91Ó°ÊÓ

A civil engineer has been studying the frequency of vehicle accidents on a certain stretch of interstate highway. Longterm history indicates that there has been an average of \(1.72\) accidents per day on this section of the interstate. Let \(r\) be a random variable that represents number of accidents per day. Let \(O\) represent the number of observed accidents per day based on local highway patrol reports. A random sample of 90 days gave the following information. $$ \begin{array}{r|rrrrr} \hline \boldsymbol{r} & 0 & 1 & 2 & 3 & 4 \text { or more } \\ \hline 0 & 22 & 21 & 15 & 17 & 15 \\ \hline \end{array} $$ (a) The civil engineer wants to use a Poisson distribution to represent the probability of \(r\), the number of accidents per day. The Poisson distribution is $$ P(r)=\frac{e^{-\lambda} \lambda^{r}}{r !} $$ where \(\lambda=1.72\) is the average number of accidents per day. Compute \(P(r)\) for \(r=0,1,2,3\), and 4 or more. (b) Compute the expected number of accidents \(E=90 P(r)\) for \(r=0,1,2,3\), and 4 or more. (c) Compute the sample statistic \(\chi^{2}=\Sigma \frac{(O-E)^{2}}{E}\) and the degrees of freedom. (d) Test the statement that the Poisson distribution fits the sample data. Use a \(1 \%\) level of significance.

Step by step solution

01

Calculate Poisson Probabilities

The Poisson distribution is given by \( P(r) = \frac{e^{-\lambda} \lambda^r}{r!} \) where \( \lambda = 1.72 \). Let's compute \( P(r) \) for \( r = 0, 1, 2, 3, 4 \). - For \( r = 0 \): \[ P(0) = \frac{e^{-1.72} \cdot 1.72^0}{0!} = e^{-1.72} \approx 0.1791 \]- For \( r = 1 \): \[ P(1) = \frac{e^{-1.72} \cdot 1.72^1}{1!} = 1.72 \cdot e^{-1.72} \approx 0.3081 \]- For \( r = 2 \): \[ P(2) = \frac{e^{-1.72} \cdot 1.72^2}{2!} = \frac{1.72^2}{2} \cdot e^{-1.72} \approx 0.2650 \]- For \( r = 3 \): \[ P(3) = \frac{e^{-1.72} \cdot 1.72^3}{3!} = \frac{1.72^3}{6} \cdot e^{-1.72} \approx 0.1518 \]- For \( r = 4 \text{ or more} \): \[ P(\geq 4) = 1 - (P(0) + P(1) + P(2) + P(3)) \approx 0.0960 \]

02

Calculate Expected Number of Accidents

Use the probabilities from Step 1 to compute the expected number of days, \( E = 90 \times P(r) \), for \( r = 0, 1, 2, 3, 4 \text{ or more} \).- \( E(0) = 90 \times 0.1791 \approx 16.12 \)- \( E(1) = 90 \times 0.3081 \approx 27.73 \)- \( E(2) = 90 \times 0.2650 \approx 23.85 \)- \( E(3) = 90 \times 0.1518 \approx 13.66 \)- \( E(\geq 4) = 90 \times 0.0960 \approx 8.64 \)

03

Compute Chi-Squared Statistic

To calculate \( \chi^2 = \Sigma \frac{(O - E)^2}{E} \), we use the observed \( O \) and expected \( E \) values from the previous steps.- For \( r = 0 \): \( \frac{(22 - 16.12)^2}{16.12} \approx 2.141 \)- For \( r = 1 \): \( \frac{(21 - 27.73)^2}{27.73} \approx 1.635 \)- For \( r = 2 \): \( \frac{(15 - 23.85)^2}{23.85} \approx 3.289 \)- For \( r = 3 \): \( \frac{(17 - 13.66)^2}{13.66} \approx 0.798 \)- For \( r = 4 \text{ or more} \): \( \frac{(15 - 8.64)^2}{8.64} \approx 4.673 \)Summing all these gives \( \chi^2 \approx 12.536 \).

04

Determine Degrees of Freedom and Test of Significance

The degrees of freedom \( df = \text{number of categories} - 1 = 5 - 1 = 4 \). With \( \alpha = 0.01 \), the critical value for \( \chi^2 \) with 4 degrees of freedom is approximately 13.28.Since our calculated \( \chi^2 = 12.536 \) is less than 13.28, we do not reject the null hypothesis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chi-Squared Test

The Chi-Squared test is a statistical method used to determine if there is a significant difference between the observed frequencies and the expected frequencies under a specific hypothesis. It is particularly useful for categorical data and assessing how well a theoretical distribution fits observed data points.

When applying the Chi-Squared test, follow these main steps:

• Calculate the expected frequencies for each category. This is usually done by applying the theoretical distribution to determine what would be expected under a true hypothesis.
• Determine the observed frequencies, which are typically collected from your sample data.
• Use the Chi-Squared formula: \[\chi^2 = \sum \frac{(O - E)^2}{E}\]Where \( O \) is the observed frequency and \( E \) is the expected frequency.
• Compare the calculated Chi-Squared value with the critical value from the Chi-Squared distribution table.

This test helps in validating the assumption about the distribution of data and is widely used in fields like biology, social sciences, and engineering.

Degrees of Freedom

Degrees of freedom essentially refer to the number of values in a calculation that are free to vary. This is critical in statistics because it influences the shape of the Chi-Squared distribution.

For a Chi-Squared test, the degrees of freedom are determined by:

• Subtracting one from the number of categories.
• In the context of a contingency table, it can also be calculated as: (number of rows - 1) × (number of columns - 1).

In the given exercise, since there are 5 categories of accident frequencies (0, 1, 2, 3, 4 or more), we calculate the degrees of freedom using:

\[df = k - 1 = 5 - 1 = 4\]
This concept helps to understand how spread out the data is and how much variability there is in the data that doesn’t lead to increased constraints.

Significance Level

The significance level, often denoted by \( \alpha \), is a measure used to decide whether to reject the null hypothesis in a hypothesis test. It represents the probability of making a Type I error, which is rejecting a true null hypothesis.

The significance level is usually set at common values like 0.05, 0.01, or 0.10:

• A 0.05 significance level indicates a 5% risk of concluding that a difference exists, even when there is no actual difference.
• A 0.01 significance level, as used in the original exercise, is more stringent, indicating a 1% risk.

In hypothesis testing, the significance level is used to compare against the \( p \)-value or the critical value from the test statistic to make a decision. If the test statistic exceeds the critical value or if the \( p \)-value is less than \( \alpha \), the null hypothesis is rejected.

Choosing an appropriate significance level depends on the field of study and the potential consequences of making errors in testing hypotheses.