A survey of 280 homeless persons showed that 63 were veterans. Construct a confidence interval for the proportion of homeless persons who are veterans.
A.
B.
C.
D.

Ask by Mccoy Nguyen. in the United States

Dec 13,2024

Question

A survey of 280 homeless persons showed that 63 were veterans. Construct a confidence interval for the proportion of homeless persons who are veterans.
A.
B.
C.
D.

Ask by Mccoy Nguyen. in the United States

Dec 13,2024

UpStudy AI · Accepted Answer

To construct a $$ 90\% $$ confidence interval for the proportion of homeless persons who are veterans, we can use the formula for the confidence interval for a proportion:

$$
\hat{p} \pm z \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
$$

where:
- $$\hat{p}$$ is the sample proportion,
- $$z$$ is the z-score corresponding to the desired confidence level,
- $$n$$ is the sample size.

### Step 1: Calculate the sample proportion $$\hat{p}$$

The sample proportion $$\hat{p}$$ is calculated as follows:

$$
\hat{p} = \frac{x}{n} = \frac{63}{280}
$$

Calculating this gives:

$$
\hat{p} = \frac{63}{280} \approx 0.225
$$

### Step 2: Find the z-score for a $$90\%$$ confidence level

For a $$90\%$$ confidence level, the z-score (from the standard normal distribution) is approximately $$1.645$$.

### Step 3: Calculate the standard error (SE)

The standard error (SE) of the proportion is calculated using the formula:

$$
SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.225(1 - 0.225)}{280}}
$$

Calculating this gives:

$$
SE = \sqrt{\frac{0.225 \times 0.775}{280}} \approx \sqrt{\frac{0.174375}{280}} \approx \sqrt{0.000622} \approx 0.0249
$$

### Step 4: Calculate the confidence interval

Now we can calculate the confidence interval:

$$
\hat{p} \pm z \cdot SE = 0.225 \pm 1.645 \cdot 0.0249
$$

Calculating the margin of error:

$$
1.645 \cdot 0.0249 \approx 0.0409
$$

Now we can find the confidence interval:

$$
(0.225 - 0.0409, 0.225 + 0.0409) = (0.1841, 0.2659)
$$

### Step 5: Round the interval

Rounding to three decimal places gives us:

$$
(0.184, 0.266)
$$

### Conclusion

The $$90\%$$ confidence interval for the proportion of homeless persons who are veterans is:

$$
\boxed{(0.184, 0.266)}
$$

Thus, the correct answer is **D. $$ (0.184, 0.266) $$**.
The $$90\%$$ confidence interval for the proportion of homeless veterans is $$ (0.184, 0.266) $$.

A survey of 280 homeless persons showed that 63 were veterans. Construct a confidence interval for the proportion of homeless persons who are veterans.
A.
B.
C.
D.

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A survey of 280 homeless persons showed that 63 were veterans. Construct a confidence interval for the proportion of homeless persons who are veterans. A. B. C. D.

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A survey of 280 homeless persons showed that 63 were veterans. Construct a confidence interval for the proportion of homeless persons who are veterans.
A.
B.
C.
D.