Using top secret University records, Dr. Rasmussen determines that 68.2% of MEEN
ID: 2923493 • Letter: U
Question
Using top secret University records, Dr. Rasmussen determines that 68.2% of MEEN students have IQs between 120 and 150, with an average of 135. Dr. Rasmussen also tests 4 randomly selected MEEN 260 students and finds the average IQ is 140 with a sample deviation of 20. We are 90% certain that the average MEEN 260 student has an IQ between "a" and "b" (i.e. a < IQ < b). What is the correct value for "a"?
140 + (t0.05, 3)(10)
135 - (z0.475)(10)
140 - (t0.05, 3)(10)
140 + (z0.45)(10)
135 - (t0.10, 3)(20)
135 + (z0.45)(15)
140 - (t0.10, 3)(10)
135 + (z0.45)(10)
140 - (z0.45)(15)
The correct answer is not listed
140 - (t0.05, 3)(20)
135 - (z0.45)(20)
140 + (t0.10, 3)(10)
135 - (z0.475)(15)
140 - (t0.05, 3)(15)
135 - (z0.45)(15)
A.140 + (t0.05, 3)(10)
B.135 - (z0.475)(10)
C.140 - (t0.05, 3)(10)
D.140 + (z0.45)(10)
E.135 - (t0.10, 3)(20)
F.135 + (z0.45)(15)
G.140 - (t0.10, 3)(10)
H.135 + (z0.45)(10)
I.140 - (z0.45)(15)
J.The correct answer is not listed
K.140 - (t0.05, 3)(20)
L.135 - (z0.45)(20)
M.140 + (t0.10, 3)(10)
N.135 - (z0.475)(15)
O.140 - (t0.05, 3)(15)
P.135 - (z0.45)(15)
Explanation / Answer
TRADITIONAL METHOD
given that,
sample mean, x =140
standard deviation, s =20
sample size, n =4
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 20/ sqrt ( 4) )
= 10
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 3 d.f is 2.353
margin of error = 2.353 * 10
= 23.53
III.
CI = x ± margin of error
confidence interval = [ 140 ± 23.53 ]
= [ 116.47 , 163.53 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =140
standard deviation, s =20
sample size, n =4
level of significance, = 0.1
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
from standard normal table, two tailed value of |t /2| with n-1 = 3 d.f is 2.353
confidence interval = [ 140 ± t a/2 ( 20/ Sqrt ( 4) ] = [ 140 ± t a/2 ( 20/ Sqrt ( 4) ] =
[ 140 - (t0.10, 3)(10) < X < 140 + (t0.10, 3)(10) ]
= [ 140-(2.353 * 10) , 140+(2.353 * 10) ]
= [ 116.47 , 163.53 ]
[ANSWER]
140 - (t0.05, 3)(10)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.