Question

Friedman and Rosenman (1974) classified people into two categories: Type A personalities and Type B personalities.  Type A's are hard-driving, competitive, and ambitious. Type B's are more relaxed, easy going people.  One factor that differentiates these two groups is the   chronically high level of frustration experienced by Type A's, especially when doing statistical analyses.  (Oftentimes Type A's have been known to suffer from numerical alexithymia.) To verify this aspect separate samples of Type A's and Type B's are obtained. The following frustration scores were obtained (note: the higher the score the greater the degree of frustration)

Type A                                    Type B

25                                            21

   23                                            16

   19                                            19

                                                    18    

Question: Calculate the estimate of the standard error of mean differences ___________

Please show all work so I may check my steps

Answer 1

TRADITIONAL METHOD
given that,
mean(x)=22.3333
standard deviation , s.d1=3.0551
number(n1)=3
y(mean)=18.5
standard deviation, s.d2 =2.0817
number(n2)=4
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((9.334/3)+(4.333/4))
= 2.048
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 2 d.f is 4.303
margin of error = 4.303 * 2.048
= 8.813
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (22.3333-18.5) ± 8.813 ]
= [-4.98 , 12.646]
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DIRECT METHOD
given that,
mean(x)=22.3333
standard deviation , s.d1=3.0551
sample size, n1=3
y(mean)=18.5
standard deviation, s.d2 =2.0817
sample size,n2 =4
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 22.3333-18.5) ± t a/2 * sqrt((9.334/3)+(4.333/4)]
= [ (3.833) ± t a/2 * 2.048]
= [-4.98 , 12.646]
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interpretations:
1. we are 95% sure that the interval [-4.98 , 12.646] contains the true population proportion
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 proportion

Answer:
stanadard error = sqrt((9.334/3)+(4.333/4))
= 2.048