A random sample of 400 U.S. adults finds that (i.e. ) regularly used Facebook as a news source.
Use the findings to construct a confidence interval for the population proportion of adults who
regularly use Facebook as a news source.
Give your answer, rounded to 3 decimal places, in the form of a compound inequality.
G

Question

A random sample of 400 U.S. adults finds that (i.e. ) regularly used Facebook as a news source.
Use the findings to construct a confidence interval for the population proportion of adults who
regularly use Facebook as a news source.
Give your answer, rounded to 3 decimal places, in the form of a compound inequality.
G

UpStudy AI · Accepted Answer

Given:
- Sample proportion: $$ \widehat{p} = 0.75 $$
- Sample size: $$ n = 400 $$
- Confidence level: $$ 95\% $$ (so the critical value is $$ z^* = 1.96 $$)

**Step 1. Compute the Standard Error (SE):**

$$
\text{SE} = \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}} = \sqrt{\frac{0.75 \times 0.25}{400}} = \sqrt{\frac{0.1875}{400}} = \sqrt{0.00046875} \approx 0.02165
$$

**Step 2. Compute the Margin of Error (ME):**

$$
\text{ME} = z^* \times \text{SE} = 1.96 \times 0.02165 \approx 0.04241
$$

**Step 3. Construct the Confidence Interval:**

Lower bound:
$$
\widehat{p} - \text{ME} = 0.75 - 0.04241 \approx 0.70759
$$

Upper bound:
$$
\widehat{p} + \text{ME} = 0.75 + 0.04241 \approx 0.79241
$$

Rounded to three decimal places, the interval is:
$$
0.708 < p < 0.792
$$
$$ 0.708 < p < 0.792 $$

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