Question

construct a confidence interval of the population proportion at the given level of confidence
x = 860, n = 1200, 95% confidence
click here to view the standard normal distribution table (page 1).
click here to view the standard normal distribution table (page 2).
the lower bound of the confidence interval is
(round to three decimal places as needed.)

Explanation:

Step1: Calculate sample proportion $\hat{p}$

The formula for sample proportion is $\hat{p}=\frac{x}{n}$.
Substituting $x = 860$ and $n=1200$, we get $\hat{p}=\frac{860}{1200}\approx0.717$.

Step2: Find $z$-value for 95% confidence level

For a 95% confidence level, the significance level $\alpha=1 - 0.95=0.05$. Then $\frac{\alpha}{2}=0.025$.
Looking up in the standard - normal distribution table, $z_{\frac{\alpha}{2}}=z_{0.025}=1.96$.

Step3: Calculate the margin of error $E$

The formula for the margin of error for a proportion is $E = z_{\frac{\alpha}{2}}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$.
Substitute $\hat{p}=0.717$, $n = 1200$, and $z_{\frac{\alpha}{2}}=1.96$ into the formula:

$$
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$$

Step4: Calculate the lower bound of the confidence interval

The formula for the confidence interval for a proportion is $\hat{p}-EThe lower bound is $\hat{p}-E$.
Substitute $\hat{p}=0.717$ and $E = 0.025$: $0.717-0.025 = 0.692$.

Answer:

$0.692$