Question

Use the given degree of confidence and sample data to construct a confidence interval for the population mean μ. Assume that the population has a normal distribution.

n =30,x bar =84.6, s=10.5, 90% confidence

.

A.79.32 <μ <89.88

B.81.34 <μ <87.86

C. 80.68 <μ <88.52

D.81.36 <μ <87.84

Answer #1

solution

)Given that,

= 84.6

s =10.5

n = 30

Degrees of freedom = df = n - 1 =30 - 1 = 29

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t
/2,df = t0.05,29 = **1.699** ( using
student t table)

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

= **1.699** * (10.5 /
30) = 3.26

The 90% confidence interval estimate of the population mean is,

- E < < + E

84.6 - 3.26 < < 84.6+ 3.26

B.81.34 <μ <87.86

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