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 0.692
(round to three decimal places as needed.)
the upper bound of the confidence interval is
(round to three decimal places as needed.)

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$. Given $x = 860$ and $n=1200$, so $\hat{p}=\frac{860}{1200}\approx0.717$.

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

For a 95% confidence interval, the significance level $\alpha=1 - 0.95=0.05$, and $\alpha/2=0.025$. From the standard - normal distribution table, $z_{\alpha/2}=z_{0.025}=1.96$.

Step3: Calculate margin of error

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

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Step4: Calculate upper bound

The upper bound of the confidence interval is $\hat{p}+E$. Substitute $\hat{p}=0.717$ and $E = 0.025$ into the formula, we get $0.717+0.025 = 0.742$.

Answer:

The upper bound of the confidence interval is $0.742$.