11.Determine a conclusion. Choose the correct answer below. (1 point) a) Reject Ho. There is insufficient evidence to conclude that the two variables are not independent b) Reject Ho. There is sufficient evidence to conclude that the two variables are not independent. c) Do not reject Ho. There is insufficient evidence to conclude that the two variables are not independent d) Do not reject Ho. There is sufficient evidence to conclude that the two variables are not independent.2. Write the appropriate null and alternative hypothesis (5 points): 3. How many different populations (groups)(2 points)4. Write the degree of freedom for the denominator (1 point): Write degree of freedom for denominator (1 point): 5. Write the decision rule if Fc=F2,27=3.35 at alpha=0.05 6. Decision and Conclusion7-8 A contingency analysis table has been constructed from data obtained in a survey of high school students in which respondents were asked to indicate whether they had ever smoked cigarettes or if they had ever smoked marijuana. Use the contingency table shown to answer questions bellow: Marijuana Cigarettes Yes No Yes 900 573 No 47 747 7.Determine the null hypothesis (Ho) and the alternative hypothesis (Ha). Choose the correct answer below.(1 points) a) Ho: Cigarette use and marijuana use are independent. Ha: Cigarette use and marijuana use are not independent. b) .Ho: Cigarette use and marijuana use are not independent. Ha: Cigarette use and marijuana use are independent. c) Ho: Cigarette use and marijuana use have equal proportions. Ha: Cigarette use and marijuana use have non-equal proportions. d) Ho: Cigarette use and marijuana use have non-equal proportions. G Ha: Cigarette use and marijuana use have equal proportions.8. Calculate the chi-square test statistic. (5 points) 9.Determine the chi-square critical value for a=0.05. (2 points) 10. Write the Decision rule: (2 points)1.Complete the one way ANOVA table below: (5 points) Source of Sum of Degrees of Mean Square F.ratio Variation squares freedom Errors Between Group 2 Within Group 1,206 Total 1612 29 SSB= MSB= MSW= F=