Question

1. When do we use a z test? Give an original example that is relevant to criminology or criminal justice.

2. Calculate the z score using the following data that reflect the number of support meetings attended weekly by current and historic mental health court participants. Show your work (three points).

Data: Sample size is 50, sample mean is 12, sample standard deviation is 1.9

Population size is 295, population mean is 10, population standard deviation is 2

3. Is the z score you found for Question 2 significant at the .05 level? Why or why not (three points)?

4. Write your findings for Question 2 in standard reporting style

Answer 1

Solution:-

2)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u = 10
Alternative hypothesis: u 10

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), and z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 0.2687

z = (x - u) / SE

z = 7.44

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the z statistic less than -7.44 or greater than 7.44.

Thus, the P-value = less than 0.0001.

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we have to reject the null hypothesis.

3)  The z score you found for Question 2 is significant at the .05 level, becuase it is lies in rejection region.

z_critical = + 1.96

Rejection region is - 1.96 < z < 1.96.