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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.

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Most popular questions from this chapter

The accompanying data consists of prices (\$) for one sample of California cabernet sauvignon wines that received ratings of 93 or higher in the May 2013 issue of Wine Spectator and another sample of California cabernets that received ratings of 89 or lower in the same issue.

\(\begin{array}{*{20}{c}}{ \ge 93:}&{100}&{100}&{60}&{135}&{195}&{195}&{}\\{}&{125}&{135}&{95}&{42}&{75}&{72}&{}\\{ \le 89:}&{80}&{75}&{75}&{85}&{75}&{35}&{85}\\{}&{65}&{45}&{100}&{28}&{38}&{50}&{28}\end{array}\)

Assume that these are both random samples of prices from the population of all wines recently reviewed that received ratings of at least 93 and at most 89 , respectively.

a. Investigate the plausibility of assuming that both sampled populations are normal.

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a. Assuming that \({\sigma _1} = 1.2\) and \({\sigma _2} = 1.1\), test the relevant hypotheses using \(\alpha = .001\).

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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

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

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Determine the number of degrees of freedom for the two-sample t test or CI in each of the following.

\(\begin{array}{l}a. m = 10, n = 10, {s_1} = 5.0, {s_2} = 6.0\\b. m = 10, n = 15,{s_1} = 5.0, {s_2} = 6.0\\c. m = 10, n = 15, {s_1} = 2.0, {s_2} = 6.0\\d. m = 12, n = 24, {s_1} = 5.0, {s_2} = 6.0\\\\\end{array}\)

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