Question

Scores this year for students taking the SAT Math test for the first time are believed to be Normally distributed with mean Mi. For students taking the test for the second time, this year's scores are also believed to be Normally distributed, but with a possibly different mean M2. We wish to estimate the difference M2 - Mi. A random sample of the SAT Math scores of 100 students who took the test for the first time this year was obtained, and the mean of these 100 scores was ĉ1 = 504.5 while the standard deviation was s1 = 100. A random sample of the SAT Math scores of 30 students who took the test for the second time this year was also obtained, and the mean of these 30 scores was 72 = 539.1 while the standard deviation was $2 = 90. Using the conservative value for the degrees of freedom, a 95% confidence interval for M2 – Mj is: O (1.92, 67.28). O (-4.74, 73.94). O (-3.10, 72.30).

Answer 1

Solution - X =104.5 Xz - 539.1 95% confidence Si=100 S2-go m. = 100 n2 = 30 interval: vo na 1904.5 – 539.1) + 1.96 1002 + go = - 34.6 I 1.96 (19.2354) 2 - 34,6 + 37.70 lower limit = +34.6 - 37.70 = -3.10 upper Limit = +34.6 + 37.70 = 72.30 7 95% is considence Interval (-3.10, 72.30) ES