Based on a poll of 1000 residents, a newspaper article claims that 62% of the residents in town favor the development of a recreational park on the west side of town. A community action group interested in preserving the environment claims that 45% of the town’s residents favor the development of a recreational park.

To determine whether the sample supports the population proportion, a simulation of 100 trials is run, each with a sample of 200, using the point estimate of the population. The minimum sample proportion from the simulation is 0.46 and the maximum sample proportion is 0.76.

(a) What is the point estimate of the population?

(b) The margin of error of the population proportion is found using an estimate of the standard deviation. What is the interval estimate of the true population proportion?

(c) The margin of error of the population proportion is found using the half the range.

What is the interval estimate of the true population proportion?

(d) Is the community action group’s claim likely based on either interval estimate of the true population proportion? Explain.

Guest Jun 24, 2021

#1**+3 **

(a) What is the point estimate of the population?

the point estimate will be equal to the population proportion which is $ 0.62 $ just as stated.

(b) The margin of error of the population proportion is found using an estimate of the standard deviation. What is the interval estimate of the true population proportion?

just use the standard error formula;

$z \times \sqrt{\frac{p \times (1-p)}{n}} $

where $p=$point estimate ; $n=$sample size ; $z=1.96$

thus:

$1.96 \times \sqrt{\frac{0.62 \times (0.38)}{200}} $

$ 1.96 \times 0.0343 $

which gives us

$ 0.0672$ which you can write as $ 6.72 \% $

(c) The margin of error of the population proportion is found using half the range.

What is the interval estimate of the true population proportion?

the simulation tells us that the minimum sample percentage is 0.46 and the maximum sample proportion is 0.76

MOE uses half the range(difference) so:

$ \frac{(0.76-0.46) }{2} $

that gives us exactly $0.15$, and that is the answer.

(d) Is the community action group’s claim likely based on either interval estimate of the true population proportion? Explain.

yes thats true, as we can see the confidence interval value is greater than 0.45

UsernameTooShort Jun 24, 2021

#1**+3 **

Best Answer

(a) What is the point estimate of the population?

the point estimate will be equal to the population proportion which is $ 0.62 $ just as stated.

(b) The margin of error of the population proportion is found using an estimate of the standard deviation. What is the interval estimate of the true population proportion?

just use the standard error formula;

$z \times \sqrt{\frac{p \times (1-p)}{n}} $

where $p=$point estimate ; $n=$sample size ; $z=1.96$

thus:

$1.96 \times \sqrt{\frac{0.62 \times (0.38)}{200}} $

$ 1.96 \times 0.0343 $

which gives us

$ 0.0672$ which you can write as $ 6.72 \% $

(c) The margin of error of the population proportion is found using half the range.

What is the interval estimate of the true population proportion?

the simulation tells us that the minimum sample percentage is 0.46 and the maximum sample proportion is 0.76

MOE uses half the range(difference) so:

$ \frac{(0.76-0.46) }{2} $

that gives us exactly $0.15$, and that is the answer.

(d) Is the community action group’s claim likely based on either interval estimate of the true population proportion? Explain.

yes thats true, as we can see the confidence interval value is greater than 0.45

UsernameTooShort Jun 24, 2021