Arecent poll of employees working for a national fast-food chain claims that 58% of its employees eat at the fast-food chain. a sample of 200 people shows that only 42% of the employees eat at the fast-food chain. to determine whether this sample supports the population proportion of 0.58, a simulation of 200 trials is run, each with a sample size of 50 and a point estimate of 0.42. the minimum sample proportion from the simulation is 0.31, and the maximum sample proportion from the simulation is 0.53. what is the margin of error of the population proportion using half the range? ±0.07 ±0.11 ±0.13 ±0.21

±0.11

Step-by-step explanation:

I just took the test!

Option B 0.11

Step-by-step explanation:

Given that  the population proportion of 0.58, a simulation of 200 trials is run, each with a sample size of 50 and a point estimate of 0.42. The minimum sample proportion from the simulation is 0.31, and the maximum sample proportion from the simulation is 0.53.

This gives an idea that confidence interval is

(0.31, 0.53) with mean being in the middle of this interval

Margin of error = 1/2 (length of this interval)

Length of confidence interval =

Margin of error =± 1/2(0.22) = ±0.11

Hence option B is correct

y=mx+b

step-by-step explanation:

m slope

b y-intercept