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How big a survey is needed to be certain that a 95% confidence interval has a margin of error no bigger than 0.01? How does this change if you are asked for an 80% confidence interval? in r studios

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Final answer:

To determine the sample size needed for a 95% confidence interval with a margin of error no bigger than 0.01, we can use a formula and estimate the population standard deviation. Plugging in the values, the sample size needed is 9604. For an 80% confidence interval, the sample size needed is 10240.

Step-by-step explanation:

To determine the sample size needed for a 95% confidence interval with a margin of error no bigger than 0.01, we need to use the formula:

n = (z * σ / E) ^ 2

Where:

n = sample size

z = z-score for the desired confidence level

σ = standard deviation of the population

E = margin of error

As we don't have the population standard deviation, we can use a conservative estimate by setting σ to 0.5 (maximum possible value). For a 95% confidence interval, the z-score is approximately 1.96.

Plugging these values into the formula, we get:

n = (1.96 * 0.5 / 0.01) ^ 2 = 9604

Therefore, a survey of at least 9604 participants is needed to be certain that a 95% confidence interval has a margin of error no bigger than 0.01.

If we are asked for an 80% confidence interval, the z-score changes to approximately 1.28. Plugging the new value into the formula, we get:

n = (1.28 * 0.5 / 0.01) ^ 2 = 10240

Therefore, a survey of at least 10240 participants is needed to be certain that an 80% confidence interval has a margin of error no bigger than 0.01.

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