Problem 4 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

To investigate the hypothesis that a horse's chances of winning an eighthorse race on a circular track are affected by its position in the starting lineup the starting position of each of 144 winners was recorded ([30]). It turned out that 29 of these winners had starting position one (closest to the rail on the inside track). We model the number of winners with starting position one by a random variable \(T\) with a \(\operatorname{Bin}(144, p)\) distribution. We test the hypothesis \(H_{0}: p=1 / 8\) against \(H_{1}: p>1 / 8\) at level \(\alpha=0.01\) with \(T\) as test statistic. a. Argue whether the test procedure involves a right critical value, a left critical value, or both. b. Use the normal approximation to compute the critical value(s) corresponding to \(\alpha=0.01\), determine the critical region, and report your conclusion about the null hypothesis.

Step-by-Step Solution

Answer

Use a right critical value; reject the null hypothesis because 29 > 28.

Step 1: Identify Type of Test

The hypothesis test is set up to determine if there is a significant increase in the probability of winning from position one compared to the expected probability under the null hypothesis. Since the alternative hypothesis is \(H_{1}: p > \frac{1}{8}\), this indicates a one-tailed test focusing on whether the true probability is greater than the null hypothesis value. This means we will use a right-sided critical value.

Step 2: Determine Normal Approximation Parameters

For a binomial distribution \(\operatorname{Bin}(n, p)\), we use the normal approximation when \(n\) is large. The mean \(\mu\) and variance \(\sigma^2\) of \(T\) are given by: \(\mu = np = 144 \times \frac{1}{8} = 18\) and \(\sigma^2 = np(1-p) = 144 \times \frac{1}{8} \times \frac{7}{8} = 15.75\). Therefore, the standard deviation \(\sigma = \sqrt{15.75} \approx 3.97\).

Step 3: Find Z-Score for Critical Value

The significance level \(\alpha = 0.01\) for a right-tailed test corresponds to finding a critical value where the probability to its right is 0.01. The Z-score for \(0.01\) in a standard normal table is approximately \(2.33\).

Step 4: Calculate Critical Value

Convert the Z-score to a binomial critical value using the normal approximation: \(T_{crit} = \mu + Z \cdot \sigma \approx 18 + 2.33 \times 3.97 \approx 27.26\). Since \(T\) must be a whole number, the critical value is 28. This means if the observed value of \(T\) is greater than or equal to 28, we reject \(H_0\).

Step 5: Make Decision Using Statstical Test

We have \(T = 29\) from the data, which is greater than our critical value of 28. Therefore, we reject the null hypothesis \(H_0\) at the 0.01 significance level, suggesting that the position one has a statistically significant advantage.

Key Concepts

Binomial Distribution

In statistics, a binomial distribution is a common way to model the number of successes in a sequence of independent experiments. Each experiment results in a binary outcome: success or failure. Its defining parameters are the number of trials, denoted as \(n\), and the probability of success for each trial, denoted as \(p\).

In the problem of the horses, the binomial distribution models the number of winners (successes) starting from position one. Here, \(n = 144\) represents the total number of races, and \(p = \frac{1}{8}\) is the hypothesized probability that any winner starts from position one.

By using a binomial distribution, we can determine how likely it is to observe a certain number of winners from this position, given pure chance.

• Success: A horse wins from start position one.
• Trials: 144 horse races.
• Probability, \(p = \frac{1}{8}\).

Normal Approximation

When working with a binomial distribution, especially with a large number of trials, it is often approximated by a normal distribution for convenience. This process is known as normal approximation. It is used because normal distributions are mathematically simpler to work with.

For example, the binomial distribution \(\operatorname{Bin}(144, \frac{1}{8})\) can be approximated by a normal distribution with mean \(\mu = np = 18\) and variance \(\sigma^2 = np(1-p) = 15.75\).

• Mean \(\mu = 18\)
• Variance \(\sigma^2 = 15.75\)

The normal distribution helps identify critical values in hypothesis testing by converting the binomial problem into a standard normal distribution problem.

This is particularly useful when determining the probability of observing a certain count of successes.

Critical Value

In hypothesis testing, a critical value is a threshold that determines the decision boundary for rejecting the null hypothesis. This value is derived from the significance level, \(\alpha\), which in this problem is 0.01.

For a right-tailed test, the critical value is the point beyond which the probability of observing the test statistic under the null hypothesis is less than the significance level. In our exercise, this is determined using the Z-score corresponding to a significance level of 0.01, which is approximately \(2.33\).

• Z-score at \(0.01\) level is approximately \(2.33\).
• Critical value for the test statistic \(T_{crit} \approx 27.26\).
• Since we need a whole number, the critical value is 28.

The decision criterion becomes clear: if the observed test statistic is greater than or equal to 28, we reject the null hypothesis.

P-value

The P-value in hypothesis testing provides a measure of the evidence against the null hypothesis. It represents the probability of observing the test statistic, or something more extreme, under the assumption that the null hypothesis is true.

In this case, we look at the P-value to assess the significance of finding 29 winners from starting position one. A low P-value, smaller than our significance level \(\alpha = 0.01\), suggests that such an outcome is unlikely under the null hypothesis.

The P-value calculates:

• The probability that the outcome, or one more extreme, could occur if \(p = \frac{1}{8}\).
• If the P-value < 0.01, we have strong evidence against \(H_0\).

Here, the observed 29 winners leads to a P-value less than 0.01, indicating a significant deviation from the expected distribution, leading to the rejection of the null hypothesis.