**For the given interval $\bar{x}\ \pm\ 2.81\left(\dfrac {\sigma}{\sqrt n}\right)$ find the confidence level?****For the given interval $\bar{x}\ \pm\ 1.44\left(\dfrac {\sigma}{\sqrt n}\right)$ find the confidence level?**

The aim of the question is to find the **Confidence Level** of given equations.

The basic concept behind this question is **Confidence Level** CL, which can be expressed as:

\[ c = 1 – \alpha \]

Here:

$c = Confidence\ Level$

$\alpha$ = no unknown population parameter

$\alpha$ is the area of the **normal distribution curve** which is divided into equal parts that is $\frac{\alpha}{2}$ for each side. It can be written as:

\[ \alpha = 1- CL \]

$z-score$ is the required **Confidence Level** that we select and can be calculated from the **standard normal probability** table. It is located at the right of $\dfrac{\alpha}{2}$ and it is expressed as $Z_{\dfrac{\alpha}{2}}$.

Such as when:

\[Confidence\ Level= 0.95\]

\[\alpha=0.05\]

\[\frac{\alpha}{2}=0.025\]

Which represents that $0.025$ is on the right side of $Z_{0.025}$

Then we can write it as follows:

\[ Z_{\dfrac{\alpha}{2}}=Z_{0.025}\]

and to the left of $Z_{0.025}$ we have:

\[=1-\ 0.025\]

\[=0.975\]

Now by using the **standard normal probability** table we will get the value of $ Z_{\dfrac{\alpha}{2}}=Z_{0.025}$:

\[ Z_{\dfrac{\alpha}{2}}=Z_{0.025}= 01.96\]

For the **confidence interval** we have the following formula:

\[\bar{X}\ -\ EBM\ ,\ \bar{X}\ +EBM\]

Or it can also be written as:

\[\bar{X}\ -\ Z_\alpha\left(\dfrac{\sigma}{\sqrt n}\right)\ \le\ \mu\ \le\ \bar{X}\ +\ Z_\alpha\left(\dfrac{\sigma}{\sqrt n}\right)\ \]

## Expert Answer

From the given formula $\bar{x}\ \pm\ 2.81\left(\dfrac {\sigma}{\sqrt n}\right)$ we have the value of $Z_{\dfrac{\alpha }{2}}$:

\[Z_{\dfrac{\alpha}{ 2}}\\ =\ 2.81 \]

Now by using the **standard normal probability table,** we will get the value of $ Z_{\frac{\alpha}{2}}$:

\[\frac{\alpha}{2}=\ 0.0025\]

\[\alpha\ =\ 0.002\ \times\ 2\]

\[\alpha\ =\ 0.005\]

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

\[c=1-\ \alpha\]

\[c=1-\ 0.005\]

\[c=\ 0.995\]

In terms of percentage, we have the **Confidence Level**:

\[Confidence\ Level=99.5 \% \]

Now for this part from the given formula $\bar{x}\ \pm\ 1.44\left(\dfrac {\sigma}{\sqrt n}\right)$ we have the value of $Z_{\dfrac{\alpha}{2}}$:

\[Z_{\dfrac{\alpha}{2}}=\ 1.44\]

Now by using the **standard normal probability table,** we will get the value of $ Z_{\dfrac{\alpha}{2}}$:

\[\frac{\alpha}{2}=\ 0.0749\]

\[\alpha\ =\ 0.0749\ \times\ 2\]

\[\alpha\ =\ 0.1498\]

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

\[c=1-\ \alpha\ \]

\[c=1-\ 0.1498\]

\[c=\ 0.8502\]

In terms of percentage, we have the **Confidence Level**:

\[ Confidence\ Level=85.02 \%\]

## Numerical Results

For the given interval $\bar{x}\ \pm\ 2.81\left(\dfrac {\sigma}{\sqrt n}\right)$ the **confidence level**:

\[Confidence\ Level=99.5 \% \]

For the given interval $\bar{x}\ \pm\ 1.44\left(\dfrac {\sigma}{\sqrt n}\right)$ the **confidence level** is:

\[ Confidence\ Level=85.02 \% \]

## Example

For the given interval $\bar{x}\ \pm\ 1.645 \left(\dfrac {\sigma}{\sqrt n} \right)$, find the **confidence level.**

**Solution**

\[Z_{\frac {\alpha} { 2}}=\ 1.645\]

Now by using the **standard normal probability table,** we will get the value of $ Z_{\dfrac{\alpha}{2}}$:

\[\ \frac{\alpha}{2}=\ 0.05\]

\[\alpha\ =\ 0.1\]

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

\[c=1-\ \alpha\ \]

\[c=1-\ 0.1\]

\[c=\ 0.9\]

In terms of percentage, we have the **Confidence Level**:

\[ Confidence\ Level=90 \% \]