Question

1. A psychologist has developed a personality test on which scores can range from 0 to 200. The mean score of well-adjusted people is 100. It is assumed that the less well-adjusted an individual is, the more the individual’s score will differ from the mean value. The Dean of Students at XYZ University is interested in discovering whether the students at XYZ are well-adjusted, on average. Fifteen students are randomly sampled. Their scores are: 80, 60, 120, 140, 200, 70, 30, 180, 70, 150, 20, 50, 170, 90, 130.
   1. Estimate the magnitude of the effect using Cohen’s d. What does your number mean?
   2. Construct the 95% confidence interval to estimate mean adjustment. State the outcome in English.
   3. In what way is the question tested in this problem not a sensible question?

Answer 1

a.
cohen's d size = (sample mean- population mean)/S.D
cohen's d size =(104-100)/54.258
cohen's d size = 0.0737
the magnitude of the effect is small effect
sample mean, x =104
standard deviation, s =54.258
sample size, n =15
b.
TRADITIONAL METHOD
given that,
sample mean, x =104
standard deviation, s =54.258
sample size, n =15
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 54.258/ sqrt ( 15) )
= 14.009
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 14 d.f is 2.145
margin of error = 2.145 * 14.009
= 30.05
III.
CI = x ± margin of error
confidence interval = [ 104 ± 30.05 ]
= [ 73.95 , 134.05 ]
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DIRECT METHOD
given that,
sample mean, x =104
standard deviation, s =54.258
sample size, n =15
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 14 d.f is 2.145
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 104 ± t a/2 ( 54.258/ Sqrt ( 15) ]
= [ 104-(2.145 * 14.009) , 104+(2.145 * 14.009) ]
= [ 73.95 , 134.05 ]
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interpretations:
1) we are 95% sure that the interval [ 73.95 , 134.05 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean
c.
Given that,
population mean(u)=100
sample mean, x =104
standard deviation, s =54.258
number (n)=15
null, Ho: μ=100
alternate, H1: μ!=100
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.145
since our test is two-tailed
reject Ho, if to < -2.145 OR if to > 2.145
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =104-100/(54.258/sqrt(15))
to =0.2855
| to | =0.2855
critical value
the value of |t α| with n-1 = 14 d.f is 2.145
we got |to| =0.2855 & | t α | =2.145
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 0.2855 ) = 0.7794
hence value of p0.05 < 0.7794,here we do not reject Ho
ANSWERS
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null, Ho: μ=100
alternate, H1: μ!=100
test statistic: 0.2855
critical value: -2.145 , 2.145
decision: do not reject Ho
p-value: 0.7794
we do not have enough evidence to support the claim that the less well-adjusted an individual is, the more the individual’s score will differ from the mean value