Subject: Mathematics    / Statistics

Provided is a data set that shows the number of arrests from a group of male juveniles in DYS custody, and group of male adults in DOC custody.

Juveniles: 04 05 05 07 08 10 04 06 07 08 07 03

Mean = 6.17 Variance = 3.81 Standard Deviation = 1.95

Adults: 12 09 06 05 02 06 10 05 11 14 02 05 10 09 04 02

Mean = 7 Variance = 13.38 Standard Deviation = 3.66

Using the steps provided below and the Worked Example as a guide, test the null hypothesis that the number of arrests for male juveniles is equal to the number of arrests for male adults.

What is the standard error of the distance between means?

What is the value for the test statistic (t-value)?

The critical value at the .05 level of significance (95% confidence) is 2.056

Do you reject the null hypothesis or fail to reject the null hypothesis?

The critical value at the .01 level of significance (99% confidence is) is 2.779

Do you reject the null hypothesis or fail to reject the null hypothesis?

Substantively, what do the results of the hypothesis test tell us about our data?

Explain what it would mean if we committed a Type II error with this example. Reference the variable specifically in your explanation.

Steps to test the null hypothesis that the mean number of arrests for juveniles is equal to the mean number of arrests for adults:

Find the standard error of the difference between means using:

S x?1- x?2=

Compute the test statistic (t-value) by dividing the difference between means by the standard error of the difference between means using:

T=

Determine the critical value (based on alpha level and degrees of freedom).

df=(N1+N2-2)

Compare our T -value and our critical value. Interpret. (If t-value exceeds critical; reject the null hypothesis)

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