Q. 13 - The Practice of Statistics for AP

Question

 In the United States, there is a strong relationship between education and smoking: well-educated people are less likely to smoke. Does a similar relationship hold in France? To find out, researchers recorded the level of education and smoking status of a random sample of 459 French men aged 20 to 60 years. 11 The two-way table below displays the data.

(a) Is the relationship between smoking status and educational level statistically significant? Give appropriate evidence to support your answer.

(b) Which cell in the table contributes most to the relationship in part (a)? Justify your answer.

Step-by-Step Solution

Answer

a) There is sufficient evidence to support the claim of an association. 

b)  "Heavy"–"University" 

Step 1: Given Information(Part a)

Need to find whether there is sufficient evidence to reject the distributor's claim.

Step 2: Explanation(Part a)

Given counts: 

The expected counts are the product of the column and row total, divided by the sample size of n=107 :

The value of the test statistic is thus:

${\chi }^2=\frac{(56-59.4815{)}^2}{59.4815}+\dots +\frac{(16-24.3399{)}^2}{24.3399}=13.305$

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of Table C containing the t-value in the row $$\begin{gathered} d f=(r-1)(c-1) \\ =(3-1)(4-1) \\ =6 \end{gathered}$$

$0.025<P<0.05$

If the P-value is less than or equal to the significance level, then the null hypothesis is rejected:

$P<0.05\Rightarrow \text{ Reject }{H}_0$

There is sufficient evidence to support the claim of an association.

Step 1: Given Information(Part b)

Need to find Which cell in the table contributes most to the relationship in part (a) 

Step 2: Explanation(Part b)

The expected counts are the product of the column and row total, divided by the sample size of  n=459

The value of the test statistic is thus:

$$\begin{gathered} {\chi }^2=\frac{(56-59.4815{)}^2}{59.4815}+\dots +\frac{(16-24.3399{)}^2}{24.3399} \\ =0.204+0.186+0.044+0.092+1.177+0.700+0.641+1.693+2.704+1.865+1.141+2.858 \\ =13.305 \end{gathered}$$

The highest contribution is$2.858$ which is of the "Heavy"-"University" group.