Question

Construct the confidence interval for the population mean p. c=0.90, x = 6.5, 0 =0.9, and n=53 A 90% confidence interval for p is ( OD (Round to two decimal places as needed.)

Answer 1

ANSWER:

For n>30 Confidence interval of x , we have μ- z*σ/sqrt(n) ,μ+z*σ/sqrt(n)

From normal table, for 90 % confidence interval, we check where p(z) =.95 as 90 percent is symmetrically distributed about mean and other 10 % is also symmetrically distributed. Thus we check for 90% + 5% = .95   We have

90% ci z= 1.65

as E(X/μ) =X =6.5

Confidence interval of μ is given by E(X/μ) - z*σ/sqrt(n) ,E(X/μ) + z*σ/sqrt(n)

Thus required C.I. is 6.5-1.65*0.9/sqrt(53),

=6.5+1.65*0.9/sqrt(53)

= (6.3 , 6.7)

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