 # Determine zα for the following of α. (Round your answers to two decimal places.)

-(a) $\alpha = 0.0089$

-(b) $\alpha = 0.09$

-(c) $\alpha = 0.707$

In this question, we have to find the value of $Z_{ \alpha }$ for all the three parts where the value of $\alpha$ is given already.

The basic concept behind this question is the knowledge of Confidence Level, standard normal probability table, and $Z_{\dfrac{\alpha}{2}}$.

In mathematics Confidence Level $CL$ is expressed as:

$c = 1 – \alpha$

where:

$c = Confidence\ Level$

$\alpha$ = no unknown population parameter

$\alpha$ is the area of the normal distribution curve which is $\frac{\alpha }{ 2 }$ for each side and can be expressed mathematically as:

$\alpha = 1- CL$

(a) Given the value of $\alpha$, we have:

$\alpha\ =\ 0.0089$

Now putting the value of given $\alpha$ in the central limit formula:

$c = 1 -\ \alpha$

$c = 1 -\ 0.0089$

$c =\ 0.9911$

In terms of percentage, we have the Confidence Level:

$Confidence\ \space Level = 99.5 \%$

Now to find the value of $Z_{ \alpha }$ we will use the help of an excel sheet and put excel function $normsinv(c)$ to get the value of corresponding $Z- value$

$Z_{ \alpha }= normsinv(c)$

$Z_{ \alpha }= normsinv(0.9911)$

$Z_{ \alpha }= 2.37$

(b) Given the value of $\alpha$ we have:

$\alpha\ =\ 0.09$

Now putting the value of given $\alpha$ in the central limit formula:

$c = 1 -\ \alpha$

$c = 1 -\ 0.09$

$c =\ 0.91$

In terms of percentage, we have the Confidence Level:

$Confidence\ \space Level = 91 \%$

Now to find the value of $Z_{ \alpha }$ we will use the help of an excel sheet and put excel function $normsinv(c)$ to get the value of corresponding $Z- value$:

$Z_{ \alpha }= normsinv(c)$

$Z_{ \alpha }= normsinv(0.91)$

$Z_{ \alpha }= 1.34$

(c) Given the value of $\alpha$ we have:

$\alpha\ =\ 0.707$

Now putting the value of given $\alpha$ in the central limit formula:

$c = 1 -\ \alpha$

$c = 1 -\ 0.707$

$c =\ 0.293$

In terms of percentage, we have the Confidence Level:

$Confidence\ \space Level = 29.3 \%$

Now to find the value of $Z_{ \alpha }$ we will use the help of an excel sheet and put excel function $normsinv(c)$ to get the value of corresponding $Z- value$:

$Z_{ \alpha }= normsinv(c)$

$Z_{ \alpha }= normsinv(0.293)$

$Z_{ \alpha }= -0.545$

## Numerical Results

$Z_{\alpha}= 2.37$

$Z_{\alpha}= 1.34$

$Z_{\alpha}= -0.545$

## Example

Find the confidence level when:

$\frac{\alpha}{2}=0.0749$

Solution

$\alpha=0.0749 \times 2$

$\alpha=0.1498$

$c=1- \alpha$

$c=0.8502$

$Confidence\ \space Level = 85.02 \%$

5/5 - (9 votes)