Answer Happy

1991, a study was conducted to determine the education levels of people who declare bankruptcy. The percentages are shown in the accompanying table Also shown in the table is the observed frequency for these education levels from a random of individuals who fes for bankruptcy in 2017. Use these data to complete parts fough Education level No high school diploma High school diploma Some colege College diploma Advanced degree Total Probability 1991 (5) Observed Frequency 2017 11 37 65 17 5. 135 21.6% 30.2% 35.0% DON 34% 100% Using-0025, perform a chi-square best to determine the probability distribution for the education levels of individus who fee for tankruptcy changed between 1901 and 2017 What is the thesis, ₂ OA. The dution of education levels is 11 mo high school ploma, 37 high school diploma, 65 some colege, 17 coleg doma, and and gree OB The disibution of education levels differs from the claimed or expected distribution OC. The detition of education levels is 21.5% no high school deema, 30.2% high school diploma 35 0% sime cage, % colage diploma, and 30% advanced de OD. The distribution of education levels Solows the normal de

In 1991, a study was conducted to determine the education levels of people who declare bankruptcy. The perc of individuals who files for bankruptcy in 2017. Use these data to complete parts a through c. Probability 1991 (%) 21.6% 30.2% 35.0% 9.8% Education level No high school diploma High school diploma Some college College diploma Advanced degree Total 3.4% 100% Observed Frequency 2017 11 37 65 17 5 135 a. Using a = 0.025, perform a chi-square test to determine if the probability distribution for the education levels of F What is the null hypothesis, Ho? OA. The distribution of education levels is 11 no high school diploma, 37 high school diploma, 65 some college. OB. The distribution of education levels differs from the claimed or expected distribution. OC. The distribution of education levels is 21.6% no high school diploma 30 20/ hinh t hamn 25 nor com

What is the null hypothesis? Calculate the test statistic. Find the expected frequencies. Solve for the test statistics X^2 Determine the p-value and interpret the meaning. What conclusion can be drawn about the motivation behind the use of a credit card by college students between the two surveys?