The population proportion is assumed to be 0.16 . If the desired standard error is 0.0127 or less, what is the minimum sample size needed?

Question

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We start with the formula for the standard error (SE) of the sample proportion:

$$
\text{SE} = \sqrt{\frac{p(1-p)}{n}}
$$

Given that $$ p = 0.16 $$ and the desired standard error is at most $$ 0.0127 $$, we set up the inequality:

$$
\sqrt{\frac{0.16(1-0.16)}{n}} \leq 0.0127
$$

First, compute the product $$ p(1-p) $$:

$$
0.16(1-0.16) = 0.16 \times 0.84 = 0.1344
$$

The inequality becomes:

$$
\sqrt{\frac{0.1344}{n}} \leq 0.0127
$$

Square both sides to remove the square root:

$$
\frac{0.1344}{n} \leq (0.0127)^2
$$

Calculate $$ (0.0127)^2 $$:

$$
(0.0127)^2 = 0.00016129
$$

Thus, the inequality is:

$$
\frac{0.1344}{n} \leq 0.00016129
$$

Solve for $$ n $$ by rearranging:

$$
n \geq \frac{0.1344}{0.00016129}
$$

Now, compute the division:

$$
n \geq 833.3
$$

Since $$ n $$ must be a whole number, we round up to the next integer:

$$
n = 834
$$

Thus, the minimum sample size needed is $$ 834 $$.
The minimum sample size needed is 834.

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