Question

the school want the team to be formed. out of the 80 surveyed, 8 said they wanted the team to be formed. to test the significance of the survey, a simulation was done assuming 20% of the students wanted the team, each with a sample size of 80, repeated 300 times. create an interval containing the middle 95% of the data based on the data from the simulation, to the nearest hundredth, and state whether the observed proportion is within the margin of error of the simulation results. mean = 0.198 sd = 0.045 proportion of students in favor of team (note: dont use percents for the interval or sample proportion.) answer attempt 2 out of 2 the interval based on the simulation is (0.11, 0.29), so the observed proportion in the sample survey of 0.01 is not within the margin of error of the simulations

Explanation:

Step1: Recall confidence - interval formula for normal distribution

For a normal distribution, the middle 95% of the data lies within $\mu\pm 2\sigma$, where $\mu$ is the mean and $\sigma$ is the standard - deviation.

Step2: Calculate the lower bound of the interval

The mean $\mu = 0.198$ and the standard deviation $\sigma=0.045$. The lower bound $L=\mu - 2\sigma=0.198-2\times0.045=0.198 - 0.09 = 0.108\approx0.11$.

Step3: Calculate the upper bound of the interval

The upper bound $U=\mu + 2\sigma=0.198+2\times0.045=0.198 + 0.09 = 0.288\approx0.29$.

Step4: Calculate the observed proportion

The observed proportion in the sample survey is $\frac{8}{80}=0.1$.

Step5: Check if the observed proportion is within the interval

Since $0.1<0.11$, the observed proportion is not within the margin of error of the simulation results.

Answer:

The interval based on the simulation is $(0.11, 0.29)$, and the observed proportion in the sample survey of $0.1$ is not within the margin of error of the simulation's results.