Respuesta :
Answer:
a) Sample size = 822
b) Margin of error = 0.03407
Step-by-step explanation:
We are given the following in the question:
p = 74% = 0.74
a) Sample size is required to obtain margin of error of 0.03
Formula:
[tex]\text{Margin of error} = z_{\text{statistic}}\times \sqrt{\dfrac{p(1-p)}{n}}}[/tex]
[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]
Putting values, we get,
[tex]0.03 = 1.96\times \sqrt{\dfrac{0.74(1-0.74)}{n}}\\\\n = (\dfrac{1.96}{0.03})^2(0.74)(1-0.74)\\\\n = 821.24 \approx 822[/tex]
Thus, the sample size must be approximately 822 to obtain a 95% confidence interval with an approximate margin of error of 0.03
b) Margin of error for the 95% confidence interval
p = 54% = 0.54
Formula:
[tex]\text{Margin of error} = z_{\text{statistic}}\times \sqrt{\dfrac{p(1-p)}{n}}}[/tex]
[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]
Putting values, we get,
[tex]\text{Margin of error} = 1.96\times \sqrt{\dfrac{0.54(1-0.54)}{822}}\\\\=0.03407[/tex]
The margin of error now will be 0.03407.
Based on the sampling information given, the sample size will be 822.
Sampling
The margin of error is given as 0.03. Therefore, the sampling size will be:
= (1.96/0.03)² × 0.74 × (1 - 0.74)
= 822
The margin of error for a 95% confidence interval will be:
= 1.96 × ✓0.54 × ✓0.46 × ✓822
= 0.3407
Learn more about samples on:
https://brainly.com/question/17831271
