Question

Consider a normal population distribution with the value of \(\sigma\) known.

(a) What is the confidence level for the interval \(\bar{x} \pm 2.88 \sigma / \sqrt{n} ?\) (Round your answer to one decimal place.)

\(\%\)

(b) What is the confidence level for the interval \(\bar{x} \pm 1.47 \sigma / \sqrt{n} ?\) (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.)

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

\(z_{\alpha / 2}=\)

(d) Answer the question posed in part (c) for a confidence level of \(78 \%\). (Round your answer to two decimal places.) \(z_{\alpha / 2}=\)

Answer 1

1)

Solution :

(a)

Z_/2 = 2.88

/2 = 0.002

= 0.002 * 2 = 0.004

c = 1 - = 1 - 0.004 = 0.996

Confidence level = 99.6%

(b)

Z_/2 = 1.47

/2 = 0.0708

= 0.0708 * 2 = 0.1416

c = 1 - = 1 - 0.142 = 0.858

Confidence level = 85.8%

(c)

At 99.7% confidence level the z is ,

= 1 - 99.7% = 1 - 0.997 = 0.003

/ 2 = 0.003 / 2 = 0.0015

Z_/2 = Z_0.0015 = 2.97

(d)

At 78% confidence level the z is ,

= 1 - 78% = 1 - 0.78 = 0.22

/ 2 = 0.22 / 2 = 0.11

Z_/2 = Z_0.11 = 1.23