Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

5. Crime and Strangers: The accompanying table lists survey results obtained fro

ID: 3226207 • Letter: 5

Question

5. Crime and Strangers: The accompanying table lists survey results obtained from a random sample of different crime victims. At the 0.05 significance level, test the claim that the type of crime is independent of whether the criminal is a stranger.

Homicide

Robbery

Assault

Criminal was a Stranger

12

379

727

Criminal was acquaintance or relative

39

106

642

a. Identify the explanatory and response variable.

b.    State the hypothesis (in words, in this context) to be tested with a chi-square test on the data in this table.

c.    Use technology to calculate the expected values, test statistic, and p-value. Report the test statistic and p-value.

Test statistic:                                                                      p-value:

d.    Do the sample data provide evidence that the type of crime is independent of whether the criminal is a stranger., at the 0.05 significance level?

e.    How might the results affect the strategy police officers use when they investigate crimes?

Homicide

Robbery

Assault

Criminal was a Stranger

12

379

727

Criminal was acquaintance or relative

39

106

642

Explanation / Answer

(a)

Explanatory variable: The criminal is a stranger or not

Response variable: Tyep of crime

(b)

H0: The type of crime is independent of whether the criminal is a stranger.

Ha: The type of crime is not independent of whether the criminal is a stranger.

(c)

Following is the output of the chi square test generated by excel:

d:

P-value is 0.0000

Since p-value is less than  0.05 significance level so we cannot conclude that  the type of crime is independent of whether the criminal is a stranger.

(e)

According to type of crime police officer should try to find that whether the criminal is in relation with crime victims.

Chi-square Contingency Table Test for Independence    Homicide   Robbery   Assault   Total   Criminal was a Stranger Observed   12 379 727 1118    Expected   29.93 284.64 803.43 1118.00 Criminal was acquaintance or relative Observed   39 106 642 787 Expected   21.07 200.36 565.57 787.00 Total Observed   51 485 1369 1905 Expected   51.00 485.00 1369.00 1905.00 119.33 chi-square 2 df 1.22E-26 p-value
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote