The prices of CD players are normally distributed, with a
standard deviation of $12. A random sample of 16
CD player prices is drawn. A confidence interval for the mean price
shows a $9.30 margin of error. What was
the level of confidence?
Given that, standard deviation = $12
sample size ( n ) = 16
Margin of error ( E ) = $9.30
where,
Using standard normal z-table we get,
P(Z < 3.10) = 0.9990
Therefore, we get,
Hence, level of confidence ( c ) =
Level of confidence = 99.8%
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