Question

A sample obese adult males were randomly divided into two groups in a lifestyle change program. In the experimental group, participants were involved in a vigorous 20-minute-per-day exercise program and also given instructions on the importance of diet. The control group only received the diet instructions. The hospital administrators predict that exercise plus diet instruction will reduce LDL cholesterol levels. Below are the LDL cholesterol data after two months in the program. What can be concluded with α = 0.10?

control	exprt
129 127 131 126 127 129 128 129 125 129	132 153 135 131 132 128 137 131 132 132

a) What is the appropriate test statistic?
---Select One--- (na, z-test, One-Sample t-test, Independent-Samples t-test, or Related-Samples t-test)

b)
Condition 1:
---Select One--- (LDL cholesterol, obese adult males, two months, control group, or exprt group)
Condition 2:
---Select One---  (LDL cholesterol, obese adult males, two months, control group, or exprt group)

c) Input the appropriate value(s) to make a decision about H₀.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
p-value = _____ ; Decision:  ---Select One--- (Reject H0 or Fail reject H0)

d) Using the SPSS results, compute the corresponding effect size(s) and indicate magnitude(s).
If not appropriate, input and/or select "na" below.
d = _____ ; Magnitude:  ---Select One--- "na, trivial effect, small effect, medium effect, or large effect"
r² =_____ ; Magnitude:  ---Select One--- "na, trivial effect, small effect, medium effect, or large effect"

e) Make an interpretation based on the results. Which is correct?

a)Participants that exercised with diet instructions had significantly reduced LDL cholesterol levels.

b)Participants that exercised with diet instructions had significantly increased LDL cholesterol levels.    

c)There is no significant LDL cholesterol difference between the groups.

Answer 1

Soln

a)

Independent-Samples t-test

c)

Group 1 (Control)

Mean (x₁) = 128

Std Dev (s₁) = 1.76

n₁ = 10

Group 2 (Expert)

Mean (x₂) = 134.3

Std Dev (s₂) = 6.99

n₂ = 10

alpha = 0.1

df = 10+10-2 = 18

t_Critical = 2.473

Null and Alternate Hypothesis

H₀: µ₁ = µ₂

H_a: µ₁ > µ₂

Test Statistic

Assuming, the population std deviation is not same.

t = (X_{1 ­­­}– X_{2 ­}– (µ₁ - µ₂))/ (s₁²/n₁ + s₂²/n₂ )^1/2 = -2.76

p-value = TDIST(0.05,15+14-2,1) = 0.006

Result

Since the p-value is less than 0.1, we reject the null hypothesis.

d)

Pooled Std Dev(s_p) = {{(n₁-1) *s₁² + (n₂-1) *s₂²}/ (n₁ + n₂ -2)}^1/2 = 5.10

Effect Size = (x₁ – x₂ )/ SD_pooled = (128-134.3)/5.10 = -1.24

e)

Option A ie Participants that exercised with diet instructions had significantly reduced LDL cholesterol levels.