Question

In the United States, there has historically been a strong relationship between smoking and education, with well-educated people less likely to smoke. To examine whether this pattern has changed, a sample of 459 men was selected at random from those who had visited a health center for a routine check-up over the course of the past year. Education is classified into three categories corresponding to the highest level of education achieved, and smoking status is classified into four categories.Education Nonsmoker Former Moderate Heavy TotalHigh School 56 54 41 36 187College 37 43 27 32 139Graduate School 53 28 36 16 133Total 146 125 104 84 4591. The proportion of former smokers with a university education is: A) 0.15 B) 0.34 C) 0.22 D) 0.572. The proportion of men with a high school education that are current or former smokers is: A) 0.12 B) 0.30 C) 0.78 D) 0.703. The degrees of freedom for the chi-square test for this two-way table are: A) 2 B) 6 C) 7 D) 12

Answer 1

Answer:

The proportion of former smokers with a university education is (A) 0.15

The proportion of men with a high school education that are current or former smokers is (B) 0.30

The degrees of freedom for the chi-square test for this two-way table are (B) 6

Step-by-step explanation:

The first thing to note is the two way table and ensure the proper arrangement of the figures in the table (Kindly find attached a picture of how the table should look)

Now, on to the first question on the former smokers with a university education = (43+28)/459 = 71/459 = 0.15 which is option A. [This is the total sum of former smokers with college and graduate school education].

The second question on the proportion of men with a high school education that are current or former smokers = (54+31+36)/459 = 0.285 = 0.30 (approximate value) which is option B.

The third question on the degrees of freedom for the chi-square test for this two-way table can be found with the formula DF = (r-1)(c-1) where,

DF = Degree of freedom ,

r = number of rows = 3

c = number of columns = 4 [Kindly note that you have to exempt the row and columns with the totals]

Therefore, DF = (3-1)(4-1) =2*3 = 6 which is option B.