Highway Accidents: Poisson Distribution A civil engineer has...

Highway Accidents: Poisson Distribution A civil engineer has been studying the frequency of vehicle accidents on a certain stretch of interstate highway. Long-term 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. (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.

Highway Accidents: Poisson Distribution A civil engineer has been studying the frequency of vehicle accidents on a certain stretch of interstate highway. Long-term 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.
(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.

Key Concepts

Significance Level

The significance level, often denoted by ?, is the threshold probability for rejecting the null hypothesis in a statistical test. It defines the maximum acceptable probability of committing a Type I error, and in hypothesis testing, comparing the p-value or test statistic against the critical value at this set level helps decide whether the observed data significantly deviates from what was expected under the model.

Degrees of Freedom

Degrees of freedom refer to the number of values in a statistical calculation that are free to vary when estimating parameters. In the context of the chi-square goodness-of-fit test, the degrees of freedom are typically calculated as the number of categories minus the number of estimated parameters minus one, which is critical for determining the appropriate chi-square distribution to use for significance testing.

Poisson Distribution

The Poisson distribution is a discrete probability model used to describe the number of events that occur in a fixed interval of time or space when these events happen with a known constant rate and independently of each other. It is characterized by a single parameter, ?, which represents the mean number of occurrences, and its probability mass function is given by P(r) = e^(-?) * (?^r) / (r!).

Expected Frequency

Expected frequency is the theoretical number of occurrences in each outcome category as predicted by a specified probability model. In the context of the Poisson distribution, the expected frequency over a number of intervals (such as days) is calculated by multiplying the probability of each outcome by the total number of intervals, which is essential for comparing theoretical predictions with observed data.

Chi-Square Goodness-of-Fit Test

The chi-square goodness-of-fit test is a statistical method used to assess whether observed frequencies differ significantly from expected frequencies under a particular theoretical distribution. By calculating the statistic ?² = ?((O - E)²/E) across categories, one can determine if the discrepancies between observed and expected values are due to chance or indicate a poor fit of the model.

Transcript

00:01 Solution to number 17 and this is an interesting problem with the poisson distribution and the poisson distribution is a type of discrete distribution where the events are random and rare and this in this case it's a traffic accidents daily traffic accidents and there is an average that's what we use is lambda for so lambda which is the average the mean is 1 .72 accidents per day and we're asked to find the probability that zero accidents occur and then the probability that one occurs, two occur, three occurs, and then greater than four occurs.

00:37 Now you can use the formula, but again i like to use the software.

00:42 So what i'm going to do is i'm going to go to second distribution and then i'm going to go down to the poisson pdf.

00:48 And then here it asks for mu or sometimes it'll ask for lambda and that's 1 .72.

00:53 And then the x value, i'm just going to find the probability that zero occurs.

00:58 And that gives me 017 .17.

01:01 I'm going to go ahead and round here.

01:04 So 0 .1791.

01:08 So you can do this with any type of software or you can just use the formula, although the formula can take a while.

01:14 But 1 .791.

01:16 And i'm going to do it one more time.

01:18 Just show you what it is the second vars for distribution.

01:22 And then i went to the posa pdf.

01:25 That's the probability density function.

01:27 So pdf.

01:28 And the mue, the mean is, that's the lambda 1 .72.

01:31 And this time i'm going to find the distribution or the probability that 1 occurs.

01:35 And it's about, let's say, 0 .308.

01:44 And that's what you're going to do for basically all these, the rest of them until you get to the greater than.

01:50 So 0 .2649.

01:54 I'll go ahead and give you these answers here and then 0 .1519.

02:00 So then to get something that's greater than, what you're going to do is you're going to take one minus the probability of 0, 1, 2, and 3, 1, minus.

02:08 So we want 4 or more, which would be 4, 5, 6, 7, all the way up to infinity.

02:12 Well, we can't go all the way up to infinity and save some time.

02:16 We're just going to take 1 minus the 4 that we've already found, 0, 1, 2, 3.

02:22 So there's another function in the calculator we can use is called the poisson cdf, the cumulative density function, and we're going to go up to 3.

02:30 So the cdf calculates the probability of 0 plus the probability 1 plus probability 2 plus probability of 3 whenever you do cdf of 3.

02:40 So 1 minus that.

02:42 Or you can just do 1 minus these four numbers here, whichever you like.

02:46 So 1 minus and then second distribution.

02:51 And i'm going to go to the placent cdf and the main, remember, was 1 .72.

02:57 And then the x value.

02:58 Now i'm not going to put 0123.

03:00 I'm just going to put the 3.

03:02 And it automatically calculate 0, 1, 2, and 3 combined.

03:06 And whenever you do that, that should be your answer.

03:08 So 0 .0962.

03:14 0 .962...