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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? Answer is $0.62

(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 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.

Sagot :

The probability computed shows that the point estimate of the population is 6.2%.

How to calculate the probability?

a. The point estimate of the population will be calculated thus:

= 62% = 62/100 = 0.62

b. The margin of error of the population will be:

= z value × [✓0.62 × ✓0.38 / ✓200]

= 1.96 × (0.78 × 0.61 / 14.14)

= 0.0672

Therefore, the interval estimate is 6.72%.

c. The margin of error of the population proportion is found using half the range will be:

= [0.76 - 0.46] /2

= 0.30/2

= 0.15

d.  The community action group’s claim is likely based on either interval estimate of the true population proportion since the confidence interval value is greater than 0.45.

Learn more about probability on:

https://brainly.com/question/24756209

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