Question
mean, mu=0.9050 standard deviation, s = 0.01 Sample size, n =48 Confidence interval xz(n) 0.9051.96(480.01) 0.905pm 0.002829 Lower bound = 0.905 - 0.002829 = 0.902171
mean, \mu=0.9050
standard deviation, s = 0.01
Sample size, n =48
Confidence interval
xz(n)
0.9051.96(480.01)
0.905\pm 0.002829
Lower bound = 0.905 - 0.002829 = 0.902171
Upper bound = 0.905 + 0.002829 = 0.907829
Interval from 0.9022 to 0.9078
Where did they get the 1.96 from?
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Linear Algebra A Modern Introduction
Authors: David Poole
3rd edition
9781133169574 , 978-0538735452
