To determine the 95% confidence interval for the true proportion of people who would support the increase in food tax, we follow these steps:
1. Identify the given values:
- Sample proportion ([tex]\( p \)[/tex]): [tex]\( 0.48 \)[/tex]
- Sample size ([tex]\( n \)[/tex]): [tex]\( 500 \)[/tex]
- Z-score for a 95% confidence interval: [tex]\( 1.96 \)[/tex]
2. Calculate the margin of error (E):
The margin of error formula for a population proportion is given by:
[tex]\[
E = Z \times \sqrt{\frac{p(1-p)}{n}}
\][/tex]
Plugging in the values:
[tex]\[
E = 1.96 \times \sqrt{\frac{0.48 (1 - 0.48)}{500}}
\][/tex]
3. Calculate the margin of error:
[tex]\[
E \approx 1.96 \times 0.0223478667 \approx 0.0437918568
\][/tex]
4. Determine the confidence interval:
The confidence interval is given by:
[tex]\[
\text{CI} = p \pm E
\][/tex]
So, the lower bound of the confidence interval:
[tex]\[
\text{Lower bound} = 0.48 - 0.0437918568 \approx 0.4362
\][/tex]
The upper bound of the confidence interval:
[tex]\[
\text{Upper bound} = 0.48 + 0.0437918568 \approx 0.5238
\][/tex]
Thus, the 95% confidence interval for the true proportion of people who would support the increase in food tax is approximately [tex]\([0.4362, 0.5238]\)[/tex].
Therefore, the correct answer from the given options is:
[tex]\[ 0.48 \pm 1.96 \sqrt{\frac{0.48(1-0.48)}{500}} \][/tex]
