**AutoWrecks, Inc. Driving Record Analysis**AutoWrecks, Inc. sells auto insurance and closely monitors its customers’ driving records. The company updates its rates based on the trends shown in these records. Historically, in a "typical" year:- 70% of customers have no moving violations.- 10% have exactly one moving violation.- 15% have exactly two moving violations.- 5% have three or more moving violations.**Recent Year Analysis**The driving records for a random sample of 120 customers were analyzed. The table below shows the observed and expected frequencies of violations based on historical data.### Table: Observed vs. Expected Frequencies| Violations | Observed Frequency (\(O\)) | Expected Frequency (\(E\)) | \(\frac{(O - E)^2}{E}\) ||-------------------|----------------------------|----------------------------|-------------------------|| No violations | 71 | 84.00 | 2.167 || Exactly one | 20 | 12.00 | 5.333 || Exactly two | 18 | 18.00 | 0.000 || Three or more | 11 | 6.00 | 4.167 || **Total** | 120 | 120 | |### Statistical Test (Part 2)To determine if there is a significant difference between this year’s moving violation distribution and a “typical” year, a hypothesis test is conducted using a significance level of 0.05. The following steps guide the test:1. **Test Statistic**: Choose the correct type of test statistic (e.g., chi-square test).2. **Value of Test Statistic**: Calculate the test statistic value (round to two or more decimal places).3. **p-value**: Determine the p-value (round to three or more decimal places).4. **Conclusion**: Decide if the null hypothesis, that there is no difference in distributions, can be rejected based on the p-value.This analysis helps AutoWrecks, Inc. adjust its policies based on current customer driving behaviors, ensuring competitive and fair insurance rates.

Solution: The given table of data isPart1Part 2 a. Null hypothesis: H0: There is no difference between this year’s moving violation distribution and the distribution in a “typical” year. Alternative hypothesis: H1: There is a difference between this year’s moving violation distribution and the distribution in a “typical” year. To test the multiple proportions chi-square test for goodness of fit should be used. Thus, the type of test statistic is chi-square.b. Then, the test statistic isc. The degrees of freedom is Then, the p-value isd. Rejection rule: If the P-value is less than or equal to 0.05, then reject the null hypothesis. Conclusion: Here, the p-value is 0.0092. This is less than 0.05. By the rejection rule, reject the null hypothesis. Then, we can reject the hypothesis that there is no difference between this year’s moving violation distribution and the distribution in a “typical” year. Thus, the correct option is “Yes”.

Math

Statistics

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% at the company's customers do nat commit a maving violation, 10% commit Axactly one moving violation, 15% commit axactly twa moving violations, and 5% commit three or mare moving violations. This past ynar's driving records for a random sampla of 120 AutoWrecks customers are summarized by the first row of numbers in the table below. That row gives this year's observed frequency for each moving violation category for the sample of 120 Autowrecks customers. The second row of numbers gives the frequencies expectad 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 contains the following valun for each of the moving violation categories. (observed frequency -Expected frequency) Expected frequency Part 1 Fil in the missing values in the table. Round your responses for the expected frequencies to two or more decimal places. Round your responses to three or more decimal places. Send data to Excel No Exactly one violation Three or more Exactly two Total violations violations violations Observed frequency 71 20 18 11 120 Expected frequency 12.00 6.00 5.333 4.167 Part 2 Answer the following to summarize the test of the hypothesis that there is no difference between this year's moving violation distribution and the distribution in a "typical" year. Use the 0.05 level of significance for then test. (a) Determine the type of test statistic to use. Type of test statistic: (Choose one) (b) Find the value of the test statistic. (Round your answer to two or mare decimal places.) (c) Find the p-valun. (Round your answer to three or more decimal places.) (d) Can we reject the hypothesis that there is no difference between this year's moving violation distribution and the distribution in a "typical" year? OYes ONo O