Discussion: Prescription drug use influences arrests

Q1. You as a researcher would like to understand whether prescription drug use influences arrests. You survey 1600 individuals, asking them if they recreationally use prescription drugs and whether they have ever been arrested.

You have two events:

Event A: Drug Use
- Prescription Drug Use
- No Prescription Drug Use

Event B: Arrest Record
- 1bArrest
- No Arrest

You decide to conduct a Chi Squared Test for Independence (X²) and formally test group differences. (2.5 pts)

Identify the independent variable (The cause)

Identify the dependent variable (The effect)

Write a null hypothesis.

Write an alternative hypothesis.

Q2. Below are the Observed Cell frequencies (5 pts).

	Prescription Drug Use (1a)	No Prescription Drug Use (2a)
Arrest (1b)	657	389	1046
No Arrest (2b)	221	333	554
Total	878	722	1600

First, calculate the joint probability (complete the highlighted sections):

The probability of prescription drug use and arrested:

Probability of (1a) prescription drug use 878/1600 = .55
Probability of (1b) arrest 1046/1600 = .65
Equation (1a) X (1b)
Joint probability of 1a and 1b:

The probability of prescription drug use andnot being arrested:

Probability of (1a) prescription drug use: 878/1600 = .55
Probability of (2b) no arrest: 554/1600 = .35
Equation (1a) X (2b)
Joint probability of 1a and 2b:

The probability of no prescription drug use and arrested:

Probability of (2a) no prescription drug use 722/1600 = .45
Probability of (1b) arrest 1046/1600 = .65
Equation (2a) X (1b)
Joint probability of 2a and 1b:

The probability of no prescription drug use andnot bring arrested:

Probability of (2a) no prescription drug use 722/1600 = .45
Probability of (2b) no arrest 554/1600 = .35
Equation (2a) X (2b)
Joint probability of 2a and 2b:

Second, fill in the table with the Expected Cell Frequencies (complete the highlighted sections):

	Prescription Drug Use (1a)	No Prescription Drug Use (2a)	Row Total
Arrest (1b)	(1a) X (1b) X 1600 =	(2a) X (1b) X 1600 =
No Arrest (2b)	(1a) X (2b) X 1600 =	(2a) X (2b) X 1600 =
Column Total			1600

Q3. As the observed and expected cell frequencies are different, you proceed with your formal Chi Squared Test for Independence (X²). The alpha level is set at .05 and the critical value is identified as 5.991.

Decision Rule: If the calculated chi squared statistic is greater than 5.991, you will need to reject the null hypothesis; if the calculated chi squared statistic is not greater than 5.991, you will fail to reject the null hypothesis.

The calculated chi squared statistic is 4.864. Discussion: Prescription drug use influences arrests

Part I: (5 pts)

What is the hypothesis decision (Reject the Null Hypothesis or Fail to Reject the Null Hypothesis)?

Complete the following interpretation bullet points:

Is this finding statistically significant?
What is our formal conclusion/what does this mean?
What is the chance of error(type and percentage)?

Part II: (5 pts)

Same problem as above. The alpha level is set at .05 and the critical value is identified as 5.991.

The calculated chi squared statistic is 19.864.

What is the hypothesis decision (Reject the Null Hypothesis or Fail to Reject the Null Hypothesis)?

Complete the following interpretation bullet points:

Is this finding statistically significant?
What is the formal conclusion/what does this mean?
What is the chance of error (type and percentage)?

Complete the measure of interpretation:

The measure of association is .53

What is the measurement type (Positive or Negative)?
What is the measurement of association conclusion?
What is the strength of the relationship (weak, modest, strong)?

Q4. According to the textbook,what is the difference between observedcell frequencies and expected cell frequenciesin a Chi Squared Test for Independence (X²)? (Answer should be between 2-3 sentences in length)(2.5 pts)