Name Problem #1 In assessing the 2016-2017 junior class at Bradley University, D
ID: 3331401 • Letter: N
Question
Name Problem #1 In assessing the 2016-2017 junior class at Bradley University, Dr. Timpe has come to the conclusion that he has either gone soft or the students have become noticeably more capable. His conclusion is primarily based on the class average on his exams, formerly thought to strike terror into the hearts of all junior level Mechanical Engineering students. The data is as follows: The 20 students from the current class have performed with an average of 65 pts and a standard deviation of 12 pts. The 100 students from, the previous five years have performed with an average of 63 pts and a standard deviation of 15 pts · Approximate the confidence with which Dr. Timpe makes his conclusions. Show all work for full credit.Explanation / Answer
TRADITIONAL METHOD
given that,
mean(x)=65
standard deviation , s.d1=12
number(n1)=20
y(mean)=63
standard deviation, s.d2 =15
number(n2)=100
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((144/20)+(225/100))
= 3.074
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table, two tailedand
value of |t | with min (n1-1, n2-1) i.e 19 d.f is 2.093
margin of error = 2.093 * 3.074
= 6.434
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (65-63) ± 6.434 ]
= [-4.434 , 8.434]
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DIRECT METHOD
given that,
mean(x)=65
standard deviation , s.d1=12
sample size, n1=20
y(mean)=63
standard deviation, s.d2 =15
sample size,n2 =100
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 = [( 65-63) ± t a/2 * sqrt((144/20)+(225/100)]
= [ (2) ± t a/2 * 3.074]
= [-4.434 , 8.434]
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interpretations:
1. we are 95% sure that the interval [-4.434 , 8.434] 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
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