence level for this interval and the first interval calculated in (a) be the same as the simultaneous confidence level determined there? Explain

Suppose \({\mu _1}\) and \({\mu _2}\) are true mean stopping distances at \(50mph\) for cars of a certain type equipped with two different types of braking systems. Use the two-sample t test at significance level

t test at significance level \(.01\) to test \({H_0}:{\mu _1} - {\mu _2} = - 10\) versus \({H_a}:{\mu _1} - {\mu _2} < - 10\) for the following data: \(m = 6,\;\;\;\bar x = 115.7,{s_1} = 5.03,n = 6,\bar y = 129.3,\;\)and \({s_2} = 5.38.\)

According to the article "Modeling and Predicting the Effects of Submerged Arc Weldment Process Parameters on Weldment Characteristics and Shape Profiles" (J. of Engr. Manuf., \(2012: 1230 - 1240\)), the submerged arc welding (SAW) process is commonly used for joining thick plates and pipes. The heat affected zone (HAZ), a band created within the base metal during welding, was of particular interest to the investigators. Here are observations on depth\((mm)\)of the HAZ both when the current setting was high and when it was lower.

\(\begin{array}{*{20}{l}}{ Non - high }&{1.04}&{1.15}&{1.23}&{1.69}&{1.92}\\{}&{1.98}&{2.36}&{2.49}&{2.72}&{}\\{}&{1.37}&{1.43}&{1.57}&{1.71}&{1.94}\\{ High }&{2.06}&{2.55}&{2.64}&{2.82}&{}\\{}&{1.55}&{2.02}&{2.02}&{2.05}&{2.35}\\{}&{2.57}&{2.93}&{2.94}&{2.97}&{}\end{array}\)

a. Construct a comparative boxplot and comment on interesting features.

b. Is it reasonable to use the two-sample \(t\) test to test hypotheses about the difference between true average HAZ depths for the two conditions?

c. Does it appear that true average HAZ depth is larger for the higher current condition than for the lower condition? Carry out a test of appropriate hypotheses using a significance level of . \(01\).

The article "Effect of Internal Gas Pressure on the Compression Strength of Beverage Cans and Plastic Bottles" (J. of Testing and Evaluation, \(1993: 129 - 131\)) includes the accompanying data on compression strength (lb) for a sample of\(12 - oz\)aluminum cans filled with strawberry drink and another sample filled with cola. Does the data suggest that the extra carbonation of cola results in a higher average compression strength? Base your answer on a\(P\)-value. What assumptions are necessary for your analysis?

\(\begin{array}{l}BeverageSampleSizeSampleMeanSampleSD\\Strawberrydrink1554021\\Cola1555415\end{array}\)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.