Question

Consider a normal population distribution with the value of σ known.

(a) What is the confidence level for the interval \(\bar{x} \pm 2.81 \sigma / \sqrt{n} ?\) (Round your answer to one decimal place.) \(\%\)

(b) What is the confidence level for the interval \(\bar{x} \pm 1.43 \sigma / \sqrt{n} ?\) (Round your answer to one decimal place.) \(\%\)

(c) What value of \(z_{\alpha / 2}\) in the CI formula below results in a confidence level of \(99.7 \% ?\) (Round your answer to two decimal places.)

\(\left(\bar{x}-z_{u / 2} \cdot \frac{\sigma}{\sqrt{n}}, \bar{x}+z_{u / 2} \cdot \frac{\sigma}{\sqrt{n}}\right)\)

\(z_{a / 2}=\)

(d) Answer the question posed in part (c) for a confidence level of \(75 \%\). (Round your answer to two decimal places.) \(z_{u / 2}=\)

Answer 1

a)z(two tailed)=+/-2.81

P(z<2.81)-P(z<-2.81)=0.995 or 99.5%

So,confidence level,c=99.5%

b)z(two tailed)=+/-1.43

P(z<1.43)-P(z<-1.43)=0.847 or 84.7%

So,confidence level,c=84.7%

c)c=0.997

alpha=1-0.997=0.003

alpha/2=0.0015

z=normsinv(0.0015) or normsinv(1-0.0015)=+/-2.97

2.97

d)c=0.75

alpha=1-0.75=0.25

alpha/2=0.125

z=normsinv(0.125) or normsinv(1-0.125)=+/-1.15

1.15