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? $$ n=71 $$ (Round up to the nearest integer.) b) Suppose we want to cut the margin of error to $$ 3 \% $$. What is the necessary sample size? $$ n=500 $$ (Round up to the nearest integer.) c) What sample size would produce a margin of error of $$ 4 \% $$. $$ n=\square $$ (Round up to the nearest integer.)

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We use the formula for the sample size for proportions:
$$
n = \frac{z_{\alpha/2}^2\,p\,(1-p)}{E^2}
$$
where

- $$ z_{\alpha/2} $$ is the critical value corresponding to the confidence level;
- $$ p $$ is the estimated proportion (here, $$ p = 0.21 $$);
- $$ E $$ is the desired margin of error.

For a $$ 90\% $$ confidence level, we have:
$$
z_{\alpha/2} \approx 1.645
$$

For part (c) the desired margin of error is:
$$
E = 0.04
$$

Thus, the formula becomes:
$$
n = \frac{(1.645)^2 \times 0.21 \times (1 - 0.21)}{(0.04)^2}
$$

Step 1. Compute $$(1.645)^2$$:
$$
(1.645)^2 \approx 2.706
$$

Step 2. Compute $$ p\,(1-p) $$:
$$
0.21 \times (1 - 0.21) = 0.21 \times 0.79 \approx 0.1659
$$

Step 3. Substitute into the formula:
$$
n = \frac{2.706 \times 0.1659}{0.04^2}
$$
Since:
$$
0.04^2 = 0.0016
$$
we have:
$$
n = \frac{2.706 \times 0.1659}{0.0016}
$$

Step 4. Compute the numerator:
$$
2.706 \times 0.1659 \approx 0.4485
$$

Step 5. Compute $$ n $$:
$$
n = \frac{0.4485}{0.0016} \approx 280.3125
$$

Step 6. Round up to the nearest integer:
$$
n = 281
$$

Thus, the sample size that would produce a margin of error of $$ 4\% $$ is:
$$
n = 281
$$
To estimate the proportion of non-graduates among the 25 to 30 age group with a margin of error of 4% and 90% confidence, you need to survey **281** individuals.

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