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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:

\[ \alpha = 1- CL \]

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

\[\alpha=0.0749 \times 2\]

\[\alpha=0.1498\]

\[c=1- \alpha\]

\[c=0.8502\]

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

5/5 - (9 votes)