Discuss the following:
AutoWrecks, Inc. sells auto insurance. AutoWrecks keeps close tabs on its customers' driving records, updating its rates according to the trends indicated by these records. AutoWrecks' records indicate that, in a "typical" year, roughly 70% of the company's customers do not commit a moving violation, 10% commit exactly one moving violation, 15% commit exactly two moving violations, and 5% commit three or more moving violations.
This past year's driving records for a random sample of 120 AutoWrecks customers is summarized in the first row of numbers in Table 1 below. This row gives this year's observed frequencies for each moving violation category for the sample of 120 AutoWrecks customers. The second row of numbers gives the frequencies expected for a sample of 120 AutoWrecks customers if the moving violations distribution for this year is the same as the distribution for a "typical" year. The bottom row of numbers in Table 1 contains the values
(fo-fe)^2 ¸ fo-fe = (observed frequency - expected frequency)^2 ¸ expected frequency
Fill in the missing values of Table. Then, using the 0.05 level of significance, perform a test of the hypothesis that there is no difference between this year's moving violation distribution and the distribution in a "typical" year.
Round responses for the expected frequencies in the following table to at least 2 decimals.
|
|
Number of Violations
|
Exactly 1 violation
|
Exactly 2 violations
|
3 or more violations
|
Total
|
|
fo
|
68
|
18
|
26
|
8
|
120
|
|
fe
|
???
|
12.00
|
???
|
6.00
|
|
(fo-fe)^2 ¸ fo-fe
|
???
|
3.000
|
???
|
0.667
|
|
Summary of the Hypothesis test:
1. Choose one type of test statistic.: Z, t, Chi-square, or F
2. What is the value of the test statistic? Round answer to at least 3 decimals.
3. What is the p-value? Round answer to at least 3 decimals.
4. Can we reject the hypothesis that there is no difference between this year's mving violation distribution and the distribution in a "typical" year? Use the 0.05 level of significance.