It's believed that as many as $$ 21 \% $$ of adults over 50 never graduated from high school. We wish to see if this percentage is the same among the 25 to 30 age group. a) How many of this younger age group must we survey in order to estimate the proportion of non-grads to within $$ 8 \% $$ with $$ 90 \% $$ confidence? (Round up to the nearest integer.)

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**Step 1. Identify the Appropriate Formula**

To estimate a proportion with confidence, we use the formula

$$
n = \frac{z_{\alpha/2}^2 \, \hat{p}(1-\hat{p})}{E^2}
$$

where
- $$ z_{\alpha/2} $$ is the critical value for the confidence level,
- $$ \hat{p} $$ is the estimated proportion,
- $$ E $$ is the margin of error.

---

**Step 2. Identify the Given Values**

- Confidence level: $$90\%$$ which implies $$\alpha = 0.10$$ and $$\alpha/2 = 0.05$$. The critical value is

$$
  z_{0.05} \approx 1.645.
  $$

- Estimated proportion: $$ \hat{p} = 0.21 $$ (since $$21\%$$ do not graduate).
- Margin of error: $$ E = 0.08 $$.

---

**Step 3. Substitute the Values into the Formula**

Plugging in the values, we have

$$
n = \frac{(1.645)^2 \times 0.21 \times (1 - 0.21)}{(0.08)^2}.
$$

First, compute the product inside the fraction:

$$
\hat{p}(1-\hat{p}) = 0.21 \times 0.79 = 0.1659.
$$

Next, square the critical value:

$$
(1.645)^2 \approx 2.706.
$$

Now, substitute these into the formula:

$$
n = \frac{2.706 \times 0.1659}{0.0064}.
$$

---

**Step 4. Calculate the Numerator and Denominator**

Calculate the numerator:

$$
2.706 \times 0.1659 \approx 0.4487.
$$

The denominator is

$$
0.0064.
$$

Thus,

$$
n \approx \frac{0.4487}{0.0064} \approx 70.11.
$$

---

**Step 5. Round Up to the Nearest Integer**

Since the sample size must be a whole number and to ensure the margin of error is met, we round up:

$$
n = 71.
$$

---

**Final Answer**

We must survey at least $$ \boxed{71} $$ individuals from the younger age group.
We need to survey at least 71 individuals from the younger age group to estimate the proportion of non-graduates within 8% with 90% confidence.

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