A random sample of 48 students took an SAT preparation cours

A random sample of 48 students took an SAT preparation course prior to taking the SAT. The sample mean of their quantitative SAT scores was 560 with a s.d. of 80, and the sample mean of their verbal SAT scores was 520 with a s.d. of 110.

a) Construct 95% confidence intervals for the mean quantitative SAT and the mean verbal SAT scores of all students who take this course.

b) What sample size would be needed to estimate the mean verbal SAT score with 95% confidence and with error of no more than 15 if it is assumed that the s.d. is no more than 120?

c) Suppose the mean scores for all students who took the SAT at that time was 525 for the quantitative and 500 for the verbal? Do the means for students who take this course differ from the means for all students at the 10% level of significance?

Solution

a)
mean quantitative SAT scores

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
Mean(x)=560
Standard deviation( sd )=80
Sample Size(n)=48
Confidence Interval = [ 560 ± t a/2 ( 80/ Sqrt ( 48) ) ]
= [ 560 - 2.012 * (11.547) , 560 + 2.012 * (11.547) ]
= [ 536.767,583.233 ]

mean verbal SAT scores
Mean(x)=520
Standard deviation( sd )=110
Sample Size(n)=48
Confidence Interval = [ 520 ± t a/2 ( 110/ Sqrt ( 48) ) ]
= [ 520 - 2.012 * (15.877) , 520 + 2.012 * (15.877) ]
= [ 488.055,551.945 ]

b)
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )
Standard Deviation ( S.D) = 120
ME =15
n = ( 1.96*120/15) ^2
= (235.2/15 ) ^2
= 245.862 ~ 246  

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