/*! 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} Q12 E The accompanying summary data on... [FREE SOLUTION] | 91影视

91影视

The accompanying summary data on total cholesterol level (mmol/l) was obtained from a sample of Asian postmenopausal women who were vegans and another sample of such women who were omnivores (鈥淰egetarianism, Bone Loss, and Vitamin D: A Longitudinal Study in Asian Vegans and Non-Vegans,鈥 European J. of Clinical Nutr., 2012: 75鈥82)

Diet sample sample sample

Size mean SD

\(\overline {\underline {\begin{array}{*{20}{l}}{ Vegan }&{88}&{5.10}&{1.07}\\{ Omnivore }&{93}&{5.55}&{1.10}\\{}&{}&{}&{}\end{array}} } \)

Calculate and interpret a \(99\% \) \(CI\) for the difference between population mean total cholesterol level for vegans and population mean total cholesterol level for omnivores (the cited article included a \(95\% \)\(CI\)). (Note: The article described a more sophisticated statistical analysis for investigating bone density loss taking into account other characteristics (鈥渃ovariates鈥) such as age, body weight, and various nutritional factors; the resulting CI included 0, suggesting no diet effect.

Short Answer

Expert verified

the solution is

\(( - 0.8654, - 0.0346)\)

The genuine difference between the population mean total cholesterol level for vegans and the population mean total cholesterol level for omnivores is between\( - 0.8654mmol/l\)and\( - 0.0346mmol/l\), according to our\(99\)percent confidence level.

Step by step solution

01

given

\(\begin{array}{l}{{\bar x}_1} = 5.10\\{{\bar x}_2} = 5.55\\{s_1} = 1.07\\{s_2} = 1.10\end{array}\)

\(\begin{array}{l}{n_1} = 88\\{n_2} = 93\\c = 99\% = 0.99\end{array}\)

02

confidence interval

Determine \({z_{\alpha /2}} = {z_{0.005}}\) using the normal probability table (search up \(0.005\) in the table, then the \(z\)-score is the discovered \(z\)-score with opposite sign) with confidence level \(1 - \alpha = 0.99\):

\({z_{\alpha /2}} = 2.575\)

Because \(0.005\) is exactly in the middle of \(0.0049\) and \(0.0051.\), we choose the average of \(2.57\)and \(2.58\).

As a result, the error margin is:

\(E = {z_{\alpha /2}} \times \sqrt {\frac{{\sigma _1^2}}{{{n_1}}} + \frac{{\sigma _2^2}}{{{n_2}}}} = 2.575 \times \sqrt {\frac{{1.0{7^2}}}{{88}} + \frac{{1.1{0^2}}}{{93}}} \approx 0.4154\)

The endpoints of the confidence interval for \({\mu _1} - {\mu _2}\)

\(\begin{array}{l}\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E = (5.10 - 5.55) - 0.4154 = - 0.45 - 0.4116 = - 0.8654\\\left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E = (5.10 - 5.55) + 0.4154 = - 0.45 + 0.4116 = - 0.0346\end{array}\)

The genuine difference between the population mean total cholesterol level for vegans and the population mean total cholesterol level for omnivores is between \( - 0.8654mmol/l\) and \( - 0.0346mmol/l\), according to our \(99\) percent confidence level.

03

conclusion

\(( - 0.8654, - 0.0346)\)

The genuine difference between the population mean total cholesterol level for vegans and the population mean total cholesterol level for omnivores is between \( - 0.8654mmol/l\)and \( - 0.0346mmol/l\), according to our \(99\)percent confidence level.

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 91影视!

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

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.

Example\(7.11\)gave data on the modulus of elasticity obtained\(1\)minute after loading in a certain configuration. The cited article also gave the values of modulus of elasticity obtained\(4\)weeks after loading for the same lumber specimens. The data is presented here. \(1\begin{array}{*{20}{c}}{Observation}&{1 min}&{4 weeks}&{Difference}\\1&{10,490}&{9,110}&{1380}\\2&{16,620}&{13,250}&{3370}\\3&{17,300}&{14,720}&{2580}\\4&{15,480}&{12,740}&{2740}\\5&{12,970}&{10,120}&{2850}\\6&{17,260}&{14,570}&{2690}\\7&{13,400}&{11,220}&{2180}\\8&{13,900}&{11,100}&{2800}\\9&{13,630}&{11,420}&{2210}\\{10}&{13,260}&{10,910}&{2350}\\{11}&{14,370}&{12,110}&{2260}\\{12}&{11,700}&{8,620}&{3080}\\{13}&{15,470}&{12,590}&{2880}\\{14}&{17,840}&{15,090}&{2750}\\{15}&{14,070}&{10,550}&{3520}\\{16}&{14,760}&{12,230}&{2530}\end{array}\)

Calculate and interpret an upper confidence bound for the true average difference between\(1\)-minute modulus and\(4\)-week modulus; first check the plausibility of any necessary assumptions.

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?

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

The degenerative disease osteoarthritis most frequently affects weight-bearing joints such as the knee. The article "Evidence of Mechanical Load Redistribution at the Knee Joint in the Elderly When Ascending Stairs and Ramps" (Annals of Biomed. Engr., \(2008: 467 - 476\)) presented the following summary data on stance duration (ms) for samples of both older and younger adults.

\(\begin{array}{*{20}{l}}{Age\;\;\;\;\;\;\;Sample Size\;\;Sample Mean\;\;Sample SD}\\{\;Older\;\;\;\;\;28\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;801\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;117}\\{Younger\;16\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;780\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;72}\end{array}\)

Assume that both stance duration distributions are normal.

a. Calculate and interpret a\(99\% \)CI for true average stance duration among elderly individuals.

b. Carry out a test of hypotheses at significance level\(.05\)to decide whether true average stance duration is larger among elderly individuals than among younger individuals.

An experiment to compare the tension bond strength of polymer latex modified mortar (Portland cement mortar to which polymer latex emulsions have been added during mixing) to that of unmodified mortar resulted in \(\bar x = 18.12kgf/c{m^2}\)for the modified mortar \((m = 40)\) and \(\bar y = 16.87kgf/c{m^2}\) for the unmodified mortar \((n = 32)\). Let \({\mu _1}\) and \({\mu _2}\) be the true average tension bond strengths for the modified and unmodified mortars, respectively. Assume that the bond strength distributions are both normal.

a. Assuming that \({\sigma _1} = 1.6\) and \({\sigma _2} = 1.4\), test \({H_0}:{\mu _1} - {\mu _2} = 0\) versus \({H_a}:{\mu _1} - {\mu _2} > 0\) at level . \(01.\)

b. Compute the probability of a type II error for the test of part (a) when \({\mu _1} - {\mu _2} = 1\).

c. Suppose the investigator decided to use a level \(.05\) test and wished \(\beta = .10\) when \({\mu _1} - {\mu _2} = 1. \). If \(m = 40\), what value of n is necessary?

d. How would the analysis and conclusion of part (a) change if \({\sigma _1}\) and \({\sigma _2}\) were unknown but \({s_1} = 1.6\)and \({s_7} = 1.4?\)

See all solutions

Recommended explanations on Math Textbooks

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.