3. In a study designed to test the effectiveness of magnets for treating back pa

Question

3. In a study designed to test the effectiveness of magnets for treating back pain,

40

patients were given a treatment with magnets and also a sham treatment without magnets. Pain was measured using a scale from 0 (no pain) to 100 (extreme pain). After given the magnet treatments, the

40

patients had pain scores with a mean of

8.0

and a standard deviation of

2.5

After being given the sham treatments, the

40

patients had pain scores with a mean of

6.1

and a standard deviation of

2.7

Complete parts (a) through (c) below.

a. Construct the

90%

confidence interval estimate of the mean pain score for patients given the magnet treatment.

What is the confidence interval estimate of the population mean

mu ?

___ l< < ___

(Round to one decimal place as needed.)

b. Construct the

90%

confidence interval estimate of the mean pain score for patients given the sham treatment.

What is the confidence interval estimate of the population mean

mu ?

___ < < ____

(Round to one decimal place as needed.)

c. Compare the results. Does the treatment with magnets appear to be effective?

A.

Since the confidence

intervalsnbsp overlap,

it appears that the magnet treatments are

lessless

effective than the sham treatments.

B.

Since the confidence

intervalsnbsp do not nbsp do not overlap,

it appears that the magnet treatments are

moremore

effective than the sham treatments.

C.

Since the confidence

intervalsnbsp do not nbsp do not overlap,

it appears that the magnet treatments are

lessless

effective than the sham treatments.

D.

Since the confidence

intervalsnbsp overlap,

it appears that the magnet treatments are

no moreno more

effective than the sham treatments.

Explanation / Answer

3.

a.
TRADITIONAL METHOD
given that,
standard deviation, =2.5
sample mean, x =8
population size (n)=40
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 2.5/ sqrt ( 40) )
= 0.4
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
since our test is two-tailed
value of z table is 1.645
margin of error = 1.645 * 0.4
= 0.7
III.
CI = x ± margin of error
confidence interval = [ 8 ± 0.7 ]
= [ 7.3,8.7 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =2.5
sample mean, x =8
population size (n)=40
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
since our test is two-tailed
value of z table is 1.645
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 8 ± Z a/2 ( 2.5/ Sqrt ( 40) ) ]
= [ 8 - 1.645 * (0.4) , 8 + 1.645 * (0.4) ]
= [ 7.3,8.7 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [7.3 , 8.7 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean
[ANSWERS]
best point of estimate = mean = 8
standard error =0.4
z table value = 1.645
margin of error = 0.7
confidence interval = [ 7.3 , 8.7 ]
confidence interval estimate of the mean pain score for patients given the magnet treatment

b.
TRADITIONAL METHOD
given that,
standard deviation, =2.7
sample mean, x =6.1
population size (n)=40
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 2.7/ sqrt ( 40) )
= 0.4
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
since our test is two-tailed
value of z table is 1.645
margin of error = 1.645 * 0.4
= 0.7
III.
CI = x ± margin of error
confidence interval = [ 6.1 ± 0.7 ]
= [ 5.4,6.8 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =2.7
sample mean, x =6.1
population size (n)=40
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
since our test is two-tailed
value of z table is 1.645
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 6.1 ± Z a/2 ( 2.7/ Sqrt ( 40) ) ]
= [ 6.1 - 1.645 * (0.4) , 6.1 + 1.645 * (0.4) ]
= [ 5.4,6.8 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [5.4 , 6.8 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean
[ANSWERS]
best point of estimate = mean = 6.1
standard error =0.4
z table value = 1.645
margin of error = 0.7
confidence interval = [ 5.4 , 6.8 ]
confidence interval estimate of the mean pain score for patients given the sham treatment.

c.
confidence interval = [ 7.3 , 8.7 ]
confidence interval estimate of the mean pain score for patients given the magnet treatment

confidence interval = [ 5.4 , 6.8 ]
confidence interval estimate of the mean pain score for patients given the sham treatment.
compare results,option :B
Since the confidence intervals nbsp do not overlap,it appears that the magnet treatments are moreeffective than the sham treatments.

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