Question

A parole board wants to predict potential parolees who will successfully readjust to life outside of prison. A study was instituted in order to examine the relationship between prior arrests and readjustment to life outside of prison. The maximum readjustment to life score is 10 – meaning they are perfectly well-adjusted to life after prison. The results of the study are summarized in the table below:

	Prior Arrests	Readjustment
Minimum value	1	2
Maximum value	5	9
Mean	1.63	6.25
Standard Deviation	1.40	2.25

Correlation Coefficient

-0.91

1. Calculate the RMSE. Show your work
2. Imagine that parole is granted to individuals who have readjustment scores of 5 and above. Given the interpretation of the RMSE (i.e. what the value of RMSE is), would you feel confident granting parole to someone with 2 prior arrests using only this prediction model? Explain.

Answer 1

a)

se^2 = sy^2 (1-r^2) (n-1)/(n-2)

= 2.25^2 * ( 1 - 0.91^2)     {since n is not given assuming n is large so n-1 approx equal to n-2}

= 0.87024375

hence

RMSE = sqrt( 0.87024375 ) = 0.932868

b)

y^ = 8.6339 - 1.4625 * x

y^ = 8.6339 - 1.4625 * 2

= 5.7089

I would not feel confident granting parole to someone with 2 prior arrests using only this prediction model

since difference between 5.7089 and is 0.7089
which is less than RMSE