Question

A psychiatrist wants to know what the impact of using a specific grounding technique will be on his patients who have panic attacks. Dr Lessanxious divided his patients into two groups. One group of patients was taught to use the specific grounding technique and the second group was not taught the grounding technique. After being taught the grounding technique (or not) all patients were asked to rate the intensity of their panic attacks on a scale of 1-20 with higher numbers indicating higher intensity. Answer the questions below based on the test results given below the questions, alpha = .05.

a. What type of test is this? Be specific.

b. Write the null hypothesis.

c. Write the alternative hypothesis.

d. How many total participants were there in the study?

e. What is the value of the test statistic?

f. What is the critical value with alpha = .05? (need the table in the book)

g. Is the result statistically significant?

h. Do you need to calculate Cohen’s d? If yes state the value, if no explain why not.

i. Write the result in correct APA statistical notation.

	Grounding Technique	No Grounding Technique
Mean	10.756	14.864
Variance	4.17717269	10.03765461
Observations	250	250
Pooled Variance	7.1074136
df	498
t stat	-17.2277930

j. Summarize the results in one or two sentences.

Answer 1

a.

The type of test is Pooled Samples t test.

b.

Null hypothesis H0: The difference between mean intensity rate of panic attacks of patients who were taught specific grounding technique and mean intensity rate of panic attacks of patients who were not taught specific grounding technique is zero.

c.

Alternative hypothesis Ha: The difference between mean intensity rate of panic attacks of patients who were taught specific grounding technique and mean intensity rate of panic attacks of patients who were not taught specific grounding technique is not zero.

d.

The total participants were there in the study = 2 * 250 = 500

e.

The value of the test statistic is -17.2277930

f.

Critical value of t with alpha = .05 and df = 498 are -1.965 and 1.965

g.

Since the test statistic is less than the critical value -1.965, we reject the null hypothesis H0.

h.

The results are significant, so we should calculate Cohen’s d.

Cohen’s d = Mean difference / Pooled Std deviations = (14.864 - 10.756) /

= 1.54

i.

There was a significant difference in the ratings for patients who were taught specific grounding technique (M = 10.756, Variance = 4.17717269) and patients who were not taught specific grounding technique (M = 14.864, Variance = 10.03765461) Conditions ; t(498) = -17.2277930 , p 0.0000

j.

Based on the sample data, there is a significant difference between mean intensity rate of panic attacks of patients who were taught specific grounding technique and mean intensity rate of panic attacks of patients who were not taught specific grounding technique is not zero.