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Thirty randomly selected students took the calculus final. If the sample mean was 95 and the standard deviation was 6.6, construct a 99% confidence interval for the mean score of all students
A: 91.68 < μ <98.32
B: 92.95 < μ < 97.05
C: 91.69 < μ < 98.31

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shikamaru1147

Answer: A: 91.68 < μ <98.32

solution:

Sample standard error = 6.6/SQRT(30) = 1.205

t-value for the 99% CI = 2.756

Margin of error = 2.756 x 1.205 = 3.321

So, the 99% CI = (95 - 3.321 , 95 + 3.321) = (91.679 , 98.321)

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