Question

In: Statistics and Probability

Utilizing the data shown in the contingency table 1 below, determine whether the relationship between types...

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

    (           %)

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

    Types of Residential Area

    Rural

    Town

    City

    Total

    Own Gun

    555

    No Gun

    989

    Total

    593

    585

    366

    1544

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

    Types of Residential Area

    Rural

    Town

    City

    Total

    Own Gun

    No Gun

    Total

Solutions

Expert Solution

a) Null and alternative hypothesis:

Ho: Types of residential area and gun ownership are independent.

Ha: Types of residential area and gun ownership are dependent.

b) Percentage:

555/1544 *100 = 35.95%

989/1544 *100 = 64.05%

1544/1544*100 = 100%

c) Expected frequencies:

Expected Frequencies
Rural Town City Total
Own Gun 593 * 555 / 1544 = 213.1574 585 * 555 / 1544 = 210.2817 366 * 555 / 1544 = 131.5609 555
No Gun 593 * 989 / 1544 = 379.8426 585 * 989 / 1544 = 374.7183 366 * 989 / 1544 = 234.4391 989
Total 593 585 366 1544

d)

(fo-fe)²/fe
Own Gun (218 - 213.1574)²/213.1574 = 0.11 (206-210.2817)²/210.2817= 0.0872 (131-131.5609)²/131.5609 = 0.0024
No Gun (375-379.8426)²/379.8426= 0.0617 (379-374.7183)²/374.7183= 0.0489 (235-234.4391)²/234.4391 = 0.0013

Test statistic:  

χ² = ∑ ((fo-fe)²/fe) = 0.3116

df = (r-1)(c-1) =    2

p-value = CHISQ.DIST.RT(0.3116, 2) = 0.8557  

Decision:  

p-value > α, Do not reject the null hypothesis.  


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