2. REASON QUANTITATIVELY A local television station reports that a sample poll indicated that $$ 46 \% $$ to $$ 58 \% $$ of the registered voters in a city plan to vote for Candidate A. a. What is the sample mean of the population? What does this mean? b. What is the margin of error? What does this mean? c. How many registered voters were in the sample? d. How can the television station reduce the margin of error to be more confident in applying the sample mean to the population of registered voters? der

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a. The sample mean is the midpoint of the interval. We calculate it as  
$$
\text{Mean} = \frac{46\% + 58\%}{2} = \frac{104\%}{2} = 52\%.
$$  
This means that based on the sample, $$52\%$$ of the registered voters plan to vote for Candidate A.

b. The margin of error is half the length of the confidence interval. The length of the interval is  
$$
58\% - 46\% = 12\%.
$$  
Thus, the margin of error is  
$$
\frac{12\%}{2} = 6\%.
$$  
This indicates that the poll estimate can be expected to be within $$ \pm6\% $$ of the true population proportion.

c. To determine the sample size, we use the margin of error formula for estimating a proportion:
$$
E = z \sqrt{\frac{p(1-p)}{n}},
$$
where  
- $$ E = 0.06 $$ (using decimals for percentages),  
- $$ p $$ is the sample proportion (here, $$ p = 0.52 $$ from the sample mean),  
- $$ z $$ is the critical value corresponding to the confidence level. Assuming a 95\% confidence level, $$ z = 1.96 $$.

Solving for $$ n $$:
$$
0.06 = 1.96 \sqrt{\frac{0.52 \times 0.48}{n}}.
$$
Square both sides:
$$
0.06^2 = 1.96^2 \times \frac{0.52 \times 0.48}{n}.
$$
That is,
$$
0.0036 = 3.8416 \times \frac{0.2496}{n}.
$$
Solve for $$ n $$:
$$
n = \frac{3.8416 \times 0.2496}{0.0036}.
$$
First, compute the numerator:
$$
3.8416 \times 0.2496 \approx 0.959.
$$
Then,
$$
n \approx \frac{0.959}{0.0036} \approx 266.4.
$$
Thus, approximately $$ n = 267 $$ registered voters were in the sample.

d. To reduce the margin of error and gain more confidence in applying the sample mean to the population, the television station can increase the sample size. A larger sample decreases the standard error because  
$$
\text{Standard Error} = \sqrt{\frac{p(1-p)}{n}},
$$
so as $$ n $$ increases, the standard error (and thus the margin of error) decreases.
a. The sample mean is 52%, which means 52% of the registered voters plan to vote for Candidate A. b. The margin of error is 6%, indicating the poll estimate is within ±6% of the true population proportion. c. There were approximately 267 registered voters in the sample. d. To reduce the margin of error and be more confident, the television station should increase the sample size.

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