Question 866629: The population of the scores of all high school seniors that took the SAT-M test (mathematics component of the SAT test) last year followed a Normal distribution, with mean mu and standard deviation = 100. You read a report that says, “On the basis of a simple random sample of 500 high school seniors that took the SAT-M test this year, a confidence interval for the population (mu) is 512.00 ± 11.52.”
A 95% confidence interval for the population (mu) would be
Question 4 options:
503.24 to 520.77
500.48 to 523.52
316 to 708
None of the above
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! The lower bound L and the upper bound U are equal to
L = xbar - ME
U = xbar + ME
where xbar is the sample mean and ME is the margin of error. In this case,
xbar = 512
ME = 11.52
These are both given when they report the confidence interval (CI) of 512.00 ± 11.52
So,
L = xbar - ME
L = 512 - 11.52
L = 500.48
and
U = xbar + ME
U = 512 + 11.52
U = 523.52
Making the final answer to be 500.48 to 523.52
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