Respuesta :
Answer:
The minimum sample size required is 1068.
Step-by-step explanation:
The complete question is:
A researcher wishes to estimate, with 95% confidence, the population proportion of adults who eat fast food four to six times per week. Her estimate must be accurate within 3% of the population proportion.
No preliminary estimate is available. Find the minimum sample size needed.
Solution:
The (1 - α)% confidence interval for the population proportion is:
[tex]CI=\hat p\pm z_{\alpha/2}\cdot\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
The margin of error for this interval is:
[tex]MOE= z_{\alpha/2}\cdot\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
The information provided is:
[tex]MOE=0.03\\\text{Confidence level}=95\%[/tex]
Assume that the sample proportion is 50%.
The critical value of z is, 1.96.
Compute the minimum sample size required as follows:
[tex]MOE= z_{\alpha/2}\cdot\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
[tex]n=[\frac{z_{\alpha/2}\cdot \sqrt{\hat p(1-\hat p)}}{MOE}]^{2}[/tex]
[tex]=[\frac{1.96\times\sqrt{0.50(1-0.50)}}{0.03}]^{2}\\\\=1067.1111\\\\\approx 1068[/tex]
Thus, the minimum sample size required is 1068.
