SOLUTION: Based on a sample of 200 textbooks at the store, you find an average of 88.75 and a standard deviation of 12.8.
The 95 % confidence interval (use z*) is:
_____ to ____ (to 3
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-> SOLUTION: Based on a sample of 200 textbooks at the store, you find an average of 88.75 and a standard deviation of 12.8.
The 95 % confidence interval (use z*) is:
_____ to ____ (to 3
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Question 1119962: Based on a sample of 200 textbooks at the store, you find an average of 88.75 and a standard deviation of 12.8.
The 95 % confidence interval (use z*) is:
_____ to ____ (to 3 decimals) Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! the half interval is z(0.975)*sigma/sqrt(n)
This is 1.96 *12.8/sqrt(200)
=1.774
the CI is 1.774 on either side of the mean
(86.976, 90.524)
