CHAPTER 12

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CHAPTER 12

SOLUTIONS TO PROBLEMS

An independent variable is the variable whose effect or influence it is you want to measure on the dependent variable. In causal terms, the independent variable is the cause and the dependent variable is the effect. Since low self-control is taken to affect one's involvement in crime, self-control is the independent variable and involvement in crime is the dependent variable.
An independent sample t test should be used whenever the two samples have been selected independently of one another. In an independent sample t test, the sample elements are not related to one another. An example of independent sample would be a random sample of 100 residents from a rural area and a completely different and unrelated random sample from an urban area. In this case, both samples have been independently and randomly selected. In a dependent sample or matched groups t test, however, the sample elements are not independent, but are instead related to one another. An example of a dependent sample would be when the same sample elements or persons are measured at two different points in time as in a "before and after" experiment. In this case the two samples are not independent, but are actually the same person simply measured at two points in time. A second common type of dependent sample is a matched-groups design. In this case, after one sample has been selected, the elements of the second sample are deliberately selected so that they "match" with a member of the first sample on important characteristics. For example, a sample of first-time offenders might be matched with a sample of recidivists in terms of their age, race, gender, years of education, and any other characteristic thought by the researcher to be important. In this type of sample, the idea is not to have a second, independent, sample, but a sample that is as comparable to the first as possible.
Since it is a very useful routine, let's go through the steps of a hypothesis test. First, we state the null and alternative hypotheses: H₀: m₁ = m₂H₁: m₁ < m₂ Second, we must determine the correct test statistic and the sampling distribution of that test statistic. To determine this, let's write down what we know about this problem. We know that we have two independent samples, and that we can assume that the two unknown population standard deviations are equal (s₁ = s₂). We also know that the hypothesis test involves a difference of population means. This is all the information we need to know to conclude that the correct test statistic is the pooled variance independent samples t test. The independent samples t test is correct because we have independent samples. Since the two unknown population standard deviations are presumed to be equal we can use the pooled variance test. Hence, the correct test is the pooled variance independent samples t test, and our sampling distribution is the student's t distribution.

Third, our alpha level is .01, and we have a one-tailed test since we have assumed a directional alternative hypothesis. To use the t table, however, we need to know the correct degrees of freedom. With independent samples and equal population variances, the number of degrees of freedom is n₁ + n₂ - 2 = 25 + 40 - 2 = 63. We have 63 degrees of freedom, a=.01 for a one-tailed test. If we go to the t table with 60 df we can see that the critical level of t is -2.390. The sign of the critical t is negative remember because our alternative hypothesis predicted that m₁ < m₂. The critical region includes any obtained t value less than or equal to -2.390. We will reject the null hypothesis then if t_obt £ -2.390.

Fourth, we have to calculate our test statistic and compare the obtained value to the critical value. The obtained value of t is:

Notice that in the calculation of t the formula calls for the variance of each sample. To obtain this we had to square the value of the provided standard deviations.

Finally, we are ready to make our decision. Since our obtained value of t is less than the critical value, and falls into the critical region, we decide to reject the null hypothesis of equal means. Our conclusion is that those whose co-workers would disapprove of them stealing from their employer steal things less frequently than those whose co-workers are more tolerant of theft.
This problem involves a hypothesis test about two population proportions.

Step 1 is to state the null and alternative hypotheses:

H₀: p₁ = p₂

H₁: p₁ ¹ p₂

The null hypothesis states that the proportion of persons who are rearrested is the same for those who are punished with a fine and those who are punished with imprisonment. The alternative hypothesis simply states that the two population proportions are different.

Step 2 is to select the test statistic and sampling distribution. Since this is a problem involving two population proportions, our test statistic is the z test, and our sampling distribution is the z or standard normal distribution.

Step 3 is to determine the critical value and critical region of the test statistic. We have a alpha level of .05 and a two-tailed hypothesis test (the alternative hypothesis did not predict direction). This means that we have a/2 = .05/2 = .025 of the normal curve in each of our two critical regions. Going to the z table, we see that the corresponding z score is ±1.96. The critical region, therefore, is the area to the left of a z score of -1.96 and the area to the right of a z score of 1.96. Our decision rule is to reject the null hypothesis if z_obt £ -1.96 or z_obt ³ 1.96.

