Question

1. Much has been made about the overall intelligence of criminal offenders in general, and violent offenders particular. There is an assumption that violent offenders are inherently less intelligent than non-violent offenders. To test this idea, data were collected from two separate groups of offenders (violent offenders and property offenders). Intelligence scores were obtained for every offender in the two samples. 2. Please submit your solution (step-by-step calculations), and the conclusion here. Descriptive Statistics appear below: Property Offenders = Average IQ = 96, sd = 8.5, n=225 Violent Offenders = Average IQ = 88, sd = 7.5, n=200 Use t to test the following set of hypotheses at the 95% level of confidence: H_a: Violent offenders are less intelligent than property offenders H_o: Violent offenders and property offenders do not differ significantly with regard to intelligence scores

Answer 1

With regard to intelligence scores, at 95% confidence level, there is significant evidence to conclude that violent offenders are less intelligent than property offenders

Explanation:

H_a: Violent offenders are less intelligent than property offenders

H_o: Violent offenders and property offenders do not differ significantly with regard to intelligence scores

Test statistic can be found using the formula

z=
`\frac{X-Y}{\sqrt{(s(x)^2)/(N(x))+(s(y)^2)/(N(y))}}` where

• X is the mean IQ score of property offenders (96)

• Y is the mean IQ score of violent offenders (88)

• s(x) is the sample standard deviation of property offenders (8.5)

• s(y) is the sample standard deviation of violent offenders (7.5)

• N(x) is the sample size for property offenders (225)

• N(y) is the sample size for violent offenders (200)

then z=
`\frac{96-88}{\sqrt{(8.5^2)/(225)+(7.5^2)/(200)}}` ≈10.3

One tail p-value of the test statistic is < 0.00001

Since p-value<0.05 (significance level) we reject the null hypothesis.