Answer:
D) 6328
Step-by-step explanation:
To find the minimum sample size required to estimate the population proportion p with a given margin of error and confidence level when [tex]\hat{p}[/tex] and [tex]\hat{q}[/tex] are unknown, we can use the formula:
[tex]n = \dfrac{z^2 \cdot \hat{p} \cdot \hat{q}}{\mathcal{E}^2}[/tex]
where:
- n is the sample size.
- z is the z-score corresponding to the confidence level.
- [tex]\hat{p}[/tex] is the estimated population proportion.
- [tex]\hat{q} = 1 - \hat{p}[/tex]
- [tex]\mathcal{E}[/tex] is the margin of error.
When the true population proportion p is unknown, it is common practice to use [tex]\hat{p}=0.5[/tex] as the estimated population proportion. This choice maximizes the sample size needed for a given margin of error.
For a 92% confidence level, the z-score is approximately 1.75.
Therefore:
- [tex]z=1.75[/tex]
- [tex]\hat{p}=0.5[/tex]
- [tex]\hat{q}=1-\hat{p}=1-0.5=0.5[/tex]
- [tex]\mathcal{E}= 0.011[/tex]
Substitute the values into the formula and solve for n:
[tex]n = \dfrac{1.75^2 \cdot 0.5 \cdot0.5}{0.011^2}\\\\\\\\n = \dfrac{3.0625 \cdot 0.5 \cdot0.5}{0.000121}\\\\\\\\n = \dfrac{0.765625}{0.000121}\\\\\\\\n=6327.47933884...[/tex]
Since we cannot have a fraction of a person in a sample, we round up to the nearest whole number. So, the minimum sample size required is approximately 6328.
