Problem 1: The psychologists are interested in whether depressed people undergoi
ID: 3242033 • Letter: P
Question
Problem 1:
The psychologists are interested in whether depressed people undergoing group therapy will perform a different number of activities of daily living after group therapy. Therefore, the psychologists have randomly selected 12 depressed clients to undergo a six-week group therapy program. Use the five steps of hypothesis testing to determine whether the average number of activities of daily living (shown below) obtained after therapy is significantly different than a mean number of activities of 14 that is typical for similar depressed people. Test the difference at the .05 level of significance (and, for practice, at the .01 level). In Step 2, show all calculations. As part of Step 5, indicate whether the psychologists should recommend group therapy for all depressed people based on evaluation of the null hypothesis at both levels of significance and calculate the effect size.
Client--After therapy
A--17
B--15
C--12
D--21
E--16
F--18
G--17
H--14
I--13
J--15
K--12
L--19
Problem 2:
The psychologists are interested in whether depressed people undergoing group therapy will perform a different number of activities of daily living before and after group therapy. Therefore, the psychologists have randomly selected 8 depressed clients in a 6-week group therapy program. Use the five steps of hypothesis testing to determine whether the observed differences in numbers of activities of daily living (shown below) obtained before and after therapy are statistically significant at the .05 level of significance (and, for practice, at the .01 level). In Step 2, show all calculations. As part of Step 5, indicate whether the psychologists should recommend group therapy for all depressed people based on the evaluation of the null hypothesis at both levels of significance and calculate the effect size.
Client--Before therapy--After therapy
A--12--17
B--7--15
C--10--12
D--13--21
E--9--16
F--8--18
G--14--17
H--11--8
Application 1: t Test for a Single Sample
Open SPSS.
Enter the number of activities of daily living performed by the depressed clients studied in Problem 1 in the Data View Window.
In the Variable View window, change the variable name to “adl” and set the decimals to zero.
*click* Analyze
*click* Compare means
*click* One-Sample T Test
*click* the name of the variable (adl) and *click* the arrow to move the variable to the Variable(s) window
Enter the population mean (14) in the “Test Value” box
*click* OK
Compare these results to those you obtained when you calculated the test “by hand.” Note that the 95% confidence interval includes zero, so the possibility that there is no difference between the sample mean and the population mean cannot be eliminated, which confirms the result of the point estimate—that the difference is not statistically significant at the .05 level of significance.
Application 2: t Test for Dependent Means
Open SPSS.
Enter the number of activities of daily living performed by the depressed clients studied in Problem 2 in the Data View window. Be sure to enter the “before therapy” scores in the first column and the “after therapy” scores in the second column.
In the Variable View window, change the variable name for the first variable to “adlpre” and the variable name for the second variable to “adlpost.” Set the decimals for both variables to zero.
*click* Analyze
*click* Compare means
*click* Paired-Samples t Test
*click* the first variable (adlpre) and *click* the second variable (adlpost) to highlight both variables. The *click* the arrow to move both variables to the Paired Variable(s) Window.
*click* OK
Compare these results to those you obtained when you calculated the test “by hand.” The t score in the Paired Samples Test output box is negative because the number of activities after therapy, which increased, was subtracted from the number of activities before therapy. Your t score would have been positive if you followed the general practice of subtracting the “before” score from the “after” score. however, examining the means shows that therapy had the desired effect of increasing the number of activities performed by the clients. Note that the 95% confidence interval does not include zero, which confirms the result of the point estimate—that the difference is statistically significant at the .05 level of significance. Examination of the actual probability (.012) indicates that the change in the number of activities was almost large enough to be statistically significant at the .01 level.
Question 5
What was the effect size for problem 1?
small
medium
large
Question 6
In problem 2, you calculated the standard deviation of the means. What was the standard deviation?
3.78
2.06
1.49
1.01
Question 7
In problem 2, you estimated the variance of the population of difference scores from the sample of difference scores. What the population variance?
