Chapter 9, Problem 21CAP

(a)

To determine

Check whether the consumption of saccharin-flavored water differ in groups or not.

Find the test statistic value and make a decision to retain or reject the null hypothesis

Answer to Problem 21CAP

The test statistic value is 4.27.

The decision is to reject the null hypothesis.

There is significant difference between the consumption of saccharin-flavoured water of two groups.

Explanation of Solution

Calculation:

A sample of 8 rats is considered and they are rats feel sick and control group rats. The claim is the there is difference between the consumption of saccharin-flavoured water of two groups. This represents the alternative hypothesis. The level of significance is 0.05.

Let ${\mu }_1$ is average amount consumed by rats were made sick, and ${\mu }_2$ is average amount consumed by control rats.

Null hypothesis:

$H_0:{\mu }_1={\mu }_2$

That is, there is no difference between the consumption of saccharin-flavoured water of two groups.

Alternative hypothesis:

$H_1:{\mu }_1\ne {\mu }_2$

That is, there is difference between the consumption of saccharin-flavoured water of two groups.

The degrees of freedom for t distribution is,

$\begin{array}{c}df=\left(n_1-1\right)+\left(n_2-1\right)\\ =\left(4-1\right)+\left(4-1\right)\\ =3+3\\ =6\end{array}$

The test is two tailed, the degrees of freedom are 6, the level of significance is 0.05.

From the Appendix B: Table B.2 The t Distribution:

• Locate the value 6 in the degrees of freedom (df) column.
• Locate the 0.05 in the proportion in Two tails combined row.
• The intersecting value that corresponds to the 6 with level of significance 0.05 is 2.447.

The critical value for a two-tailed test with $\alpha =0.05$ is ±2.447.

Decision rules:

• If the positive test statistic value is greater than the positive 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.

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

$t_{\text{obt}}=\frac{\left(M_1-M_2\right)-\left({\mu }_1-{\mu }_2\right)}{s_{M_1-M_2}}$

In the formula, $M_1$ is first sample mean, $M_2$ is second sample mean, ${\mu }_1$ is first population mean, ${\mu }_2$ is second population mean, $s_{M_2-M_2}$ is estimated standard error of difference given as,

$s_{M_2-M_2}=\sqrt{\frac{s_p^2}{n_1}+\frac{s_p^2}{n_2}}$ and $s_p^2=\frac{s_1^2\left(d{f}_1\right)+s_2^2\left(d{f}_2\right)}{d{f}_1+d{f}_2}$.

In the formula, $s_1^2$ is the sample variance of the first sample, $s_2^2$ is the sample variance of the second sample, $d{f}_1$ are the degrees of freedom for first sample and $d{f}_2$ are the degrees of freedom for second sample.

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 names as Amounts, and Types of Rats.
• Choose Data view, enter the data.
• Choose Analyze>Compare means>Independent Samples T Test.
• In Test variables, enter the column as Amounts.
• In Grouping variables, enter thecolumn asTypes of Rats.
• Select Define Groups; enter Group 1 as 1 and Group 2 as 2.
• Click OK.

Output using SPSS software is given below:

Thus, the test statistic value is 4.27.

Justification: The value of test statistic is 4.27 and critical value is 2.447. The test statistic value is greater than the critical value. The test statistic value fall under critical region and hence the null hypothesis is rejected.

Thus, there is significant difference between the consumption of saccharin-flavoured water of two groups.

(b)

To determine

Find the effect size by using eta-squared.

Answer to Problem 21CAP

The value of eta-squared is 0.75.

The effect size is large.

Explanation of Solution

Eta-square:

The proportion of variance is measured using eta-square. But eta-square is biased because it over estimates the proportion of variance that is explained by treatment. It is denoted by $\left({\eta }^2\right)$.

${\eta }^2=\frac{t^2}{t^2+df}$

In the formula, t is the value of test statistic and df is the corresponding degrees of freedom.

The description of effect size using eta-square:

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

Calculation:

From the SPSS output, the test statistic value is 4.27 and $df$ is 6. The value of Cohen’s d is,

$\begin{array}{c}{\eta }^2=\frac{{\left(4.27\right)}^2}{{\left(4.27\right)}^2+6}\\ =\frac{18.2329}{18.2329+6}\\ =\frac{18.2329}{24.2329}\\ =0.75\end{array}$

The value of estimated eta-square is 0.75. This value is greater than 0.25. Hence the eta-square has a large effect size.