A researcher conducts a study examining the effectiveness of a group exercise program at an assisted living facility for elderly adults. One group of residents is selected to participate in the program, and a second group serves as a control. After the program was completed, the researcher records a combined score measuring balance and strength for each individual. The data are as follows:

Control Exercise n = 10 n = 15 M = 12 M = 15.5 SS = 120.5 SS = 190.0

Required:
a. Does the exercise program have a significant effect? Use an alpha level of .05, two tails.
b. Compute Cohen's d to measure the size of the treatment effect.
c. Calculate the two-sample t-test. To receive fiull credit, please clearly label your pooled variance, estimated standard error, and final t-score.
d. What t values form the boundary for the critical region?

Respuesta :

Answer:

Check the explanation

Step-by-step explanation:

Hypotheses are:

[tex]H_{0}:\mu_{1}=\mu_{2},H_{1}:\mu_{1}\neq \mu_{2}[/tex]

Following is the output of t test:

Hypothesis Test: Independent Groups (t-test, pooled variance)

X1                     X2    

12                     15.5                   mean  

3.6591            3.6839               std. dev.  

10                     15                        n

 

                       23                        df  

                  -3.5000          difference (X1 - X2)  

                  13.4999          pooled variance  

                  3.6742            pooled std. dev.  

                  1.5000            standard error of difference

                  0                     hypothesized difference

                  -2.33                     t  

                  .0287               p-value (two-tailed)

The p-value is:

p-value = 0.0287

Since p-value is less than 0.05 so we reject the null hypothesis. That is we can conclude that the  exercise program have a significant effect.

(b)

The cohen's d is

[tex]d=\left |\frac{\bar{x}_{1}-\bar{x}_{2}}{s_{p}} \right |=\left |\frac{12-15.5}{3.6742 } \right |=0.95[/tex]