Question

question 10
2 pts
a total of 80,000 people reside in the city of riverbank. a random sample of 250 riverbank residents are surveyed about whether they support the building of a new shopping mall, and, of those who are surveyed, 32% are in favor of plans to build the shopping mall. based on what you have learned so far in stat 1350, and specifically from the coverage presented in chapter 3, if a 95% confidence interval is to be constructed based on this data, what margin of error would we report?
1% or lower
10% or higher
5.8%
6.3%
5.0%

Explanation:

Step1: Identify the formula for margin of error

For a proportion in a large - sample situation, the formula for the margin of error $E$ of a 95% confidence interval is $E = z\sqrt{\frac{p(1 - p)}{n}}$, where $z$ is the z - value for a 95% confidence interval, $p$ is the sample proportion, and $n$ is the sample size. The $z$ - value for a 95% confidence interval is $z=1.96$, $p = 0.32$ (32% expressed as a decimal), and $n = 250$.

Step2: Calculate $(1 - p)$

$1-p=1 - 0.32=0.68$

Step3: Calculate $\frac{p(1 - p)}{n}$

$\frac{p(1 - p)}{n}=\frac{0.32\times0.68}{250}=\frac{0.2176}{250}=0.0008704$

Step4: Calculate $\sqrt{\frac{p(1 - p)}{n}}$

$\sqrt{\frac{p(1 - p)}{n}}=\sqrt{0.0008704}\approx0.0295$

Step5: Calculate the margin of error $E$

$E = 1.96\times0.0295\approx0.058$ or 5.8%

Answer:

5.8%