Question

1. A prison administrator is interested in examining the relationships between the type of prison security (maximum, medium, and minimum) and the number of previous offenses committed by an inmate. He believes that maximum security prisons have inmates with many prior offenses, medium security with not so many offenses, and minimum security with almost no prior offenses. He randomly selects eight new inmates from each of the three security levels and compares the number of offenses for which they have ever been charged.

a) What is the independent variable?              

b) What is the dependent variable?

c) What is the null hypothesis?

d) What is the research hypothesis?

e) Interpret the results.

Maximum	Medium	Minimum
8	4	2
6	4	2
4	3	3
5	6	2
3	5	4
7	6	1
6	3	2
9	3	2

Answer 1

Part (A and B)

Here the independent variable is the type of prison security and dependent variable is the number of previous offenses committed by an inmate. The independent variable is a categorical ordinal variable and dependent variable is nominal variable.

Part C),D

Here given the significance level is 0.01. Here we need to test the groups are equal against at least one is different. So clearly we need to apply here one way ANOVA. Here we can we R software to get the desired results.The R code is given below:

y=c(8,6,4,5,3,7,6,9,4,4,3,6,5,6,3,3,2,2,3,2,4,1,2,2)
x=as.factor(rep(1:3,each=8))
anova=aov(y ~ x)
summary(anova)

The result of the summary table as follows:

Df Sum Sq Mean Sq F value Pr(>F)
x 2 56.33 28.167 13.14 0.000199 ***
Residuals    21 45.00 2.143   
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

The critical value of F statistics is given by: qf(0.99,2,21) = 5.780416. So here the observed F statistics value is 13.14 which is greater than critical value, so we can conclude that at 1% level of significance the groups are different.

e) Now we have to find which groups are different from each other. For this we will apply Tukey's Honest Difference Test in R. The R code is given as:

TukeyHSD(anova,conf.level = 0.99)

The result is as follows:

Fit: aov(formula = y ~ x)

$`x`
diff lwr upr p adj
2-1 -1.75 -4.137015 0.6370148 0.0650195
3-1 -3.75 -6.137015 -1.3629852 0.0001278
3-2 -2.00 -4.387015 0.3870148 0.0320870

From the above table we can clearly observe that there are differences between means of group 1 and 2 and between means of group 2 and 3 as the interval contains 0 value.