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Suppose that prior to conducting a coin-flipping experiment, we suspect that the coin is fair. How many times would we have to flip the coin to obtain a 98% confidence interval of width?

User Alanmars
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Final answer:

To obtain a 98% confidence interval width, you need to calculate the required sample size using the formula n = (Z^2 * p * (1-p)) / E^2. Substitute the values into the formula and round up the result to the nearest whole number.

Step-by-step explanation:

To obtain a 98% confidence interval width, you need to determine the sample size required to estimate the population proportion with a specified margin of error. The formula to calculate the sample size is:

n = (Z^2 * p * (1-p)) / E^2

In this case, since we suspect the coin is fair, we can assume p = 0.5 (50%). Z is the z-score corresponding to the desired confidence level, which is 98%. E is the desired margin of error, which you can choose based on how precise you want the estimate to be. Plug these values into the formula to calculate the required sample size.

For example, if we choose a margin of error of 0.02, the formula would be:

n = (Z^2 * 0.5 * (1-0.5)) / 0.02^2

Once you compute the sample size, round it up to the nearest whole number since you can't have a fraction of a coin flip. So if the calculated sample size is 78.4, for example, you would need to flip the coin 79 times to obtain a 98% confidence interval of width.

User Pflevy
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