Question

A school psychologist is interested in the effect a popular new hypnosis technique on anxiety. The psychologist collects a sample of 30 students and gives them the hypnosis once a week for two months. Afterwards the students fill out an anxiety inventory in which their mean score was 55.08. Normal individuals in the population have an anxiety inventory mean of 50 with a variance of 7.29. What can be concluded with α = 0.10?

a) What is the appropriate test statistic?
---Select--- na, z-test, one-sample t-test, independent-samples t-test, related-samples t-test

b)
Population:
---Select--- normal individuals, students receiving hypnosis, new hypnosis, technique two months
Sample:
---Select--- normal individuals, students receiving hypnosis, new hypnosis, technique two months

c) Obtain/compute the appropriate values to make a decision about H₀.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
critical value = _________; test statistic = ______________
Decision:  ---Select--- Reject H0, Fail to reject H0

d) If appropriate, compute the CI. If not appropriate, input "na" for both spaces below.
[ _________ ,___________ ]

e) Compute the corresponding effect size(s) and indicate magnitude(s).
If not appropriate, input and select "na" below.
d = __________;   ---Select--- na, trivial effect, small effect, medium effect, large effect
r² = ___________ ;   ----Select--- na, trivial effect, small effect, medium effect, large effect

f) Make an interpretation based on the results.

A) The population has significantly lower anxiety than students that underwent hypnosis.

B) The population has significantly higher anxiety than students that underwent hypnosis.    

C) The new hypnosis technique does not significantly effect anxiety.

Answer 1

(a) Z - Test. ( as population standard deviation is given)

(b) Population = normal individuals

Sample = students recieving hypnosis

(c) Hypothesis testing

Step 1: Null hypothesis states that anxiety inventory mean is 50

Ho :   = 50

Ha : 50

Step 2: Test Statistics

sample size = n = 30

population mean = =  50

population standard deviation = = sqrt (variance) = sort (7.29) = 2.70

sample mean = = 55.08

level of significance  = 0.10

Assuming that the population is normally distributed.

z= 10.305

This is a two tailed test since the anxiety inventory can be more or less than the mean.

The z-critical values for a two-tailed test, for a significance level of α=0.10

zc ​= −1.64 and zc ​=1.64

Since the Z value (10. 305) is greater than Z critical value (1.65) we reject the Null hypothesis.

Decision - Reject the Null hypothesis.

(d) Confidence interval

Now we will find confidence interval.

z value for confidence interval 90% = 1.645

E = 0.811

Hence CI is mean +/- E = 50+/-0.838 = (49.189 , 50.811)

As 55.08 is not in the range, we reject the Null hypothesis.