Question

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 100 Auto Wrecks 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 100 AutoWrecks customers. The second row of numbers gives the frequencies expected for a sample of 100 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 BO (fo-fr)2 (Jo J (Observed frequency - Expected frequency)2 Expected frequency for each of the moving violation categories. Fill in the missing values of Table 1. Then, using the 0.10 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. Then complete Table 2. (fo-fe2 Round your responses for the expected frequencies in Table 1 to at least two decimal places. Round your in table to ease no decina puces. Round your opin ve - responses in Table 1 to at least three E decimal places. Round your responses in Table 2 as specified. Ei send data ed to Excel Table 1: Information about the Sample
Please answer my problem ASAP

Thank you.

Answer 1

THe missing values are Table 1 are as follows

	No Violations	Exactly one violation	Exactly two violations	Three or more Violations	Total
Observed Frequency (fO)	58	13	21	8	100
Expected Frequency (fE)	70	10	15	5	100
(fO - fE)2/fE	2.057	0.900	2.400	1.800	7.157

The type of the test statistic is Chi-square ( $\chi ^{2}$ ) test for goodness of fit

The value of the test statistic is 7.157 $\chi ^{2}=\sum \frac{(f_{O}-f_{E})^{2}}{f_{E}}=7.157$

The critical value of the Chi-Square  ( $\chi ^{2}$ ) test at 0.10 level of significance and (4-1)=3 df is    $\chi ^{2}$ =6.25

Since the test statisti value is greater than the critical value, i.e., 7.157>6.25, we can reject the hypothesis that there is no difference between this year's moving violation distribution and the distribution in a "typical" year at 0.10 significance lvel

The correct answer is YES