The number of automobile accidents per day in a particular city is believed to have a Poisson distribution. A sample of 80 days during the past year gives the following data.
Compute the value of the test statistic. Use α=0.05.
Number of Occurrences /Observed Frequency (Days)
0 34
1 25
2 11
3 7
4 3
To compute the value of the test statistic, we need to perform a goodness-of-fit test using the observed frequencies and compare them to the expected frequencies. The expected frequencies will be calculated based on the assumption that the number of automobile accidents per day follows a Poisson distribution.
First, let's calculate the expected frequencies for each category. The formula for the expected frequency of a Poisson distribution is given by λ^k * e^(-λ) / k!, where λ is the mean of the distribution and k is the number of occurrences.
For this problem, we need to find the mean (λ) of the Poisson distribution. We can calculate it by taking the sum of all the observed frequencies multiplied by their corresponding values, and then dividing it by the total number of days:
λ = (0*34 + 1*25 + 2*11 + 3*7 + 4*3) / 80.
Calculating this expression gives us λ = 0.825.
Now, we can calculate the expected frequencies for each category using the Poisson distribution formula. Here are the expected frequencies:
Expected Frequency (Days)
0: λ^0 * e^(-λ) / 0! = 0.439
1: λ^1 * e^(-λ) / 1! = 0.361
2: λ^2 * e^(-λ) / 2! = 0.149
3: λ^3 * e^(-λ) / 3! = 0.040
4: λ^4 * e^(-λ) / 4! = 0.008
Now, we can compare the observed frequencies to the expected frequencies using a goodness-of-fit test statistic. For this problem, we will use the chi-square test statistic.
The formula for the chi-square test statistic is given by:
χ² = Σ((Observed Frequency - Expected Frequency)² / Expected Frequency)
We need to calculate this expression for each category and then sum them up:
χ² = ((34-0.439)² / 0.439) + ((25-0.361)² / 0.361) + ((11-0.149)² / 0.149) + ((7-0.040)² / 0.040) + ((3-0.008)² /0.008).
Calculating this expression gives us χ² ≈ 43.069.
Finally, we need to compare this test statistic with the critical value of the chi-square distribution at a significance level of α=0.05. Since there are 4 categories, we have (4-1) = 3 degrees of freedom.
Looking up the critical value for a chi-square distribution with 3 degrees of freedom and α=0.05, we find it to be 7.815.
Since our test statistic χ² (43.069) is greater than the critical value (7.815), we reject the null hypothesis at the 0.05 significance level. Therefore, we have evidence to suggest that the number of automobile accidents per day in the city does not follow a Poisson distribution with λ = 0.825.