To be able to use a normal distribution to find the margin of error for a data set with a sample proportion [tex]\(\hat{p}\)[/tex] and a sample size [tex]\(n\)[/tex], we look at the conditions required to approximate a binomial distribution by a normal distribution. This is particularly useful when we want to apply the Central Limit Theorem and use the properties of the normal distribution for inferential statistics.
The important criteria to check for using a normal approximation in this context are:
1. The sample size [tex]\(n\)[/tex] should be large enough.
2. Both [tex]\(n\hat{p}\)[/tex] and [tex]\(n(1-\hat{p})\)[/tex] should be sufficiently large to ensure that the sampling distribution of the sample proportion [tex]\(\hat{p}\)[/tex] is approximately normal.
Specifically, the rule of thumb states that both:
[tex]\[ n\hat{p} \geq 10 \][/tex]
and
[tex]\[ n(1-\hat{p}) \geq 10 \][/tex]
should hold true.
Among the options provided:
A. [tex]\(n \hat{p} \leq 10\)[/tex] – This indicates a condition that would typically not meet the criteria and therefore is not correct.
B. [tex]\(n \hat{p} \geq 30\)[/tex] – This is a stricter criterion than necessary and not the standard rule of thumb.
C. [tex]\(n(1-\hat{p}) \geq n \hat{p}\)[/tex] – This essentially requires [tex]\( (1-\hat{p}) \geq \hat{p} \)[/tex], which is not directly related to the criteria needed.
D. [tex]\(n(1-\hat{p}) \geq 10\)[/tex] – This aligns perfectly with the necessary conditions alongside [tex]\(n \hat{p} \geq 10\)[/tex].
Hence, the correct answer, which ensures that we can use a normal distribution to find the margin of error for the sample proportion, is:
[tex]\[ \boxed{D} \][/tex]
where [tex]\(n(1-\hat{p}) \geq 10\)[/tex].
