Question

A recent community survey of 100 citizens showed that 55% of respondents approved of water rationing. The data was simulated 500 times. In those simulations, the lowest sample was p^ =45 and the highest sample was p^ =66. Use the standard deviation method for determining the error margin with 95% confidence.

A. What is the error margin?

B. What is the interval estimate for the true population proportion?

C. From this estimate, can you say for certain that the majority of the community supports water rationing?

Answer 1

The interval estimate for the true population proportion is 45.25% to 64.75% and we can conclude that the majority of the community supports water rationing

How to determine the error margin?

The given parameters are:

• p = 55% = 0.55
• n = 100
• Confidence level = 95%

The error margin is calculated using:

$E = z * \sqrt{\frac{p* (1 - p)}{n}}$

Where z is the critical value.

At 95% confidence level, the critical value is:

z = 1.96

So, we have:

$E = 1.96 * \sqrt{\frac{0.55 * (1 - 0.55)}{100}}$

Evaluate

$E = 1.96 * \sqrt{0.002475}$

Evaluate

E = 0.0975

Hence, the error margin is 0.0975

How to determine the interval estimate for the true population proportion?

This is calculated using:

I = p ± E

So, we have:

I = 0.55 ± 0.0975

Expand

I = (0.55 - 0.0975, 0.55 + 0.0975)

Evaluate

I = (0.4525, 0.6475)

Express as percentage

I = (45.25%, 64.75%)

Hence, the interval estimate for the true population proportion is 45.25% to 64.75%

How to interpret the estimate?

From the simulation,

• Lowest sample p = 45%
• Highest sample p =66%

The values are approximately the interval estimates.

i.e.

45.25% ≈ 45% and 64.75% ≈ 65%

Hence, we can conclude that the majority of the community supports water rationing

Read more about confidence intervals at:

https://brainly.com/question/15712887

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