Chapter 9, Problem 32PR

(a)

To determine

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

Answer to Problem 32PR

The test statistic value is –3.035.

The decision is to reject the null hypothesis.

Explanation of Solution

Calculation:

A sample of 11children’s each of group is considered and the first group is shaken baby syndrome and without syndrome. The claim is the there is difference between the mean scoresof two groups. This represents the alternative hypothesis. The level of significance is 0.05.

Let ${\mu }_1$ is mean score children with shaken baby syndrome, and ${\mu }_2$ is mean score children with shaken baby without syndrome.

Null hypothesis:

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

That is, there is no difference between the mean scoresof two groups.

Alternative hypothesis:

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

That is, there is difference between the mean scoresof two groups.

The degrees of freedom for t distribution is,

$\begin{array}{c}df=\left(n_1-1\right)+\left(n_2-1\right)\\ =\left(11-1\right)+\left(11-1\right)\\ =10+10\\ =20\end{array}$

The test is two tailed, the degrees of freedom are 10, the level of significance is 0.05.

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

• Locate the value 20 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 20.0 with level of significance 0.05 is 2.086.

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

The first group sample size is 11, sample mean is 86.36, and the standard deviation is 15.16. The second group sample size is 11, sample mean is 104.09, and the standard deviation is 12.10.

The pooled sample variance is,

$\begin{array}{c}{s}_p^2=\frac{{\left(15.16\right)}^2\left(10\right)+{\left(12.10\right)}^2\left(10\right)}{10+10}\\ =\frac{2,298.256+1,464.1}{20}\\ =\frac{3,762.356}{20}\\ =188.12\end{array}$

The test statistic value is,

$\begin{array}{c}{t}_{\text{obt}}=\frac{\left(86.36-104.09\right)-0}{\sqrt{\frac{188.12}{11}+\frac{188.12}{11}}}\\ =\frac{-17.73}{\sqrt{34.2036}}\\ =\frac{-17.73}{5.8484}\\ =-3.035\end{array}$

Thus, the test statistic value is –3.035.

Justification: The value of test statistic is –3.035 and critical value is –2.086. The test statistic value is less than the critical value. The test statistic value fall under critical region and hence the null hypothesis is rejected.

(b)

To determine

Find the value of estimated Cohen’s d to compute the effect size.

Answer to Problem 32PR

The value of estimated Cohen’s d is –1.29.

There is large effect size.

Explanation of Solution

Calculation:

Cohen’s d:

The Cohen’s d is the measure to estimate the effect size which is used most frequently for t test. In t test the value of population standard deviation is not known, hence this value is replaced with sample standard deviation for estimation the effect size as,

The Cohen’s d for atwo-independent-sample t test is,

$\text{Cohen's}d=\frac{M_1-M_2}{\sqrt{s_p^2}}$

The description of effect size using Cohen’s d:

• If value of Cohen’s d is less than 0.2, then effect size is small.
• If value of Cohen’s d is in between 0.2 and 0.8, then effect size is medium.
• If value of Cohen’s d is greater than 0.8, then effect size is large.

The value of $M_1$ is 86.36, the value of $M_2$ is 104.09 and $s_p^2$ is 188.12. The value of Cohen’s d is,

$\begin{array}{c}d=\frac{M_1-M_2}{\sqrt{s_p^2}}\\ =\frac{86.36-104.09}{\sqrt{188.12}}\\ =\frac{-17.73}{13.716}\\ =-1.29\end{array}$

$\left|d\right|=1.29$

The absolute value of estimated Cohen’s d is 1.29. This absolute value is greater than 0.8. Hence the estimated Cohen’s d has a large effect size.

However, the value of estimated Cohen’s d is –1.29.

(c)

To determine

Find the value of eta-squared to compute the proportion of variance.

Answer to Problem 32PR

The value of eta-squared is 0.32.

There is large effect size.

Explanation of Solution

Calculation:

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.

The test statistic value is –3.035 and $df$ is 20. The value of Cohen’s d is,

$\begin{array}{c}{\eta }^2=\frac{{\left(-3.035\right)}^2}{{\left(-3.035\right)}^2+20}\\ =\frac{9.2112}{9.2112+20}\\ =\frac{9.2112}{29.2112}\\ =0.32\end{array}$

The value of estimated eta-square is 0.32. This value is greater than 0.25. Hence the eta-square has a large effect size.