Question

6 multiple choice 4 points construct the requested confidence interval. a survey of 280 homeless persons showed that 63 were veterans. construct a 90% confidence interval for the proportion of homeless people who are veterans. (0.161, 0.289) (0.184, 0.266) (0.176, 0.274) (0.167, 0.283)

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 63$ (number of successes) and $n=280$ (sample size).
$\hat{p}=\frac{63}{280}=0.225$

Step2: Find $z -$ value

For a $90\%$ confidence interval, the significance level $\alpha=1 - 0.90=0.10$, and $\alpha/2=0.05$. The $z -$ value $z_{\alpha/2}=z_{0.05}\approx1.645$ (from standard normal distribution table).

Step3: Calculate margin of error

The margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$.
Substitute $\hat{p}=0.225$, $n = 280$, and $z_{\alpha/2}=1.645$ into the formula:

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Step4: Construct confidence interval

The confidence interval is $\hat{p}-ESubstitute $\hat{p}=0.225$ and $E = 0.041$:
$0.225-0.041 < p<0.225 + 0.041$
$0.184 < p<0.266$

Answer:

(0.184, 0.266)