Question

A sample proportion of 0.44 is found. To determine the margin of error for this statistic, a simulation of 100 trials is run, each with a sample size of 100 and a point estimate of 0.44. The minimum sample proportion from the simulation is 0.32, and the maximum sample proportion from the simulation is 0.50. What is the margin of error of the population proportion using an estimate of the standard deviation?

Answer 1

Answer:

margine of error = $\pm$ 0.13

Step-by-step explanation:

given data

sample proportion = 0.44

simulation trials = 100

sample size = 100

point estimate = 0.44

minimum sample proportion  = 0.32

maximum sample proportion = 0.50

solution

we will get here first z score that is express as

$z = \frac{x-\mu }{\sigma }$  ............1

here x = 0.32and 0.50

$\mu = 0.44\\\sigma = \frac{x(max) - x(min)}{4} \\\sigma = \frac{0.50 - 0.32)}{4} = 0.045$

so z will be

$z1 = \frac{0.50 - 0.44 }{0.045 } = 1.35\\z2 = \frac{0.32 - 0.44 }{0.045 } = -2.66$

so now we get here margin of error that is express as

margin of error = $\pm \ z \times \sqrt{\frac{p(1-p)}{n}}$         ................2

we use here z heighervalue  

margin of error = $\pm \ 2.66 \times \sqrt{\frac{0.44(1-0.44)}{100}}$

margine of error = $\pm$ 0.13

Answer 2

Answer:

±0.06

Step-by-step explanation:

To find the margin of error using the standard deviation method, use the equation $2(\frac{maximum-minimum}{6})$.

In this situation, it would look like this: $2(\frac{0.50-0.32}{6})$. Using this equation, you can find the margin of error by using the standard deviation method.

$2(\frac{0.50-0.32}{6})$

$2(\frac{0.18}{6})$

$2(0.03)$

$0.06$

Hope this helps!

(I know this is right because its what I answered on the test, and got 100%)