How do you find the margin of error for a poll, assuming that $95\%$ confidence and $\text{pi}$ = $0.4$ for $n$ =$180$ ?

Hint: To solve the question we need to know the steps to find the margin error. The first step is to find the Significance level which is used to measure the strength of the sample. The second step is to find the critical z- value which is evaluated with different values of confidence given. The third step is to find the standard error and ultimately finding the margin of error.

Complete step-by-step answer:
The question asks us to find the margin error for a poll which has $95\%$ confidence and $\text{pi = \hat{p}}$ = $0.4$ for when $n$ =$180$ is given. Margin of error is defined as the product of confidence coefficient and standard error of p. In this question the value of $\text{pi}$ and $n$ is given which are $0.4$ and $180$ respectively. To find the margin of error the first step is to find the significance level which is the difference between $1$ and confidence. So mathematically it will be represented as:
Significance level = $\alpha $ = $\text{1- confidence}$
The value of the confidence given in the question is $95\%$ . We need to convert the percentage into decimal form which could be done by dividing the number by $100$.
$= \dfrac{95}{100}=0.95$
On putting the decimal value we get:
$= \text{1- 0}\text{.95}$
$= 0.05$
The second step is to find the critical z- value = ${{z}_{\dfrac{\alpha }{2}}}$
$= {{z}_{\dfrac{0.05}{2}}}$
$= {{z}_{0.025}}$
$= \text{1}\text{.96 (From z table)}$
Standard error of $\hat{p}$ : SE = $\sqrt{\dfrac{\hat{p}\times \left( 1-\hat{p} \right)}{n}}$
On putting the values given in the question we get
$= \sqrt{\dfrac{0.4\times \left( 1-0.4 \right)}{180}}$
On subtracting $0.4$ from $1$ we get $0.6$.
$= \sqrt{\dfrac{0.4\times 0.6}{180}}$
Multiplying the values in the numerator together we get:
$= \sqrt{\dfrac{0.24}{180}}$
On dividing the numerator from denominator we get:
$= \sqrt{0.00133}$
Now we need to find the square root of $0.00133$, on finding the value we get:
$= 0.03651483$
On rounding off upto 3 decimal place we get:
$= 0.037$
Margin of Error = ${{z}_{\dfrac{\alpha }{2}}}\times \sqrt{\dfrac{\hat{p}\times \left( 1-\hat{p} \right)}{n}}$
On substituting the value we get:
$= 1.96\times 0.037$
On multiplying the two value the product we get is:
$= 0.0725$
$\therefore $ The margin of error for a poll, assuming that $95\%$ confidence and $\text{pi}$ = $0.4$ for $n$ =$180$ is $0.0725$

Note: Margin of error is the degree of error which is received from random sampling surveys. Lower margin of error indicates the higher confidence levels in the produced results. The values of fractions in decimal form may come big, so we need to round it off upto 3- decimal place. We find the critical z- value we take the help of z table.