How do you solve this with order of operation?
1 answer:
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Answer:
Check the explanation
Step-by-step explanation:
(a) The appropriate test is the matched-pairs test because a student’s score on Try 1 is certainly correlated with his/her score on Try 2. Using the differences, we have xbar = 29 and s = 59.
(b) To test H0: mu=0 vs. H1 mu > 0, we compute
with df = 426. This is certainly significant, with P < 0.0005. Coached students do improve their scores on average
(a) H0: μ1 = μ2 vs. Ha: μ1 > μ2, where μ1 is the mean gain among all coached students and μ2 is the mean gain among uncoached students. H0 and Ha. Using the conservative approach, df = 426 is rounded down to df = 100 in (t table) and we obtain 0.0025 < P < 0.005. Using software, df = 534.45 and P = 0.004. There is evidence that coached students had a greater average increase.
(b) 8 ± t*(3.0235) where t* equals 2.626 (using df = 100 with (t table) ) or 2.585 (df = 534.45 with software). This gives either 0.06 to 15.94 points, or 0.184 to 15.816 points, respectively.
(c) Increasing one’s score by 0 to 16 points is not likely to make a difference in being granted admission or scholarships from any colleges.