Step 4 is to calculate the test statistic from our sample data. The first thing we have to do is estimate the value of the pooled estimate of the population proportion ():

Now we can substitute these values into our formula for z_obt:

Step 5 is to make our decision. Since z_obt is not less than -1.96 or greater than 1.96, and does not fall into the critical region, we decide not to reject the null hypothesis. We cannot, therefore, reject the notion that the proportion who are rearrested is not different between those given fines and those given prison. A fine does not appear to motivate one to commit crimes more than the experience of prison.
The problem asks you to test a null hypothesis about two population means. The sample data we have is as follows:
```
       Males                    Females
        n₁ = 50                  n₂ = 25
                         
        s₁ = 4.5                s₂ = 3.0 
```
Step 1: State the null and alternative hypotheses. The null hypothesis is that the two population means are equal, while the alternative is that the male mean is less than the female mean. The alternative, then, is one-tailed hypothesis:

H₀: m₁ = m₂

H₁: m₁ < m₂

Step 2: Select the appropriate test statistic and sampling distribution. The two samples are independent random samples, with unknown population standard deviations, and a small sample size n₁ = 50 n₂ = 25. This should tell you that the correct test statistic test will be a t test. Since the problem instructs you not to presume that the population standard deviations are equal (s₁ ¹ s₂), we cannot use a pooled variance estimate of the standard error of the difference. The correct statistical test, then, is the separate variance t test, and the sampling distribution is the student's t distribution.

Step 3: The alpha level is .05, with a one-tailed test. To use the t distribution, you need to know the correct degrees of freedom. With a separate variance t test, the correct degrees of freedom for this problem is:

Go to the z table with 74 degrees of freedom to find t_crit. Since there is no value for 74, approximate with 60 degrees of freedom. With 60 df and an alpha of .05 for a one-tailed test, the critical value of t is -1.671. The critical value is negative since the alternative hypothesis was that m₁ < m₂. The region of rejection is the area to the left of a t score of -1.671. Our decision rule, therefore, will be to reject the null hypothesis if t_obt £ -1.671.

Step 4: We now calculate the test statistic and compare it to the critical value.

Step 5: Since t_obt £ t_crit (-2.84 £ -1.671), our decision is to reject the null hypothesis of equal population means. Our conclusion is that the mean score on the domestic disturbance scale is significantly lower for males than females. In other words, males are less likely to see the fair handling of domestic disturbances as an important part of police work.
The null hypothesis in this problem is that the mean difference in the number of capital cases lost on appeal between trained and untrained judges is zero. The alternative hypothesis is that the untrained judges lose more cases on appeal than the trained judges. Step 1: The null and alternative hypotheses are:

H₀: m_D = 0

H₁: m_D > 0

Step 2: Since our sample members (the judges) were deliberately matched in order to be comparable, we have matched samples. The appropriate test statistic, then, is the dependent samples or matched groups t test, and the sampling distribution is the student's t distribution.

Step 3: Our alpha level is .01, and we have a one-tailed hypothesis test. To use the t distribution, we need to know the correct degrees of freedom. Since we do not have independent observations, but pairs of observations, the number of degrees of freedom is equal to n - 1, where n is the number of pairs of observations. In this problem, we have 15 - 1 = 14 degrees of freedom. Going to the t table with a=.01 (one-tailed test) and 10 degrees of freedom, the critical value of t is 2.624. The critical region, therefore, is the area to the right of a t score of 2.624. Our decision rule is to reject the null hypothesis if t_obt ³ 2.624.

