Question

The article "'The Effects of a Low-Fat, Plant-Based Dietary Intervention on Body Weight, Metabolism, and Insulin Sensitivity in Postmenopausal Women" (Amer. J. of Med., \(2005: 991 - 997\)) reported on the results of an experiment in which half of the individuals in a group of\(64\)postmenopausal overweight women were randomly assigned to a particular vegan diet, and the other half received a diet based on National Cholesterol Education Program guidelines. The sample mean decrease in body weight for those on the vegan diet was\(5.8\;kg\), and the sample SD was\(3.2\), whereas for those on the control diet, the sample mean weight loss and standard deviation were\(3.8\)and\(2.8\), respectively. Does it appear the true average weight loss for the vegan diet exceeds that for the control diet by more than\(1\;kg\)? Carry out an appropriate test of hypotheses at significance level .05.

Short answer

Do not reject null hypothesis.

Step-by-step solution

Find the value of normal distribution

Denote with\({\mu _1}\)the true average of the mean body mass decrease for vegan diet and with \({\mu _2}\) for the control diet.

Given two normal distributions, the random variable (standardized)

\(T = \frac{{\bar X - \bar Y - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{S_1^2}}{m} + \frac{{S_2^2}}{n}} }}\)

has approximately students \(t\) distribution with degrees of freedom\(\nu \), where \(\nu \) is

\(\nu = \frac{{{{\left( {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/m} \right)}^2}}}{{m - 1}} + \frac{{{{\left( {s_2^2/n} \right)}^2}}}{{n - 1}}}}\)

has to be rounded down to the nearest integer.

The two-sample \(t\) test for testing \({H_0}:{\mu _1} - {\mu _2} = {\Delta _0}\) uses the following value of test statistic

\(t = \frac{{\bar x - \bar y - {\Delta _0}}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }}\)

For the adequate alternative hypothesis the adequate area under the \({t_\nu }\) curve is calculated is the \(P\) value.

Determine the degree of freedom

The hypotheses of interest are \({H_0}:{\mu _1} - {\mu _2} = 1\) versus \({H_1}:{\mu _1} - {\mu _2} > 1\) (exceeds by more than 1 kg). Using the given data in the exercise, the \(t\) value is

\(t = \frac{{\bar x - \bar y - {\Delta _0}}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }} = \frac{{5.8 - 3.8 - 1}}{{\sqrt {\frac{{{{3.2}^2}}}{{32}} + \frac{{{{2.8}^2}}}{{32}}} }} = 1.33\)

and the corresponding degrees of freedom can be computed using formula

\(\nu {\rm{ }} = \frac{{{{\left( {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/m} \right)}^2}}}{{m - 1}} + \frac{{{{\left( {s_2^2/n} \right)}^2}}}{{n - 1}}}} = \frac{{{{\left( {\frac{{{{3.2}^2}}}{{32}} + \frac{{{{2.8}^2}}}{{32}}} \right)}^2}}}{{\frac{{{{\left( {{{3.2}^2}/32} \right)}^2}}}{{32 - 1}} + \frac{{{{\left( {{{2.8}^2}/32} \right)}^2}}}{{32 - 1}}}} = 60.93\)

The \(P\) value for one sided testing (upper) is

\(P(T > 1.33) = 0.094\)

where the value was computed using software (see the table in the appendix as well). Since

\(P(T > 1.33) = 0.094P = 0.094 > 0.05\)

do not reject null hypothesis

at \(0.05\)level. There are not enough evidence to support the claim that the true average weight loss for the vegan diet exceeds the true average weight loss for the control died more than 1 kilogram.