A) The American Academy of Psychological Professionals has surveyed the populati

Question

A) The American Academy of Psychological Professionals has surveyed the population of clinical psychologists treating depression and has found that clinicians report significant improvement for patients treated with cognitive-behavior therapy after an average of 20 weeks of treatment, with a standard deviation of 8 weeks. A psychologist has developed a new therapeutic approach that she believes will lead to improvement of depressive symptoms in a shorter time. She uses this new approach with 25 clients and has an independent evaluator determine the when her clients have met the criterion for improvement. The average number of weeks of treatment needed to show improvement in this group is 17 weeks. Is there evidence that this new therapeutic approach leads to an improvement in a shorter time? Test the null hypothesis at the .01 level of significance. Show all work and present a hypothesis test summary following the outline. Be sure to include all 6 elements of the hypothesis test summary – research problem, statistical hypotheses, decision rule, calculations, decision and interpretation.)

B) An alternative to the hypothesis test above might be the calculation of a confidence interval. Find the 99% confidence interval for the data listed above.

Explanation / Answer

a.
Given that,
population mean(u)=20
standard deviation, =8
sample mean, x =17
number (n)=25
null, Ho: =20
alternate, H1: >20
level of significance, = 0.01
from standard normal table,right tailed z /2 =2.326
since our test is right-tailed
reject Ho, if zo > 2.326
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 17-20/(8/sqrt(25)
zo = -1.875
| zo | = 1.875
critical value
the value of |z | at los 1% is 2.326
we got |zo| =1.875 & | z | = 2.326
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : right tail - ha : ( p > -1.875 ) = 0.9696
hence value of p0.01 < 0.9696, here we do not reject Ho
ANSWERS
---------------
null, Ho: =20
alternate, H1: >20
test statistic: -1.875
critical value: 2.326
decision: do not reject Ho
p-value: 0.9696

b.
TRADITIONAL METHOD
given that,
standard deviation, =8
sample mean, x =17
population size (n)=25
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 8/ sqrt ( 25) )
= 1.6
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
since our test is two-tailed
value of z table is 2.576
margin of error = 2.576 * 1.6
= 4.122
III.
CI = x ± margin of error
confidence interval = [ 17 ± 4.122 ]
= [ 12.878,21.122 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =8
sample mean, x =17
population size (n)=25
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
since our test is two-tailed
value of z table is 2.576
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 = [ 17 ± Z a/2 ( 8/ Sqrt ( 25) ) ]
= [ 17 - 2.576 * (1.6) , 17 + 2.576 * (1.6) ]
= [ 12.878,21.122 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [12.878 , 21.122 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean
[ANSWERS]
best point of estimate = mean = 17
standard error =1.6
z table value = 2.576
margin of error = 4.122
confidence interval = [ 12.878 , 21.122 ]

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