A heath organization wants to estimate the percentage of people who exarcise ragularly, They want to be 92 \%s confident that their estimate is within 2 percentage points of the

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Let
- $$ p $$ be the true proportion of people exercising regularly.
- The desired margin of error be $$ E = 0.02 $$ (i.e. 2 percentage points).
- The confidence level be $$ 92\% $$. This implies a confidence coefficient such that
  $$
  \alpha = 1 - 0.92 = 0.08,
  $$
  and therefore
  $$
  \alpha/2 = 0.04.
  $$
  
The corresponding critical value $$ z_{\alpha/2} $$ is the value of $$ z $$ for which
$$
P(Z \leq z_{\alpha/2}) = 1 - 0.04 = 0.96.
$$
Using a standard normal table or calculator, we find
$$
z_{0.04} \approx 1.75.
$$

Since no prior estimate for $$ p $$ is provided, we use the conservative estimate $$ p = 0.5 $$, which maximizes the product $$ p(1-p) $$.

The sample size formula to estimate a population proportion is:
$$
n = \frac{z_{\alpha/2}^2 \, p(1-p)}{E^2}.
$$

Substitute the values:
$$
n = \frac{(1.75)^2 \cdot (0.5)(0.5)}{(0.02)^2}.
$$

Now, compute step by step:

1. Calculate $$ (1.75)^2 $$:
   $$
   (1.75)^2 = 3.0625.
   $$

2. Compute $$ p(1-p) $$:
   $$
   0.5 \times 0.5 = 0.25.
   $$

3. Multiply these results:
   $$
   3.0625 \times 0.25 = 0.765625.
   $$

4. The square of the margin of error is:
   $$
   (0.02)^2 = 0.0004.
   $$

5. Divide to find $$ n $$:
   $$
   n = \frac{0.765625}{0.0004} \approx 1914.0625.
   $$

Since the sample size must be a whole number and we round up to ensure the desired margin of error, the organization should sample at least
$$
n = 1915 \text{ individuals.}
$$
The health organization should survey at least 1,915 individuals to be 92% confident that their estimate of the percentage of people who exercise regularly is within 2 percentage points of the true proportion.

A heath organization wants to estimate the percentage of people who exarcise ragularly, They want to be 92 %s confident that their estimate is within 2 percentage points of the

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