Question

QUESTION 16 When a class of 23 students took the SAT, they had a mean score of 1445 with a standard deviation of 178. Construct a 99% confidence interval for the mean SAT score of all students (). O (1368, 1522) O (1381, 1509) (1352, 1538) (1340, 1550)

Answer 1

Solutions
n=23
x̄= 1445
S=178
(1-α)%= 99%
1-α= 0.99
α= 0.01
α/2=0.005
t(α/2),(n-1)=t(0.005),(23-1)
=t(0.005),(22)= 2.819..... From students t table.
Margin of error=t(α/2),(n-1) ×( S /✓n)
= (2.819)×(178/✓23)
=104.6288
99% Confidence Interval for the mean SAT score of all students ( μ ) is given as,
x̄ ± Margin of error=(1445-104.6288,1445+104.6288)
=(1,340.3712, 1549.6288)
= (1340 ,1550)

99% Confidence Interval for the mean SAT score of all students ( μ ) is (1340 ,1550)