For a normal population with known variance σ2, what value of zα/2 in the Equation below...

For a normal population with known variance σ², what value of z_α/2 in the Equation below gives a 98% CI?

x-zα/2σn≤μ≤x+zα/2σn

Solutions

Expert Solution

solution:

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table )

Margin of error = E = Z/2 * ( / $\sqrt$ n)

At 98% confidence interval
is,

- E < < + E

- Z/2 * ( / $\sqrt$ n) < < + Z/2 * ( / $\sqrt$ n)

- 2.326 * ( / $\sqrt$ n) < < + 2.326 * ( / $\sqrt$ n)

orchestra answered 1 year ago

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