To study the effectiveness of a weight-loss training method, a random sample of

Question

To study the effectiveness of a weight-loss training method, a random sample of 40
females went through the training. After a two-month training period, the weight
change is recorded for each participant. The researcher wants to test that the mean
weight reduction is larger than 5 pounds at level = 0.05.
a. The researcher knows that the population standard deviation is approximately
7 pounds. What is the required sample size to detect the power of 80% when the
true weight reduction is 8 pounds? Is the sample size of 40 enough?
b. The average weight loss for the 40 women in the sample was 7.2 pounds with
the sample standard deviation of 4.5 pounds. Is there sufficient evidence that the
training is effective? Test using = 0.05.
c. Find the power of the test conducted in part (b) if the true mean weight reduction
is 8 pounds.

Explanation / Answer

The researcher wants to test that the mean
weight reduction is larger than 5 pounds at level = 0.05
random sample of 40
females went through the training
the population standard deviation is approximately 7 pounds
the required sample size to detect the power of 80% when the
true weight reduction is 8 pounds
a.
n= (((Zalpha +Zbeta))/(U-Uo))^2
Zalpha at 0.05 = 1.96
Zbeta at 80% = 0.84
n = ((7(1.96+0.84))/(8-5))^2
n =42.684 = 43

b.
Given that,
population mean(u)=5
sample mean, x =7.2
standard deviation, s =4.5
number (n)=40
null, Ho: =5
alternate, H1: >5
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.6849
since our test is right-tailed
reject Ho, if to > 1.6849
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =7.2-5/(4.5/sqrt(40))
to =3.092
| to | =3.092
critical value
the value of |t | with n-1 = 39 d.f is 1.6849
we got |to| =3.092 & | t | =1.6849
make decision
hence value of | to | > | t | and here we reject Ho
p-value :right tail - Ha : ( p > 3.092 ) = 0.00183
hence value of p0.05 > 0.00183,here we reject Ho
ANSWERS
---------------
null, Ho: =5
alternate, H1: >5
test statistic: 3.092
critical value: 1.6849
decision: reject Ho
p-value: 0.00183
we have enough evidence to support the claim that trainig is effective

c.
Given that,
Standard deviation, =4.5
Sample Mean, X =7.2
Null, H0: =5
Alternate, H1: >5
Level of significance, = 0.05
From Standard normal table, Z /2 =1.6449
Since our test is right-tailed
Reject Ho, if Zo < -1.6449 OR if Zo > 1.6449
Reject Ho if (x-5)/4.5/(n) < -1.6449 OR if (x-5)/4.5/(n) > 1.6449
Reject Ho if x < 5-7.402/(n) OR if x > 5-7.402/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 40 then the critical region
becomes,
Reject Ho if x < 5-7.402/(40) OR if x > 5+7.402/(40)
Reject Ho if x < 3.83 OR if x > 6.17
Implies, don't reject Ho if 3.83 x 6.17
Suppose the true mean is 8
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(3.83 x 6.17 | 1 = 8)
= P(3.83-8/4.5/(40) x - / /n 6.17-8/4.5/(40)
= P(-5.861 Z -2.572 )
= P( Z -2.572) - P( Z -5.861)
= 0.0051 - 0 [ Using Z Table ]
= 0.005
For n =40 the probability of Type II error is 0.005

power of the test is 1- beta = 1-0.005 =0.995

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