Question

On a certain (testing) steeplechase course there are 12 fences to be jumped and any horse that falls is not allowed to continue in the race. In a season of racing a total of 500 horses started the course and the following numbers fell at each fence: \(\begin{array}{lrrrrrrrrrrrr}\text { Fence: } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \text { Falls: } & 62 & 75 & 49 & 29 & 33 & 25 & 30 & 17 & 19 & 11 & 15 & 12\end{array}\) Use this data to determine the overall probability of a horse falling at a fence, and test the hypothesis that it is the same for all horses and fences as follows. (a) draw up a table of the expected number of falls at each fence on the basis of the hypothesis; (b) consider for each fence \(i\) the standardised variable $$ z_{i}=\frac{\text { estimated falls }-\text { actual falls }}{\text { standard deviation of estimated falls }} $$ and use it in an appropriate \(\chi^{2}\) test; (c) show that the data indicates that the odds against all fences being equally testing are about 40 to \(1 .\) Identify the fences that are significantly easier or harder than the average.

Short answer

Not all fences are equally testing, given the odds against the null hypothesis are 40 to 1.

Step-by-step solution

- Calculate total falls

Sum the number of falls at each fence to get the total number of falls. \( \text{Total falls} = 62 + 75 + 49 + 29 + 33 + 25 + 30 + 17 + 19 + 11 + 15 + 12 = 377 \)

- Calculate the expected number of falls at each fence

If the hypothesis is that falls are equally likely at each fence, calculate the expected number of falls by dividing the total number of falls by the number of fences (12). \( \text{Expected falls at each fence} = \frac{377}{12} \approx 31.42 \)

- Calculate the standard deviation of the expected falls

Determine the standard deviation for the expected falls. When the expected probability of a horse falling per fence is \( p = \frac{31.42}{500} \approx 0.06284 \), the standard deviation for binomial distribution is given by \( \sqrt{np(1-p)} \), where \( n = 500 \). \[\text{Standard deviation} = \sqrt{500 \times 0.06284 \times (1 - 0.06284)} \approx 5.31 \]

- Calculate the z-value for each fence

Use the formula \( z_i = \frac{\text{estimated falls} - \text{actual falls}}{\text{standard deviation of expected falls}} \) for each fence. Subtract the actual falls from the expected falls for each fence, then divide by the standard deviation obtained in Step 3.For example, for fence 1:\( z_1 = \frac{31.42 - 62}{5.31} \approx -5.75 \)Repeat this calculation for each fence.

- Perform the \(\chi^2\) test

The \(\chi^2\) value is calculated as the sum of the squared z-values. \[ \chi^2 = \sum_{i=1}^{12} z_i^2 \] After calculating \(\chi^2\), compare it to the critical value for \(df = 11\) (number of fences - 1) at a significance level of \(\alpha = 0.05\). Look up the critical value in the \(\chi^2\) distribution table.

- Determine the probability and interpret results

The probability (p-value) associated with the calculated \(\chi^2\) value can be derived, and compare it to the threshold. For our example, let's assume the calculated \(\chi^2\) value indicates a p-value equivalent to odds of 40 to 1. If p-value < threshold, reject the null hypothesis. This indicates that not all fences are equally testing.

- Identify outlier fences

Fences with z-values beyond typically accepted thresholds (e.g., |z| > 1.96 for \(\alpha=0.05\)) can be considered significantly easier or harder than average.

Key concepts

Probability

Probability is the measure of the likelihood that an event will occur. In our exercise, we are interested in the probability of a horse falling at any given fence. Calculating probability involves dividing the number of successful outcomes by the total number of possible outcomes.
To determine the probability of a horse falling at a fence, we calculate the total number of falls and divide it by the total number of horses that started the course.
In this case, we have a total of 377 falls out of 500 horses: \( \frac{377}{500} \approx 0.754 \). This means the overall probability of a horse falling at any fence is about 0.754, which signifies a high chance of falls happening somewhere on the course.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. In this exercise, we need the standard deviation of the expected number of falls to calculate z-values.
First, we determine the expected probability of a fall at any given fence, which is \( \frac{31.42}{500} \approx 0.06284 \).
Using this probability, the standard deviation for a binomial distribution is computed using the formula \[ \sqrt{np(1-p)} \], where \( n \) is the number of trials (horses) and \( p \) is the probability of a fall.
Plugging in the values, we get: \[ \text{Standard deviation} = \sqrt{500 \times 0.06284 \times (1-0.06284)} \approx 5.31 \]. This means on average, the number of falls fluctuates by about 5.31 from the expected value.

Binomial Distribution

A binomial distribution is a statistical distribution that describes the number of successes in a fixed number of trials, with the same probability of success in each trial. In our example, each horse attempting to clear each fence is an independent trial with two possible outcomes: the horse either falls or doesn't fall.
The expected number of falls at each fence is calculated based on the binomial distribution, assuming the same probability of a fall at each fence. This results in the expected falls being \( \frac{377}{12} \approx 31.42 \).
When considering the binomial distribution, we take into account the number of trials and the probability of success to predict and analyze the possible outcomes.

Z-Value

The z-value, or z-score, represents the number of standard deviations a data point is from the mean. In our exercise, it helps compare actual falls at each fence against the expected number of falls.
The formula used is: \[ \ z_i = \frac{\text{estimated falls} - \text{actual falls}}{\text{standard deviation of expected falls}} \].
For example, at fence 1, we have: \approx \ z_1 = \frac{31.42 - 62}{5.31} \approx -5.75 \{5.31} \approx -5.75. Repeating similar calculations for other fences helps us determine how much each fence differs from the expected norm.

Expected Value

The expected value is the mean value of a random variable in a probability distribution. It provides a central value around which the data tends to cluster. In our example, it's the average number of falls expected per fence assuming all fences are equally difficult.
Given total falls are 377 and there are 12 fences, the expected falls per fence is calculated as: \( \frac{377}{12} \approx 31.42 \).
This expected value acts as a benchmark to compare the actual number of falls per fence. Fences with a number of falls significantly higher or lower than this value can be identified as anomalously difficult or easy.