Step 4: Now we calculate our test statistic and compare it to our critical value. To do this we have to calculate several statistics, including the mean and standard deviation of the difference scores .
```
        Number of Cases Lost 
             On Appeal      
Judge   Untrained       Trained     D      D² 1          3             0         3      9
 2          1             3        -2      4
 3          2             4        -2      4
 4          7             4         3      9
 5          5             2         3      9
 6          4             5        -1      1
 7          6             1         5     25
 8          2             1         1      1
 9          7             0         7     49
10          5             6        -1      1
11          3             4        -1      1 
12          4             2         2      4
13          5             5         0      0
14          6             3         3      9
15          2             1         1      1 
                                 S=21   S=127
```
The mean difference score is equal to . The standard deviation of the difference scores is equal to

Now we have all the information we need to calculate our test statistic:

Step 5: Since our t_obt (2.06) is not greater than or equal to 2.624, we do not reject the null hypothesis. There is no difference in the number of capital cases lost on appeal between trained and untrained judges.
This is a hypothesis test involving two population proportions.

Step 1: The null hypothesis is that the two population proportions are equal. That the proportion of criminal parents is the same between delinquents and non-delinquents. The alternative hypothesis is that the delinquents have a higher proportion of criminal parents. This is a one-tailed alternative.

H₀: p₁ = p₂

H₁: p₁ > p₂

Step 2: Since this is a difference of proportions problem, the correct test statistic is the z test, and the sampling distribution is the z or standard normal distribution.

Step 3: The alpha level is .01 for a one-tailed test. All .01 of the critical region is in the right tail of the distribution. Find the area of the z distribution corresponding to .490. The closest to this is .4901, and the corresponding z score is 2.33. Our critical z score, then is 2.33, and the critical area is any z score to the right of 2.33. Our decision rule is to reject the null hypothesis if z_obt ³ 2.33.

Step 4: Calculate the value of the test statistic. Before we do this, we must determine the value of the pooled estimate of the population proportion ():

Step 5: Since our obtained z is greater than the critical value of z (2.33), and z_obt falls into the critical region, our decision is to reject the null hypothesis. Delinquent children have a significantly higher proportion of criminal parents than non-delinquent children.
This is a hypothesis test about the number of delinquent acts committed by a sample of youths before and after they dropped out of school.

Step 1: The null hypothesis is that the number of delinquent offenses committed before dropping out of school is the same as the number committed after dropping out. This is the same thing as saying that the mean difference between the two scores is equal to zero. The alternative hypothesis is that the mean different score is different from zero. The alternative is a two-tailed hypothesis.

H₀: m_D = 0

H₁: m_D ¹ 0

Step 2: Since the two samples are the same youths at two points in time (before and after dropping out) we have dependent samples. The correct test statistic, then, is the dependent samples or matched groups t test, and the sampling distribution is the student's t distribution.

Step 3: We have an alpha level of .05, with a two-tailed hypothesis test. To use the t distribution, we need to know the correct degrees of freedom. Since we have independent pairs of observations in a dependent samples t test, there are n - 1, degrees of freedom, where n is the number of pairs of observations. In this case, we have 11 - 1 = 10 degrees of freedom. The critical value of t for a two-tailed alpha of .05 and 10 degrees of freedom is ±2.228. The critical regions are the areas to the left of a t score of -2.228 and to the right of a t score of 2.228. Our decision rule, then, is to reject the null hypothesis if t_obt £ -2.228 or if t_obt ³ 2.228.

Step 4: Now we have to calculate our test statistic, before we do this, however, we must determine the mean and standard deviation of the difference scores:
```
            Number of Delinquent Acts
Person        Before        After     D      D²   
 1              5             7      -2      4
 2              9             5       4     16
 3              2             3      -1      1
 4              7             7       0      0
 5              8            11      -3      9
 6             11            13      -2      4
 7              8             4       4     16
 8              8            10      -2      4
 9              5             7      -2      4
10              2             1       1      1
11              9             3       6     36  
                                    S=3   S=95
```
The mean of the difference scores is equal to . The standard deviation of the difference scores is equal to:

Now we can place these values into the formula for our z test:

Step 5: Since our critical value of z is not greater than or equal to 2.228 nor less than or equal to -2.228, and does not fall into the critical region, our decision is to fail to reject the null hypothesis. We cannot reject the assumption that the number of delinquent offenses committed before dropping out is the same as the number committed after dropping out.

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