Question

this is estimating population proportion in Statistics. 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=500, x=100

Answer 1

Given that:

`\begin{gathered} n=500 \\ x=100 \end{gathered}`

You need to use the following formula in order to calculate the Margin of error E:

`E=Z_{(a)/(2)}\sqrt[]{(pq)/(n)}`

Where "p" is the probability of success, "q" is the probability of failure, "Z" is z-score, and "n" is the sample size.

In this case, since you need to use a 95% degree of confidence, by definition:

`Z_{(a)/(2)}=1.96`

By definition:

`\begin{gathered} p=(x)/(n) \\ \\ q=1-p \end{gathered}`

Substituting the values of "n" and "x" into the first formula and evaluating, you get that:

`p=(100)/(500)=0.2`

Therefore, "q" is:

`\begin{gathered} q=1-0.2 \\ q=0.8 \end{gathered}`

Knowing all those values, you can substitute them into the formula for calculating the Margin of Error:

`E=1.96\sqrt[]{((0.2)(0.8))/(500)}`

Finally, evaluating, you get:

`\begin{gathered} E=1.96\sqrt[]{(0.16)/(500)} \\ \\ E\approx0.0351 \end{gathered}`

Therefore, the answer is:

`E\approx0.0351`