Suppose n=36,¯x=42n=36,x¯=42 and σ=14.9σ=14.9. Based on this, what is the maximal margin of error associated with a 99% confidence interval for the true population mean.
Margin of error = Round to 2 decimal places.
Solution :
Given that,
Point estimate = sample mean =
= 42
Population standard deviation =
= 14.9
Sample size = n = 36
At 99% confidence level
= 1 - 99%
= 1 - 0.99 =0.01
/2
= 0.005
Z/2
= Z0.005 = 2.576
Margin of error = E = Z/2
* (
/n)
= 2.576 * ( 14.9 / 36
)
= 6.40
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