Q.49 - The Practice of Statistics for AP

Question

Regulating guns The National Gun Policy Survey asked a random sample of adults, “Do you think there should be a law that would ban possession of handguns except for the police and other authorized persons?” Here are the responses, broken down by the respondent’s level of education: 

(a) How do opinions about banning handgun ownership seem to be related to the level of education? Make an appropriate graph to display this relationship. Describe what you see. 

(b) Determine whether or not the sample provides convincing evidence that education level and opinion about a handgun ban are independent in the adult population 

Step-by-Step Solution

Answer

a). The people in each category are not supporting the law and this proportion seems to be highest in the group of "Some college" .

b). There is sufficient evidence to conclude that there is not an association between two variables. 

Part (a) Step 1: Given Information

Part (a) Step 2: Explanation

To construct the graph, the proportions are computed as:

Part (a) Step 3: Explanation

The graph could be constructed:

From the above figure, it is clear that most of the people in each category are not supporting the law and this proportion seems to be highest in the group of "Some college".

Part (b) Step 1: Given Information

Part (b) Step 2: Explanation

The appropriate null and alternative hypotheses for the provided test could be written as:

$H_0$ : Two variables are independent

$H_a$ : Two variables are dependent

The expected count is computed as:

Part (b) Step 3: Explanation

The test statistic is:

$$\begin{gathered} {\chi }^2=\sum \frac{(O-E{)}^2}{E} \\ =\frac{(58-46.9409{)}^2}{46.9409}+\frac{(58-69.0591{)}^2}{69.0591}+\dots \dots .+\frac{(99-104.7794{)}^2}{104.7794} \\ =8.525 \end{gathered}$$

The degree of freedom is calculated as:

 $$\begin{gathered} Degree of freedom=(Number of rows - 1)(Number of column - 1) \\ =(5-1)(2-1) \\ =4 \end{gathered}$$

The $p$-value using chi-square table at $4$ degree of freedom is $0.074133$.

The p -value is above significance level. The null hypothesis does not get rejected. Thus, there is sufficient evidence to conclude that there is not an association between two variables.