The items below are based on the following scenario. A counseling psychologist developing a technique to reduce procrastination has students time their procrastination for a week and uses this as a pretest measure of procrastination. Students then attend a workshop in which they are instructed to do a specific warm-up exerdse for studying by focusing on a pleasant activity. Students again time their procrastination for a week, and the time from this second week is the posttest measure.
1. If the psychologist wants to see if there is a change (increase or decrease) in procrastination by students who attended his workshop, the appropriate description of "Population 2" (the comparison population) would be people whose:.
a. posttest scores will be lower than their pretest scores.
b. difference scores will be 0 or greater than 0.
c. difference scores will be 0.
d. difference scores will be smaller than their pretest scores.
2. If the psychologist wants to see if there is a change (increase or decrease) in procrastination by 10 of the students who attended his workshop using the.05 significance level, the cutoff t score(s) would be:.
a. -.62,0,0,+2.62
b. +2.262
c. -2.262,0
d. -2.262, +2.262
3. If the psychologist finds that the sum of squared deviations from the mean of the difference scores of the sample is 135, the estimated population variance would be:.
a. 135/10 = 13.5
b. 135/9 = 15.0
c. 10/135 =.074
d. 9/135 =.067

1) Option C is correct.

Difference should be 0.

2) Option D is correct.

The cutoff t-scores are -2.262, 2.262.

t < -2.262, t > 2.262

3) Option A is correct.

The variance = (135/10) = 13.5

Step-by-step explanation:

1) For any hypothesis test, the first step is usually to define the null and alternative hypothesis.

In hypothesis testing, especially one comparing two sets of data, the null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test. It usually maintains that, with random chance responsible for the outcome or results of any experimental study/hypothesis testing, its statement is true.

The alternative hypothesis usually confirms the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test. It usually maintains that significant factors other than random chance, affect the outcome or results of the experimental study/hypothesis testing and result in its own statement.

For this test, the counselling psychologist wants to check if there is a difference, whether positive or negative between the procrastination times before and after attending the workshop.

So, the appropriate parameter to compute would be the difference in procrastination times before and after the workshop.

Then the null hypothesis would be that the difference between these two times is 0.

Alternative hypothesis would be that there is no difference between the two times.

Mathematically, if the difference between the two times is p.

The null hypothesis is represented as

H₀: p = 0

The alternative hypothesis is represented as

Hₐ: p ≠ 0

Hence, it is evident that the comparison population would be the difference scores being 0.

b) This test aims to test whether there's a negative or positive difference in the scores before and after the workshop.

So, the test will be in both directions. The cut off t-score indicating the rejection regions would be found using the significance level provided (0.05) and the degree of freedom.

degree of freedom = df = n - 1 = 10 - 1 = 9

t(9, 0.05) = ± 2.2622 (from the t-distribution tables)

c) Variance is an average of the squared deviations from the mean.

Mathematically, estimate of the population variance is given as

Variance = σ² = [Σ(x - xbar)²/N]

Where Σ(x - xbar)² = square of the deviations from the mean = 135

N = Sample Size = 135

So, variance = (135/10) = 13.5

Hope this Helps!!!