Question

1. Utilizing the data shown in the contingency table 1 below, determine whether the relationship between types of residential area and gun ownership is statistically significant. Once you have arrived at an answer for each question, please write a sentence or two interpreting the results, you may want to round the decimals to the nearest whole number (70 pts.).

1. What are the alternative and null hypotheses (5pts)?

2. Calculate the total column percentages (10pts.)

Gun Ownership	Types of Residential Area
	Rural	Town	City	Total
Own Gun	218	206	131	555 (         %)
No Gun	375	379	235	989 (           %)
Total	593	585	366	1544 (           %)

3. Calculate the expected frequency for each cell and explain the meaning of expected frequencies (15 pts).

Gun Ownership	Types of Residential Area
	Rural	Town	City	Total
Own Gun				555
No Gun				989
Total	593	585	366	1544

4. Calculate chi-square (χ ²) for each cell using to the formula (Oi − Ei )²/ Ei (15pt).

Gun Ownership	Types of Residential Area
	Rural	Town	City	Total
Own Gun
No Gun
Total

1. 5. What is the Chi-square statistic(5 pt)? And what is the critical value for this case at 5% significance level, what information did you use to find it (10pts)?
2. 6. Draw a conclusion about your null hypothesis at 5% significance level, and explain how did you arrive that conclusion (10pt)

Answer 1

SOLUTION A) NULL HYPOTHESIS H0​: The two variables are independent

ALTERNATIVE HYPOTHESIS Ha​: The two variables are dependent

SOLUTION C) SOLUTION D)

Expected Values RURAL TOWN CITY Total OWN GUN 593 x 555 = 213.157 585x 555 = 210.282 366x555 = 131.561 555 1544 1544 366989 = 234.439 NO GUN 593x989 = 379.843 585x989 = 374.718 989 Total 593 585 366 1544

SOLUTION E: The Chi-Squared statistic is computed as follows:

Squared Distances RURAL TOWN CITY OWN GUN (218-213.157) 213.157 - 0.11 (206-210.282) 210.282 0.087 (131-131.561) 131.561 0.002 NO GUN (375–379.843) 379.843 0.062 (379-374.718) 374.718 0.049 (235-234.439) 234.439 0.001

v_Y(i-bi) = 0.11+0.062 +0.087 +0.049 +0.002 +0.001=0.312

Based on the information provided, the significance level is α=0.05 , the number of degrees of freedom is df=(2−1)×(3−1)=2, Critical value of chi square is χ2= 5.991.

f) Decision about the null hypothesis: Since it is observed that χ2=0.312≤χc2​=5.991, it is then concluded that the null hypothesis is not rejected.It is concluded that the null hypothesis Ho is not rejected. Therefore, there is NOT enough evidence to claim that the two variables are dependent, at the 0.05 significance level.