SubjectIs achieving a basic skill level related to the location of the school? The results of a random sample of students by the location of school and the number of students achieving basic skill levels in three subjects is shown in the contingency table. At a = 0.05, test the hypothesis that the variables are independent. Complete parts(a) through (d).Location Reading Math ScienceUrbanSuburban464640636561(a) Identify the claim and state the null and alternative hypotheses.H: Skill level in a subject islocation.H.: Skill level in a subject isV location.Thev is the claim.(b) Determine the degrees of freedom, find the critical value, and identify the rejection region.What are the degrees of freedom?d.f. =OFind the critical value.(Round to three decimal places as needed.)Choose the correct rejection region below.O A. zx3O B. xổO D. sx3(c) Calculate the test statistic. If convenient, use technology.(Round to three decimal places as needed.)(d) Decide to reject or fail to reject the null hypothesis. Then interpret the decision in the context of the original claim.Choose the correct conclusion below.O A. Fail to reject the null hypothesis. There is not enough evidence to reject the claim that skill level in a subject is independent of location.O B. Reject the null hypothesis. There is not enough evidence to reject the claim that skill level in a subject is independent of location.OC. Fail to reject the null hypothesis. There is enough evidence to reject the claim that skill level in a subject is independent of location.O D. Reject the null hypothesis. There is enough evidence to reject the claim that skill level in a subject is independent of location.

Note: According to the guidelines, only first three subparts have been solved. Given: The contingency table is : Location Reading Math Science Total Urban 46 46 40 132 Suburban 63 65 61 189 Total 109 111 101 321a) The null hypothesis and alternative hypothesis can be framed as: H0: Skill level in a subject is independent of location.Ha: Skill level in a subject is dependent of location.b) The degrees of freedom for the chi-square test is: Number of rows (r) is 2. Number of columns (c) is 3. df=(r-1)(c-1)=(2-1)(3-1)=2 Therefore, the degrees of freedom is 2. At α=0.05 and df=2, the χ2-critical value from χ2 table is: χα,df2=χ0.05,22=5.991 Therefore, χ02=5.991. The Correct option is C, therefore the rejection region is χ2>χ02 c) The expected frequency table is: Location Reading Math Science Total Urban 109×132321=44.82 111×132321=45.64 101×132321=41.53 132 Suburban 109×189321=64.18 111×189321=65.36 101×189321=59.47 189 Total 109 111 101 321 Calculation of test statistic: χ2=∑(observed-exptected)2expected=(46-44.82)244.82+(63-64.18)264.18+(46-45.64)245.64+.....+(61-59.47)259.47=0.153 Therefore, χ2=0.153