In a population proportion, when the $$ z $$-score $$ (z) $$ increase, the sample size $$ n $$ will increase. A true B false

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Given the formula for determining the required sample size for estimating a population proportion, we have

$$
n = \frac{z^2 \times p(1-p)}{E^2}
$$

where:

- $$ z $$ is the $$ z $$-score corresponding to the desired confidence level,
- $$ p $$ is the estimated population proportion,
- $$ E $$ is the margin of error.

Notice that $$ z $$ appears squared in the numerator. This means that if the $$ z $$-score increases, the value of $$ z^2 $$ increases (often significantly due to the squaring), leading to a larger sample size $$ n $$.

Thus, the statement is:

A true
When the $$ z $$-score increases, the sample size $$ n $$ increases. So, the answer is A true.

In a population proportion, when the -score increase, the sample size will increase.
A true
B false

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In a population proportion, when the -score increase, the sample size will increase. A true B false