**-(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:

\[ \alpha = 1- CL \]

## Expert Answer

(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**

\[\alpha=0.0749 \times 2\]

\[\alpha=0.1498\]

\[c=1- \alpha\]

\[c=0.8502\]

\[ Confidence\ \space Level = 85.02 \% \]