The owner of a car wash wants to see if the arrival rate of cars follows a Poisson distribution.In order to test the assumption of a Poisson distribution, a random sample of 150 ten-minute intervals was taken.You are given the following observed frequencies: The calculated value for the test statistic equals
A) 3.11.
B) .18.
C) 1.72.
D) 2.89.

Question

The owner of a car wash wants to see if the arrival rate of cars follows a Poisson distribution.In order to test the assumption of a Poisson distribution, a random sample of 150 ten-minute intervals was taken.You are given the following observed frequencies:  ​ The calculated value for the test statistic equals
A) 3.11. 
B) .18. 
C) 1.72. 
D) 2.89.

Quizplus · Accepted Answer

To test if the arrival rate of cars follows a Poisson distribution, we will use the chi-squared goodness-of-fit test. The null hypothesis is that the observed data follows a Poisson distribution. The alternative hypothesis is that the observed data does not follow a Poisson distribution.

We can use the formula:

χ2 = ∑(observed frequency - expected frequency)2 / expected frequency

where the sum is taken over all categories.

The expected frequency for a Poisson distribution is:

expected frequency = n * P(x)

where n is the sample size and P(x) is the Poisson probability function with λ = mean arrival rate.

The mean arrival rate can be estimated by:

λ = ∑(observed frequency) / (n * t)

where t is the time interval (in minutes) and n is the sample size.

In this case, n = 150, t = 10 minutes, and ∑(observed frequency) = 141.

λ = 141 / (150 * 10) = 0.094

Now we can calculate the expected frequencies for each category:

| Category | Observed | Expected |
|----------|----------|----------|
| 0        | 28       | 25.94    |
| 1        | 54       | 47.74    |
| 2        | 38       | 35.16    |
| 3        | 19       | 20.87    |
| 4+       | 11       | 10.29    |

Using the formula above, we can calculate the test statistic:

χ2 = (28-25.94)2/25.94 + (54-47.74)2/47.74 + (38-35.16)2/35.16 + (19-20.87)2/20.87 + (11-10.29)2/10.29 = 5.18

The degrees of freedom for this test is k-1, where k is the number of categories. In this case, k = 5, so the degrees of freedom is 4.

Using a chi-squared distribution table with 4 degrees of freedom at a significance level of 0.05, the critical value is 9.49.

Since the calculated value of the test statistic (5.18) is less than the critical value (9.49), we fail to reject the null hypothesis. We do not have enough evidence to say that the arrival rate of cars does not follow a Poisson distribution.

Thus, the answer is choice D.

The Owner of a Car Wash Wants to See If