20.32
17.71
15.75
18.98
Question 8
In problem 2, you determined the t-score for the study. What was the t-score?
3.36
4.03
2.98
3.76
Question 9
In problem 2, was the null hypothesis rejected or accepted at p = .05?
rejected
accepted
small
medium
large
Explanation / Answer
Problem 1:
The psychologists are interested in whether depressed people undergoing group therapy will perform a different number of activities of daily living after group therapy. Therefore, the psychologists have randomly selected 12 depressed clients to undergo a six-week group therapy program. Use the five steps of hypothesis testing to determine whether the average number of activities of daily living (shown below) obtained after therapy is significantly different than a mean number of activities of 14 that is typical for similar depressed people. Test the difference at the .05 level of significance (and, for practice, at the .01 level). In Step 2, show all calculations. As part of Step 5, indicate whether the psychologists should recommend group therapy for all depressed people based on evaluation of the null hypothesis at both levels of significance and calculate the effect size.
Here we have to test,
H0 : mu = 14 vs H1 : mu not= 14
where mu is population mean after therapy.
Assume alpha = level of significance = 0.01
Here sample data is given and sample size is small so we use one sample t-test.
The test statistic is,
t = (Xbar - mu) / (sd/sqrt(n))
where Xbar is sample mean after therapy
sd is sample standard deviation after therapy
n is sample size
We can do one sample t-test in MINITAB.
steps :
ENTER data into MINITAB sheet --> Stat --> Basic statistics --> 1-sample t --> samples in columns : select data column --> Click on perform hypothesis test --> Hypothesized mean : 14 --> Options --> Confidence level : 99.0% --> Alternative : not equal --> ok --> ok
Output :
One-Sample T: after therapy
Test of mu = 14 vs not = 14
Variable N Mean StDev SE Mean 99% CI T P
after therapy 12 15.7500 2.8002 0.8083 (13.2395, 18.2605) 2.16 0.053
Here test statistic = 2.16
P-value = 0.053
P-value > alpha
Accept H0 at 1% level of significance.
Conclusion : There is not sufficient evidence to say that population mean after therapy is different from 14.
Effect size :
d = (Xbar - mu) / sd = (15.7500 - 14) / 2.8002 = 0.62
It represents medium effect size.
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Problem 2:
The psychologists are interested in whether depressed people undergoing group therapy will perform a different number of activities of daily living before and after group therapy. Therefore, the psychologists have randomly selected 8 depressed clients in a 6-week group therapy program. Use the five steps of hypothesis testing to determine whether the observed differences in numbers of activities of daily living (shown below) obtained before and after therapy are statistically significant at the .05 level of significance (and, for practice, at the .01 level). In Step 2, show all calculations. As part of Step 5, indicate whether the psychologists should recommend group therapy for all depressed people based on the evaluation of the null hypothesis at both levels of significance and calculate the effect size.
Now again we have to test the hypothesis that,
H0 : Mud = 0 Vs H1 : Mud not= 0
Assume alpha = 0.01
Here we use paired t-test.
We can do paired t-test in minitab.
steps :
steps :
ENTER data into MINITAB sheet --> Stat --> Basic statistics --> Paired t --> Samples in columns --> First sample : select before --> Second sample : select after --> Options --> Confidence level : 99.0 --> Test mean : 0.0 --> Alternative : not equal --> ok --> ok
Paired T-Test and CI: before, after
Paired T for before - after
N Mean StDev SE Mean
before 8 10.5000 2.4495 0.8660
after 8 15.5000 3.9641 1.4015
Difference 8 -5.00000 4.20883 1.48805
99% CI for mean difference: (-10.20740, 0.20740)
T-Test of mean difference = 0 (vs not = 0): T-Value = -3.36 P-Value = 0.012
Here test statistic = -3.36
P-value = 0.012
P-value> alpha
Accept H0 at 0.01 significance level.
Conclusion : There is sufficient evidence to say that the population mean difference may be 0.