- 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:
\[ Confidence\ Level=90 \% \]