Question

A biological psychologist wants to know if physical exercise changes reaction speed in elderly people.   A total of 18 older volunteer participants were found. One sample of n = 9 subjects received an exercise treatment, while the second sample (n = 9) of participants were given stretching lessons for the same amount of time per week. Reaction speed was tested on the computer after six weeks of treatment. Reaction speed was measured in milliseconds. Data shown below.

Why do you use this type of statistical test? _________

What is the I.V.? ___________________   What is the D.V.? ______________________

Use the data below to fill in the blanks to compute your t_calculated.

stretch group exercise group

n_stretch = 9           df_stretch = 8   n_exercise = 9              df_exercise = 8

∑X_stretch = 450        M_Stretch = 50 ∑X_exercise = 390            M _Exercise = 43.3

SS _stretch = 260    S²_stretch  = _____ SD_stretch = _____ SS_Exercise = 376   S²_Exercise = ______ SD_Exercise = ______

Use the 4 step method for hypothesis testing.

Step 1.

Step 2. alpha = .05, 2 tailed test

Step 3. Calculate your test statistic (t_calculated) (show formulas and work)

t_calculated = M_stretch – M_exercise / S_{M1– M2}

T_calculated = ___________

Step 4. State your conclusion about the null hypothesis:       Reject        or       Retain

Summarize your results in APA format.

Answer 1

Two sample t test why because we do not know population standard deviations and sample size is less than 30

Independent variable is physical exercise

Dependent variable is reaction speed in elderly people

n_stretch = 9           df_stretch = 8   n_exercise = 9              df_exercise = 8

∑X_stretch = 450        M_Stretch = 50 ∑X_exercise = 390            M _Exercise = 43.3

SS _stretch = 260    S²_stretch  = SS _stretch/(n-1) = 260/8 = 32.5 SD_stretch = SQRT(S²_stretch) = 5.7009

SS_Exercise = 376   S²_Exercise = SS_Exercise/(n-1) = 376/8 = 47 SD_Exercise = SQRT(S²_Exercise) = 6.8557

i) Hypothesis :		α=	0.05
		df	16	n1+n2-2
Ho:	μ1​ = μ2
Ha:	μ1​ not = μ2
ii) t Critical Value :
tc	2.119905299	T.INV.2T(alpha,df)	TWO
ts	< for -	tc	TWO	To reject
ts	> for +	tc	TWO	To reject
iii) Test :
Sp^2	39.75044165	((n1-1)S1^2+(n2-1)S2^2)/(n1+n2-2)
t stat	2.254291536	(X1 bar-X2 bar )/SQRT(Sp^2*(1/n1 + 1/n2))	Equal vriance
P value :
P value	0.038549013	T.DIST.2T(ts,df)	TWO
Decision :
P value	<	α	Reject H0

t stat > tc, Reject H0

iv) Conclusion:

There is enough evidence to conclude that mean reaction speed in elderly people after taking exercise treatment is different than mean reaction speed in elderly people after taking stretching lessons at 5% significance level