A random sample of 48 students took an SAT preparation course prior to taking th

Question

A random sample of 48 students took an SAT preparation course prior to taking the SAT. The s sample mean of their verbal SAT scores was 520 with a s.d. of 110 eir quantitative SAT scores was 560 with a s.d. of 80, and the o Construct 95% confidence intervals or the mean quantitative S and the mean verb SA scores ofall students who a ethis course o Construct 95% confidence intervals for the standard deviations of the SAT and VSA scores of all students who a e ths course. o What sample size would be needed to estimate the mean VSAT score with 95% confidence and wa error of no more than 15 f is assumed hat he sa no more o than 100? o Suppose the mean scores for all students who took the SAT at that time was 545 for the quantitative and 480 for the verbal? Do the means for students who take this t mecan or l course differ from the means for all students at the 10% level of significance?

Explanation / Answer

Here Sample size = 48

sample mean of quantitative SAT = 560

s.d = 80

sample mean of verbal SAT = 520

s.d = 110

(a) 95% confidence intervals for mean quantitiave SAT = xq +- t47,0.05 (s/ sqrt(n)

= 560 +- 2.01174 * [ 80/ sqrt(48)] = 560 +- 23.23 = (536.77, 583.23)

95% confidence intervals for mean Verbal SAT = xv +- t47,0.05 (s/ sqrt(n)

= 520 +- 2.01174 * [ 110/ sqrt(48)] = 520 +- 31.94 = (486.06, 551.94)

(b) 95% confidence interval for standard deviation of quantitiave SAT =

Lower bound = sqrt [(n-1) s2 / x2 0.025] = sqrt [ (48-1) * 80 * 80 / 67.82] = 66.60

Upper BOund =  sqrt [(n-1) s2 / x2 0.975] = sqrt [ (48-1) * 80 * 80 / 29.9562] = 100.21

so 95% confidence interval for standard deviation = (66.60, 100.21)

95% confidence interval for standard deviation of Verbal SAT =

Lower bound = sqrt [(n-1) s2 / x2 0.025] = sqrt [ (48-1) * 110 * 110 / 67.82] = 91.57

Upper BOund =  sqrt [(n-1) s2 / x2 0.975] = sqrt [ (48-1) * 110 * 110 / 29.9562] = 137.78

so 95% confidence interval for standard deviation = (91.57, 137.78)

(c) Here sample size = n

Here margin of error = 15

Sample standard deviation = 100

margin of error = test statistic * Standard error of the mean

15 = 2.01174 * (s/ sqrt(n)

15 = 2.01174 * (100/ sqrt(n))

sqrt(n) = 2.01174 * 100/ 15

sqrt(n) = 13.4116

n = 179.87 = 180

(4) Here Pr(  xq < 545 ; 560 ; 11.547)

Z = (545 - 560)/ 11.547 = -1.30

Pr(  xq < 545 ; 560 ; 11.547) = Pr(Z < -1.30) = 0.0968

Now for verbal score

Pr(  xq < 480; 520 ; 15.877)

Z = (480 - 520)/ 15.877 = -2.52

Pr(  xq < 480 ; 520 ; 15.877) = Pr(Z < -2.52) = 0.00587

so yes this batch of students have probability less than 0.10 so we say that mean are different from the mean for all students.

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