Question

A researcher believes that maximum security inmates have a more lengthy arrest record than minimum security inmates. She collects a sample of 25 inmates from the minimum security prison and 32 inmates from the maximum security prison and computes the following data from the samples:

                                                Mean                           Standard Deviation

Minimum:                   5.8                                         2.7

Maximum:                  6.2                                         1.9

1.Calculate the 95% CI for each group. Based on these data, is the researcher correct that maximum security inmates have a longer arrest records than minimum security inmates? Why or why not?

2. Assume the data in question #1 were population parameters (m and s) and not sample statistics. Suppose the researcher draws at random one minimum security inmate and one maximum security inmate. Both inmates report having 7 prior arrests. Which inmate is more dissimilar from the population mean from which they were drawn? Why?

Answer 1

For the minimum group,

For the maximum group,

Since the confidence interval for maximum security group is in general more than one with minimum security group, we can conclude that maximum security inmates have a longer arrest records than minimum security inmates.

If one inmate reports having 7 prior arrests, then likelihood of belonging to minimum security group is

and that to maximum security group is

Thus likelihood of the inmate with 7 prior arrests have more likelihood to belong to minimum security group

Given; mean X = 5.8; s = 2.7 and n=25 Level of significance,a = 0.01 The test statistic is given by Thus the confidence interval is given by where to 01/2.25 is such that PT < 10.01:25 = a) Thus based on the data confidence interval is (4.29.7.31)

Given; mean X = 6.2; s = 1.9 and n=25 Level of significance,a = 0.01 The test statistic is given by _ VX-) V8 . Thus the confidence interval is given by (x - ato 01/2:25–1,8 + P to 01/2:25-1) where to 01/2-25 is such that PT < 0.01:25 = a) Thus based on the data confidence interval is (5.137.7.263)

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1 (x - 5.8) c. exp. 22.72 -= 0.134