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 *H*_{0}.

(Hint: Make sure to write down the null and alternative hypotheses
to help solve the problem.)

** p-value** = _____ ;
Decision: ---

If not appropriate, input and/or select "na" below.

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 (x_{1}) = 128

Std Dev (s_{1}) = 1.76

n_{1} = 10

*Group 2 (Expert)*

Mean (x_{2}) = 134.3

Std Dev (s_{2}) = 6.99

n_{2} = 10

alpha = 0.1

df = 10+10-2 = 18

t_{Critical} = 2.473

*Null and Alternate
Hypothesis*

H_{0}: µ_{1} = µ_{2}

H_{a}: µ_{1} > µ_{2}

*Test Statistic*

Assuming, the population std deviation is not same.

t = (X_{1 }– X_{2 }– (µ_{1} -
µ_{2}))/ (s_{1}^{2}/n_{1} +
s_{2}^{2}/n_{2} )^{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}) = {{(n_{1}-1)
*s_{1}^{2} + (n_{2}-1)
*s_{2}^{2}}/ (n_{1} + n_{2}
-2)}^{1/2} = 5.10

Effect Size = (x_{1} – x_{2} )/
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.

A sample of soy bean plots were divided into three equal groups
in a completely randomized design, and each group received a
different formulation of a selective weed killer (wk). The
researchers measured the area of crop damage inflicted by each
formulation. The aim is to choose the weed killer formulation that
inflicts the least amount of crop damage. What can the researchers
conclude with an α of 0.05?
wk 1
wk 2
wk 3
11
10
13
10
15...

A sample of soy bean plots were divided into three equal groups
in a completely randomized design, and each group received a
different formulation of a selective weed killer (wk). The
researchers measured the area of crop damage inflicted by each
formulation. The aim is to choose the weed killer formulation that
inflicts the least amount of crop damage. What can the researchers
conclude with an α of 0.01?
wk 1
wk 2
wk 3
11
10
13
10
15...

How does gender and occupational prestige affect credibility?
Graduate students in a public health program are asked to rate the
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and degree associated with the author are changed. The student
raters are randomly assigned to one group from the following name
("John Lake", "Joan Lake") and degree (M.D., R.N., Ph.D.)
combination. The raters score the paper...

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