A random sample of 500 likely voters in a city is polled and 285 are found to be Democrats. Find 90, 95 and 99% approximate confidence intervals for the percentage of Democrats in the city.

#### Solution

**90% Confidence Interval for the percentage of Democrats in the city:**

Given that sample size $n = 500$, observed $X = 285$.

Thus the sample proportion is

$\hat{p}=\frac{X}{n}=\frac{285}{500}=0.57$.

#### Step 1 Specify the confidence level $(1-\alpha)$

Confidence level is $1-\alpha = 0.9$. Thus, the level of significance is $\alpha = 0.1$.

#### Step 2 Given information

Given that sample size $n =500$, observed number of successes $X=285$.

The estimate of the proportion of success is $\hat{p} =\frac{X}{n} =\frac{285}{500}=0.57$.

##### Step 3 Specify the formula

$100(1-\alpha)$% confidence interval for population proportion is

`$$ \begin{equation*} \hat{p} - E \leq p \leq \hat{p} + E. \end{equation*} $$`

where `$E=Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$`

and `$Z_{\alpha/2}$`

is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.

##### Step 4 Determine the critical value

The critical value of $Z$ for given level of significance is $Z_{\alpha/2}$.

Thus `$Z_{\alpha/2} = Z_{0.05} = 1.645$`

.

##### Step 5 Compute the margin of error

The margin of error for proportions is

`$$ \begin{eqnarray*} E & = & Z_{\alpha/2} \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\\ & =& 1.645 \sqrt{\frac{0.57*(1-0.57)}{500}}\\ & = &0.036. \end{eqnarray*} $$`

##### Step 6 Determine the confidence interval

$90$% confidence interval estimate for population proportion is

`$$ \begin{eqnarray*} \hat{p} - E & &\leq p \leq \hat{p} + E\\ 0.57 - 0.036 & & \leq p \leq 0.57 + 0.036\\ 0.5336 & & \leq p \leq 0.6064. \end{eqnarray*} $$`

Thus, $90$% confidence interval estimate for population proportion $p$ is $(0.5336,0.6064)$.

**95% Confidence Interval for the percentage of Democrats in the city:**

Confidence level is $1-\alpha = 0.95$. Thus, the level of significance is $\alpha = 0.05$.

$100(1-\alpha)$% confidence interval for population proportion is

`$$ \begin{equation*} \hat{p} - E \leq p \leq \hat{p} + E. \end{equation*} $$`

where `$E=Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$`

and `$Z_{\alpha/2}$`

is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.

The critical value of $Z$ for given level of significance is `$Z_{\alpha/2}$`

.

Thus `$Z_{\alpha/2} = Z_{0.025} = 1.96$`

.

The margin of error for proportions is

`$$ \begin{eqnarray*} E & = & Z_{\alpha/2} \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\\ & = & 1.96 \sqrt{\frac{0.57*(1-0.57)}{500}}\\ & = & 0.043. \end{eqnarray*} $$`

$95$% confidence interval estimate for population proportion is

`$$ \begin{eqnarray*} \hat{p} - E & &\leq p \leq \hat{p} + E\\ 0.57 - 0.043 & & \leq p \leq 0.57 + 0.043\\ 0.5266 & & \leq p \leq 0.6134. \end{eqnarray*} $$`

Thus, $95$% confidence interval estimate for population proportion $p$ is $(0.5266,0.6134)$.

**99% Confidence Interval for the percentage of Democrats in the city:**

Confidence level is $1-\alpha = 0.99$. Thus, the level of significance is $\alpha = 0.01$.

$100(1-\alpha)$% confidence interval for population proportion is

`$$ \begin{equation*} \hat{p} - E \leq p \leq \hat{p} + E. \end{equation*} $$`

where `$E=Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$`

and `$Z_{\alpha/2}$`

is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.

The critical value of $Z$ for given level of significance is `$Z_{\alpha/2}$`

.

Thus `$Z_{\alpha/2} = Z_{0.005} = 2.576$`

.

The margin of error for proportions is

`$$ \begin{eqnarray*} E & = & Z_{\alpha/2} \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\\ & =& 2.576 \sqrt{\frac{0.57*(1-0.57)}{500}}\\ & =& 0.057. \end{eqnarray*} $$`

$99$% confidence interval estimate for population proportion is

`$$ \begin{eqnarray*} \hat{p} - E & &\leq p \leq \hat{p} + E\\ 0.57 - 0.057 & & \leq p \leq 0.57 + 0.057\\ 0.513 & &\leq p \leq 0.627. \end{eqnarray*} $$`

Thus, $99$% confidence interval estimate for population proportion $p$ is $(0.513,0.627)$.