Question #339347

For a normal distribution, assume the 'sample' standard deviation is 8. What is the maximal margin of error associated with a 99% confidence interval for the true population mean if n = 25. ?

(Round to two decimal places)



1
Expert's answer
2022-05-10T23:40:45-0400

The critical value for α=0.01\alpha = 0.01 and df=n1=24df = n-1 = 24 is tc=z1α/2;n1=2.79694.t_c = z_{1-\alpha/2; n-1} = 2.79694.


E=tc×sn=2.79694×825=4.48E=t_c\times\dfrac{s}{\sqrt{n}}=2.79694\times\dfrac{8}{\sqrt{25}}=4.48

The maximal margin of error is 4.48.4.48.


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