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
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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