Question

The article "'Supervised Exercise Versus NonSupervised Exercise for Reducing Weight in Obese 2009: 85-90) reported on an investigation in which participants were randomly assigned to either a supervised exercise program or a control group. Those in the control group were told only that they should take measures to lose weight. After 4 months, the sample mean decrease in body fat for the 17 individuals in the experimental group was 6.2 kg with a sample standard deviation of 4.5 kg, whereas the sample mean and sample standard deviation for the 17 people in the control group were 1.7 kg and 3.1 kg, respectively. Assume normality of the two weight-loss distributions (as did the investigators).

1. Calculate a 99% lower prediction bound for the weight loss of a single randomly selected individual subjected to the supervised exercise program. Can you be highly confident that such an individual will actually lose weight?
2. Does it appear that true average decrease in body fat is more than two kg larger for the experimental condition than for the control condition? Use the accompanying Minitab output to reach a conclusion at significance level of .01. (Note: Minitab accepts such summary data as well as individual observations. Also, because the test is upper-tailed, the software provides a lower confidence bound rather than a conventional CI.)

Sample N Mean StDev SE Mean

Exptl 17 6.20 4.50 1.1

Control 17 1.70 3.10 0.75

Difference = mu (1) – mu (2)

Estimate for difference: 4.50

95% lower bound for difference : 2.25

T – test of difference = 2 (vs >) :

T – value = 1.89

P – value = 0.035 DF =28

Short answer

(a) -5.7605

We cannot be highly confident that such an individual will actually lose weight, because the lower confidence bound is negative, which indicates a weight gain.

(b) There is not sufficient evidence to support the claim that the true average decrease in body fat is more than two kg larger for the experimental condition than for the control condition.

Step-by-step solution

Step 1: To Calculate a 99% lower prediction bound for the weight loss of a single randomly selected individual subjected to the supervised exercise program.  

(a)

Given:

\(\begin{array}{l}\bar x = 6.2\\s = 4.5\\n = 17\\c = 99\% = 0.99\end{array}\)

We need to determine a lower PREDICTION bound.

Determine the t-value by looking in the row starting with degrees of freedom\(df = n - 1 = 17 - 1 = 16\)and in the column with\(\alpha = 1 - c = 0.01\)in the table of the Student's T distribution:

\({t_\alpha } = 2.583\)

The margin of error is then:

\(E = {t_{\alpha /2}} \times s\sqrt {1 + \frac{1}{n}} = 2.583 \times 4.5\sqrt {1 + \frac{1}{{17}}} \approx 11.9605\)

The boundaries of the confidence interval then become:

\(\bar x - E = 6.2 - 11.9605 = - 5.7605\)

Thus the 99 % lower confidence bound is -5.7605.

We cannot be highly confident that such an individual will actually lose weight, because the lower confidence bound is negative, which indicates a weight gain.

To find the average decrease in body fat is more than two kg larger for the experimental condition than for the control condition

(b)

Given in the output:

\(\begin{array}{l}{{\bar x}_1} = 6.20\\{s_1} = 4.50\\{{\bar x}_2} = 1.70\\{s_2} = 3.10\\{n_1} = {n_2} = 17\\\alpha = 0.01\end{array}\)

Given claim: More than 2

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain an equality.

\(\begin{array}{l}{H_0}:{\mu _1} - {\mu _2} = 2\\{H_a}:{\mu _1} - {\mu _2} > 2\end{array}\)

Determine the test statistic:

\(t = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }} = \frac{{6.20 - 1.70 - 2}}{{\sqrt {\frac{{4.5{0^2}}}{{17}} + \frac{{3.1{0^2}}}{{17}}} }} \approx 1.886\)

Determine the degrees of freedom (rounded down to the nearest integer):

\(\Delta = \frac{{{{\left( {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/{n_1}} \right)}^2}}}{{{n_1} - 1}} + \frac{{{{\left( {s_2^2/{n_2}} \right)}^2}}}{{{n_2} - 1}}}} = \frac{{{{\left( {\frac{{4.5{0^2}}}{{17}} + \frac{{3.1{0^2}}}{{17}}} \right)}^2}}}{{\frac{{{{\left( {4.5{0^2}/17} \right)}^2}}}{{17 - 1}} + \frac{{{{\left( {3.1{0^2}/17} \right)}^2}}}{{17 - 1}}}} \approx 28\)

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of Student's T distribution in the appendix containing the t-value in the row\(df = 28\):

\(0.025 < P < 0.050\)

If the P-value is less than or equal to the significance level, then the null hypothesis is rejected:

\(P > 0.01 \Rightarrow Fail to reject {H_0}\)

There is not sufficient evidence to support the claim that the true average decrease in body fat is more than two kg larger for the experimental condition than for the control condition.