Question

This is from my stat class Estimating a Population Proportion If a trial is repeated n times with x successes. In each case use a 95% degree of confidence and find the margin of error E.n=2000, x=300

Answer 1

Given the word problem, we can deduce the following information:

1. We use a 95% degree of confidence.

2. n=2000, x=300

To find the Margin of Error, we use the formula:

`ME=z\cdot\sqrt[]{(p(1-p))/(n)}`

where:

ME= Margin of Error

p=sample proportion

n=sample size

z=z-value

We can get the sample proportion, p by using the formula:

`p=(x)/(n)`

where:

x=number of successes

So,

`\begin{gathered} p=(300)/(2000) \\ \text{Calculate} \\ p=0.15 \end{gathered}`

Since the degree of confidence is 95%, the z-value is 1.96 or z=1.96.

Next, we plug in z=1.96, p=0.15, and n=2000 into the Margin of Error formula:

`\begin{gathered} ME=z\cdot\sqrt[]{(p(1-p))/(n)} \\ =(1.96)(\sqrt[]{(0.15(1-0.15))/(2000)} \\ \text{Calculate} \\ ME=0.0156 \end{gathered}`

Therefore, the Margin of Error is 0.0156.