Chapter 10, Problem 24P

Research has shown that people are more likely to show dishonest and self-interested behaviors in darkness than in a well-lit environment (Zhong, Bohns, & Gina, 2010). In one experiment, participants were given a set of 20 puzzles and were paid $0.50 for each one solved in a 5-minute period. However, the participants reported their own performance and there was no obvious method for checking their honesty. Thus, the task provided a clear opportunity to cheat and receive undeserved money. One group of participants was tested in a room with dimmed lighting and a second group was tested in a well-lit room. The reported number of solved puzzles was recorded for each individual. The following data represent results similar to those obtained in the study.

  

a. Is there a significant difference in reported performance between the two conditions? Use a two-tailed test with $\alpha$ = .01.

b. Compute Cohen's d to estimate the size of the treatment effect.

To determine

To check: Whether there is significant difference in reported performance between the two conditions using two-tailed test with $\alpha =0.01$

Answer to Problem 24P

Reject null and conclude that there is significant difference in reported performance between the two conditions

Explanation of Solution

Given info:

Two samples each of size 9 is given in the question. Two-tailed hypothesis test using $\alpha =0.01$ is to be performed.

Calculations:

Step 1: Null Hypothesis is $H_0:{\mu }_1-{\mu }_2=0$ and Alternate Hypothesis is $H_1:{\mu }_1-{\mu }_2\ne 0$

Step 2: For the given sample, the t-statistics will have degrees of freedom equals:

$\begin{array}{c}df=n_1+n_2-2\\ =9+9-2\\ =16\end{array}$

For a two-tailed test with $\alpha =0.01$ and $df=16$, the critical value (CV)  is obtained from the t-table as $t=\pm 2.921$

Step 3: The t-statistics is calculated using formula:

$t=\frac{(M_1-M_2)-({\mu }_1-{\mu }_2)}{s_{M_1-M_2}}$

Software procedure:

Step-by-step procedure to obtain the sample mean and variance for the two groups, using the SPSS software:

• Go to Variable View> Enter the name of the variable as Money and Paper.
• Go to Data View>Enter the values ofMoney and Paper.
• Choose Analyze > Descriptive Statistics> choose Descriptive.
• Select Money and Paper and move it under variable(s)> Choose Options> Select Mean , Std. Deviation and Variance
• ChooseContinue> choose OK.

Output using the SPSS software is given below:

From the SPSS output, the value of sample means, and variances are:

$\begin{array}{c}{M}_1=7.56\\ M_2=11.33\\ s_1^2=5.27\\ s_2^2=4.75\end{array}$

Next, calculate pooled variance, $s_p^2$, using formula:

$s_p^2=\frac{(d{f}_1)s_1^2+(d{f}_2)s_2^2}{d{f}_1+d{f}_2}$

Substitute 5.27 for $s_1^2$, 4.75 for $s_2^2$, 8 for both $d{f}_1$ and $d{f}_2$ in the above formula

$\begin{array}{c}{s}_p^2=\frac{(8)5.27+(8)4.75}{8+8}\\ =\frac{8(5.27+4.75)}{2(8)}\\ =\frac{10.02}{2}\\ =5.01\end{array}$

Now, the formula for estimated standard error of the difference is

$s_{M_1-M_2}=\sqrt{\frac{s_p{}^2}{n_1}+\frac{s_p{}^2}{n_2}}$

Substitute 5.01 for $s_p^2$, 9 for both $n_1$ and $n_2$ in the formula

$\begin{array}{c}{s}_{M_1-M_2}=\sqrt{\frac{5.01}{9}+\frac{5.01}{9}}\\ =\sqrt{\frac{10.02}{9}}\\ =1.055\end{array}$

Finally, substitute 1.055  for $s_{M_1-M_2}$, 7.56 for $M_1$ and 11.33 for $M_2$ in the t-statistics formula

$\begin{array}{c}t=\frac{7.56-11.33}{1.055}\\ =-\frac{3.77}{1.055}\\ =-3.57\end{array}$

Step 4: Rejection rule: Reject when $t-statistics<-2.921$ or $t-statistics>2.921$

Since $t-statistics(=-3.65)<critical\_value(=-2.921)$, reject the null hypothesis.

Step 5: Based on the results of hypothesis test, there is sufficient evidence to reject the null hypothesis.

Hence, reject null and conclude that there is significant difference in reported performance between the two conditions

To determine

To Find: The value of Cohen’s d for the given question.

Answer to Problem 24P

The value of Cohen’s d is 1.69.

Explanation of Solution

Given info:

Two samples each of size 9 is given in the question.  The sample means, and pooled variance were calculated in the part (a) of the question.

Calculations:

The formula for Cohen’s d is:

$Cohen’sd=\frac{M_1-M_2}{\sqrt{s_p{}^2}}$

Substitute 5.01 for $s_p^2$, 7.56 for $M_1$ and 11.33 for $M_2$

$\begin{array}{c}Cohen’sd=\frac{\left|7.56-11.33\right|}{\sqrt{5.01}}\\ =\frac{3.77}{2.238}\\ =1.69\end{array}$

Hence, Cohen’s d is 1.69.