Question

Chapter 6, Section 4-D, Exercise 180

Use the formula to find the standard error of the distribution of differences in sample means, x¯1-x¯2. Samples of size 115 from Population 1 with mean 90 and standard deviation 14 and samples of size 90 from Population 2 with mean 73 and standard deviation 16 Round your answer for the standard error to two decimal places.

Answer 1

TRADITIONAL METHOD
given that,
mean(x)=90
standard deviation , s.d1=14
number(n1)=115
y(mean)=73
standard deviation, s.d2 =16
number(n2)=90
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((196/115)+(256/90))
= 2.133
II.
margin of error = t a/2 * (standard 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 89 d.f is 1.987
margin of error = 1.987 * 2.133
= 4.238
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (90-73) ± 4.238 ]
= [12.762 , 21.238]
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DIRECT METHOD
given that,
mean(x)=90
standard deviation , s.d1=14
sample size, n1=115
y(mean)=73
standard deviation, s.d2 =16
sample size,n2 =90
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 = [( 90-73) ± t a/2 * sqrt((196/115)+(256/90)]
= [ (17) ± t a/2 * 2.133]
= [12.762 , 21.238]
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interpretations:
1. we are 95% sure that the interval [12.762 , 21.238] 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:
standard error = 2.13