Question

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 (“Vegetarianism, Bone Loss, and Vitamin D: A Longitudinal Study in Asian Vegans and Non-Vegans,” European J. of Clinical Nutr., 2012: 75–82)

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 (“covariates”) such as age, body weight, and various nutritional factors; the resulting CI included 0, suggesting no diet effect.

Short answer

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

given

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

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

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.

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.