Problem 5 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

24.5 Suppose we want to make a \(95 \%\) confidence interval for the probability of getting heads with a Dutch 1 Euro coin, and it should be at most \(0.01\) wide. To determine the required sample size, we note that the probability of getting heads is about \(0.5\). Furthermore, if \(X\) has a \(\operatorname{Bin}(n, p)\) distribution, with \(n\) large and \(p \approx 0.5\), then \(\frac{X-n p}{\sqrt{n / 4}}\) is approximately standard normal. a. Use this statement to derive that the width of the \(95 \%\) confidence interval for \(p\) is approximately $$ \frac{z_{0.025}}{\sqrt{n}} $$ Use this width to determine how large \(n\) should be. b. The coin is thrown the number of times just computed, resulting in 19477 times heads. Construct the \(95 \%\) confidence interval and check whether the required accuracy is attained.

Step-by-Step Solution

Answer

Required sample size is 38416; interval: [0.5052, 0.5084]; accuracy is attained.

Step 1: Understand the Confidence Interval Formula

The general formula for a confidence interval for a proportion is given by \( \hat{p} \pm z_{\alpha/2} \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \). In this problem, you have \( p \approx 0.5 \), meaning \( \hat{p} \approx 0.5 \). With the large sample size condition, \( \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \approx \frac{1}{2\sqrt{n}} \).

Step 2: Derive the Expression for Interval Width

The width of the confidence interval is the distance between the upper and lower bounds, or \( 2 \times \frac{z_{\alpha/2}}{2\sqrt{n}} = \frac{z_{\alpha/2}}{\sqrt{n}} \). With \(95\%\) confidence, \( z_{0.025} \approx 1.96 \).

Step 3: Find Required Sample Size

Given the maximum allowable width of the interval is \(0.01\), set \( \frac{1.96}{\sqrt{n}} = 0.01 \). Solve for \( n \):\[1.96 = 0.01 \sqrt{n} \implies \sqrt{n} = \frac{1.96}{0.01} = 196 \implies n = 196^2 = 38416.\]

Step 4: Construct Confidence Interval with Given Sample

Given 19477 heads in \( n = 38416 \) trials, the proportion \( \hat{p} \) is \( \frac{19477}{38416} \approx 0.5068 \). Use the interval formula: \( \hat{p} \pm 1.96 \cdot \sqrt{\frac{0.5068 \cdot (1-0.5068)}{38416}} \). Compute the margin of error: \( \approx 1.96 \cdot 0.000805 = 0.0016 \).

Step 5: Calculate the Confidence Interval

The confidence interval is \( 0.5068 \pm 0.0016 \), which gives the interval: \( [0.5052, 0.5084] \).

Step 6: Verify Required Accuracy

The calculated interval width is \( 0.5084 - 0.5052 = 0.0032 \), which is much smaller than the required \( 0.01 \) width, thus the required accuracy is achieved.

Key Concepts

Binomial Distribution

Binomial distribution is a foundational concept in statistics, often used to model the number of successes in a series of independent trials. In the context of our problem, the binomial distribution is used to describe the number of heads obtained when flipping a coin multiple times. Here,

• Each coin flip represents a single trial.
• Getting a head is considered a success, and its probability is denoted by \( p \).
• We assume \( p \approx 0.5 \), as a fair coin has an equal chance of landing heads or tails.

Given many trials, or a large sample size \( n \), the binomially distributed random variable \( X \) can be approximated by a normal distribution. This is based on the Central Limit Theorem, which states that as \( n \) becomes large, the distribution of \( X \) approaches a normal distribution. In our exercise, this is utilized to approximate the distribution of heads as being normal, which simplifies calculations for the confidence interval.

Sample Size Calculation

Determining the sample size is crucial to ensure that the confidence interval of a proportion achieves a desired precision. In our example with the Euro coin, we want the confidence interval width to be at most 0.01. The formula that relates sample size to the confidence interval width is:\[\frac{z_{0.025}}{\sqrt{n}}\]where \( z_{0.025} \approx 1.96 \) for a 95% confidence interval. We rearrange this formula to solve for \( n \):

• The width \( W \) of the confidence interval must be \( \leq 0.01 \).
• Set the expression \( \frac{1.96}{\sqrt{n}} = 0.01 \).
• Solve for \( n \) to find \( n = 38416 \).

By ensuring \( n \) is at least 38416 flips, we confirm that our confidence interval will meet the required precision.

Standard Normal Distribution

The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1, denoted by \( N(0,1) \). It is a special case of the normal distribution and plays a key role in statistics, especially in calculating probabilities and critical values that affect confidence intervals and hypothesis testing.

• In confidence interval calculations, the standard normal distribution helps determine the critical value \( z_{\alpha/2} \).
• For a 95% confidence level, \( z_{0.025} = 1.96 \).
• This critical value is used to determine the margin of error for the confidence interval.
• The approximation \( \frac{X - np}{\sqrt{np(1-p)}} \approx N(0,1) \) is used in large sample sizes to estimate \( X \) from a binomial distribution.

In this exercise, the approximate normality of the binomial distribution underpins our use of the standard normal distribution to calculate the sample size required for a precise confidence interval for the coin-tossing experiment.