Chapter 10, Problem 22CAP

1.

To determine

Test whether obese patients has consumed more calories than normal weight siblings or not.

State the test statistic value and make a decision to retain or reject the null hypothesis at 0.05, level of significance.

Answer to Problem 22CAP

The test statistic value is –1.082.

The decision is to retain the null hypothesis.

The obese patients consumed did not significantly consume more calories than their normal-weight siblings.

Explanation of Solution

The given information is that, sample of 20 obese patientsis considered in which there are two types of sibling are there normal and overweight siblings. The claim is the obese patients consumed significantlymore calories than their normal-weight siblings. This represents the alternative hypothesis. The level of significance is 0.05.

The formula of test statistic for one-sample t test is,

$t_{\text{obt}}=\frac{M_D-{\mu }_D}{s_{MD}}$

In the formula, $M_D$ denotes the sample mean difference, ${\mu }_D$ denotes the population mean difference, $s_{MD}=\frac{s_D}{\sqrt{n_D}}$ is the standard error for the difference and $n_D$ is the sample size.

Decision rules:

• If the positive test statistic value is greater than the critical value, then reject the null hypothesis or else retain the null hypothesis.
• If the negative test statistic value is less than negative critical value, then reject the null hypothesis or else retain the null hypothesis.

Let ${\mu }_D$ is the average consumption of calories between normal and overweight sibling.

Null hypothesis:

$H_0:{\mu }_D=0$

That is, the obese patients consumed did not significantly consumemore calories than their normal-weight siblings.

Alternative hypothesis:

$H_1:{\mu }_D<0$

That is, the obese patients consumed significantlyconsumemore calories than their normal-weight siblings.

The degrees of freedom for t distribution is $df=n_D-1$. Substitute, $n_D=10$.

$\begin{array}{c}df=n_D-1\\ =10-1\\ =9\end{array}$

Critical value:

The given significance level is $\alpha =0.05$.

The test is one tailed, the degrees of freedom are 9, and the alpha level is 0.05.

From the Appendix C: Table C.2 the t Distribution:

• Locate the value 5 in the degrees of freedom (df) column.
• Locate the 0.05 in the proportion in one tails combined row.
• The intersecting value that corresponds to the 9 with level of significance 0.05 is –1.833.

Thus, the critical value for $df=9$ with one-tailed test (lower) is –1.833.

Software procedure:

Step by step procedure to obtain test statistic value using SPSS software is given as,

• Choose Variable view.
• Under the name, enter the name as Times.
• Choose Data view, enter the data.
• Choose Analyze>Compare means>Paired Samples T Test.
• In Paired variables, enter the Variable 1 as Normal.
• Enter the Variable 2 as Over.
• Click OK.

Output using SPSS software is given below:

Thus, the test statistic value is –1.082.

Conclusion:

The value of test statistic is –1.082.

The critical value is –1.833.

The test statistic value is greater than the critical value.

The test statistic value does not fall under critical region.

Hence the null hypothesis is retained and obese patients consumed did not significantly consume more calories than their normal-weight siblings.

2.

To determine

Compute effect size using omega-squared.

Answer to Problem 22CAP

The effect size using omega-squaredis 0.02.

Explanation of Solution

Calculations:

From the SPSS output, the test statistic value is –1.082 and $df$ is 9.

Omega-square:

The proportion of variance is also measured using omega-square. The bias that is caused by eta-square is corrected by omega-squared. The value of omega-square is conservation and give small estimate of proportion of variance. It is denoted by $\left({\omega }^2\right)$.

${\omega }^2=\frac{t^2-1}{t^2+df}$

In the formula, t is the value of test statistic and df is the corresponding degrees of freedom. Subtracting 1 in the numerator reduces the effect size.

The description of effect size using omega-square:

• If value of omega-square is less than 0.01, then effect size is trivial.
• If value of omega-square is in between 0.01 and 0.09, then effect size is small.
• If value of omega-square is in between 0.10 and 0.25, then effect size is medium.
• If value of omega-square is greater than 0.25, then effect size is large.

Substitute, $t=-1.082$ and $df=9$, in Omega -square formula.

$\begin{array}{c}{\omega }^2=\frac{{\left(-1.082\right)}^2-1}{{\left(-1.082\right)}^2+9}\\ =\frac{1.1707-1}{1.1707+9}\\ =\frac{0.1707}{10.1707}\\ =0.02\end{array}$

Thus, the proportion of variance using omega-squared is 0.02. This value is in between 0.01 and 0.09. Hence the Omega-square has a small effect size.

3.

To determine

Explain whether the results support the researcher’s hypothesis or not.

Answer to Problem 22CAP

No, the results do not support the researcher’s hypothesis.

Explanation of Solution

Justification: The decision of the test is that the obese patients did not consume significantlymore calories than their normal-weight siblings. That is, there is no difference in the consumption of calories between obese patients and normal weight siblings.

But the researcher has claimed that ‘the obese patients consumed significantly more calories than their normal-weight siblings’. Hence, the researcher’s hypothesis is not supported.