Problem 24 - Calculus for Biology and Medicine

Question

How often should you toss a coin to be at least \(90 \%\) certain that your estimate of \(P\) (heads) is within \(0.01\) of its true value?

Step-by-Step Solution

Answer

Toss the coin 6766 times.

Step 1: Understanding the Problem

We need to determine how many times we should toss a coin to be 90% sure that our estimate of the probability of getting heads is within 0.01 of the true probability. We use the concept of statistical confidence intervals to solve this problem.

Step 2: Set Up the Confidence Interval Formula

We use the normal approximation to the binomial distribution. The formula for a confidence interval for a proportion is:\[\hat{P} \pm Z \sqrt{\frac{\hat{P}(1 - \hat{P})}{n}} \]where \(\hat{P}\) is the sample proportion, \(Z\) is the Z-score corresponding to the confidence level, and \(n\) is the number of trials (coin tosses).

Step 3: Determine the Z-score for 90% Confidence

For a 90% confidence interval, the Z-score can be found from a Z-table or standard normal distribution calculator. The Z-score for 90% confidence is approximately 1.645.

Step 4: Set Up the Equation to Solve for n

We want the margin of error to be at most 0.01, so we set up the inequality:\[1.645 \sqrt{\frac{\hat{P}(1 - \hat{P})}{n}} \leq 0.01\]Assuming the worst-case scenario where \(\hat{P}(1 - \hat{P})\) is maximized at 0.25 (when \(\hat{P} = 0.5\)).

Step 5: Solve for n

Insert the worst-case \(\hat{P} = 0.5\) into the equation:\[1.645 \sqrt{\frac{0.5 \times 0.5}{n}} \leq 0.01\]Simplify and solve for \(n\):\[1.645 \times \sqrt{\frac{0.25}{n}} = 0.01\]\[\sqrt{\frac{0.25}{n}} = \frac{0.01}{1.645}\]\[\frac{0.25}{n} = \left(\frac{0.01}{1.645}\right)^2\]\[ = \frac{0.25}{\left(\frac{0.01}{1.645}\right)^2} \approx 6765.64\]

Step 6: Round Up to Ensure the Desired Certainty

Since the number of trials must be a whole number, we round up from 6765.64 to 6766. This ensures we meet the condition of being at least 90% certain.

Key Concepts

Normal Approximation

When dealing with probabilities, sometimes it’s easier to use a normal approximation. This is especially true when working with the binomial distribution. The beauty of normal approximation is that it simplifies calculations with large numbers of trials, such as coin tosses. When the number of trials ( \( n \) ) is large, the binomial distribution can be approximated by the normal distribution. This method helps in making the calculations more manageable.

• Use normal approximation when you have large sample sizes.
• It turns a binomial problem into one that can be solved using normal distribution techniques.
• This usually requires both the expected number of successes and failures to be at least 5, but it’s especially handy in large-scale statistical problems.

In this exercise of coin tossing, we approximate using the normal distribution to simplify finding how many tosses we need for a reliable estimate.

Binomial Distribution

The binomial distribution is fundamental for understanding probability scenarios where there are two possible outcomes. This is seen in cases like coin tosses, where the outcomes are either heads or tails. The binomial distribution appropriately models scenarios where each trial is independent, and there's a constant probability of success.

• Defined by two parameters: the number of trials ( \( n \) ) and the probability of success ( \( P \) ).
• Used extensively in situations where you're repeating an experiment or process several times.
• Each trial's outcome is discrete, either a success or a failure.

In this exercise, we apply the binomial distribution to model the probability of getting heads in numerous coin tosses.

Margin of Error

Margin of error is crucial in confidence interval calculations. It tells us how much we expect our estimate to vary due to chance. It's essentially the range within which we believe the true parameter lies.

• It is determined by the Z-score and the standard error.
• Smaller margin of error gives a more precise estimate.
• You can adjust the sample size to control the margin of error—larger sample sizes usually result in smaller margins.

For the coin toss problem, we are aiming for a margin of error of 0.01, meaning we want the estimated probability of getting heads to be very close to the true probability.

Z-score

The Z-score stands as a cornerstone in statistics, offering a way to measure how far away a data point is from the mean, in terms of standard deviations. For confidence intervals, it helps determine how confident we are in our estimates.

• Derived from the standard normal distribution.
• Each confidence level has a corresponding Z-score; for instance, a 90% confidence level equates to a Z-score of approximately 1.645.
• These scores are fundamental in calculating margins of error for confidence intervals.

In our exercise, the 90% confidence interval uses a Z-score of 1.645, guiding us on how many coin tosses are necessary to remain highly confident in our estimate of heads.