Dr. Joe determines that 68.2% of college students have IQs between 120 and 150,

Question

Dr. Joe determines that 68.2% of college students have IQs between 120 and 150, with an average of 135. Dr. Joe also tests 4 randomly selected college students and finds the average IQ is 140 with a sample deviation of 20. We are 90% certain that the average student has an IQ between "a" and "b" (i.e. a < IQ < b). What is the correct value for "a"?

O A 135 (zo.45)(10) B. 140 (to.os, 3(20) O C. 140(to.os, 3)(10) D 140(to.10, 3)(10) OE 140(z0.45)(10) F 135 (z0.47510) G 140- (to.os, 3)(0) OH. 140-(to.10, 3)(10) OL 140 (zo.45)(15) OJ. 135(zo.45)(15) O K 135 - (to.10, 3)(20) OL 135 (z0.475)(15) M. 135-(z0o.45) 15) ON. 140 - (to.os, 3) (15) 135-(z0.45)(20) O P. The correct answer is not listed

Explanation / Answer

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 - (t0.05, 3)(10) < X < 140 + (t0.05, 3)(10) ]
= [ 140-(2.353 * 10) , 140+(2.353 * 10) ]
= [ 116.47 , 163.53 ]

[ANSWER]
140 - (t0.05, 3)(10)

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