Question

Assume that a sample is used to estimate a population mean μ μ . Find the margin of error M.E. that corresponds to a sample of size 10 with a mean of 17.7 and a standard deviation of 17.6 at a confidence level of 99.8%. Report ME accurate to one decimal place because the sample statistics are presented with this accuracy. M.E. =

Answer #1

Solution :

Given that,

= 17.7

s =17.6

n = Degrees of freedom = df = n - 1 =10 - 1 = 9

At 99.8% confidence level the t is ,

= 1 - 99.8% = 1 - 0.998 = 0.002

/ 2= 0.001

t
/2,df = t0.001, 9 = **4.297** ( using student t
table)

Margin of error = E = t/2,df * (s /n)

= **4.297*** ( 17.6/
10)

= 23.9

Margin of error = E =23.9

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