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In a small town of 5,832 people, the mayor wants to determine the proportion of voters who would support an increase in food tax. Assuming the conditions for inference have been met, what is the 95% confidence interval for the true proportion of people who would support the increase in food tax?

A. [tex]\(0.48 \pm 1.65 \sqrt{\frac{0.48(1-0.48)}{500}}\)[/tex]

B. [tex]\(0.48 \pm 1.96 \sqrt{\frac{0.48(1-0.48)}{500}}\)[/tex]

C. [tex]\(0.52 \pm 1.65 \sqrt{\frac{0.52(1-0.52)}{500}}\)[/tex]

D. [tex]\(0.52 \pm 1.96 \sqrt{\frac{0.52(1-0.52)}{500}}\)[/tex]



Answer :

GinnyAnswer
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]

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