CHI-SQUARE TESTS, INFERENCES FOR REGRESSION, AND ANOVAChi-square goodness-of-fit testAutoWrecks, Inc. sells auto insurance. AutoWrecks keeps close tabs on its customers' driving records, updating its rates according to the trends indicated bythese records. AutoWrecks' records indicate that, in a "typical" year, roughly 70% of the company's customers do not commit a moving violation, 10% commitexactly 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 AutoWrecks customers are summarized by the first row of numbers in the table below. That rowgives this year's observed frequency for each moving violation category for the sample of 100 AutoWrecks customers. The second row of numbers gives thefrequencies 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 contains the following value for each of the moving violation categories.(fo-ƒE)²ƒEPart 10/5(Observed frequency - Expected frequency)²Expected frequencyFill in the missing values in the table. Round your responses for the expected frequencies to two or more decimal places. Round yourto three or more decimal places.(Jo-12)²SEresponsesTiff ObservedfrequencyfoExpectedfrequencySE2(fo-fe)²ƒEPart 2Noviolations5500Exactly oneviolation1310.000.900Exactly twoviolations2615.008.067(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 more decimal places.)Three or moreviolations(c) Find the critical value. (Round your answer to two or more decimal places.)06(d) Can we reject the hypothesis that there is no difference between this year's moving violationdistribution and the distribution in a "typical" year?O Yes O No00Answer the following to summarize the test of the hypothesis that there is no difference between this year's moving violation distributidistribution in a "typical" year. Use the 0.05 level of significance for the test.TotalX100

Chi square is the type of parametric test. It used to test whether there is variable is independent or not. It used categorical variable to test its independency. There are three different type of chi square test, they are test for single variance, chi square of attribute and goodness of fit. Part I - Given : Random sample of 100 Wrecks customers are categorized as no violations, exactly one violations, exactly two violations and three or more violations. Where 70% company do not commit moving violations and 3% company commit three or more moving violations. No violations Exactly one violations Exactly two violations Three or more violations Total Observed frequency 55 13 26 6 100 Expected frequency 70 10 15 5 100 Test statistic 55-70270=3.214 0.900 8.067 6-525=0.200 3.214+0.900+8.067+0.200=12.381 Part II - Given : a) Type of test statistic - Chi square test statistic b) Value of test statistic : From the table, χ2=12.381 c) Critical value : Degree of freedom : df=n-1=4-1=3 The t critical value with 4 degree of freedom with 0.05 level of significance is 7.815 Excel code for critical value is CHISQ.INV.RT(0.05,3) d) Here, the chi square test statistic is greater than critical value. Thus, the null hypothesis is rejected. Option B is correct. Incorrect options ; Option A - It is rejected.