Question

she gives 50 item test of depression symptoms before a series of 10 group therapy sessions. scores are normally distributed. use alpha 0.05 to determine reduced depression v calculate t obtained based on differences.

Before: 50,46,43,42,41,38,36

After: 44,45,39,42,38,35,33

1. Ha establishes :,

2. t obtained round 3 decimals :

3. the degrees of freedom :

4. critical value alpha 0.05 round 3 decimals:

5. conclusion : reject or fail to reject

Answer 1

Here, we have to use paired t test.

The null and alternative hypotheses for this test are given as below:

Null hypothesis: H0: Depression does not reduce after therapy sessions.

Alternative hypothesis: Ha: Depression reduces after therapy sessions.

H0: µd = 0 versus Ha: µd > 0

This is a right tailed test.

We take difference as before minus after.

Test statistic for paired t test is given as below:

t = (Dbar - µd)/[Sd/sqrt(n)]

From given data, we have

Dbar = 2.8571

Sd = 1.9518

n = 7

df = n – 1 = 6

α = 0.05

t = (Dbar - µd)/[Sd/sqrt(n)]

t = (2.8571 – 0)/[ 1.9518/sqrt(7)]

t = 3.8730

Test statistic = 3.8730

Critical t value = 1.9432

(by using t-table)

The p-value by using t-table is given as below:

P-value = 0.0041

Test statistic > Critical value

Or

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that Depression reduces after therapy sessions.