All Crime,,,All Crime,,,Aggravated Assault,,,Burglary,

Worked Example for Week 10

Using the same data from our previous Worked Examples, along with new data, I will show you

how you can test the difference between means using some basic descriptive statistics of the two

samples being compared. As a result, you will see an example of hypothesis testing: a statistical

test that examines the possible significant difference between two groups. The process can be

done for proportions as well, but for our purposes we will focus on means.

To perform this test we will require descriptive statistics for two groups. The first group will be

the sample of prisoners we have used in all of our Worked Examples. The second group will be a

new hypothetical group from a different prison. Let?s assume the first group is male prisoners

and the second group is female prisoners. Thus, we are testing to see if there is a significant

difference in the mean number of months incarcerated for males and females.

Our null hypothesis is always written that the two means are equal. To reject this, we need to

find a significant difference. Similar to ?innocent until proven guilty?, we assume equal means

until proven different. The null hypothesis can be written as follows:

H0: ?1 = ?2 or 1 = 2

Our alternative hypothesis is written that the two means are not equal (significantly different).

H1: ?1 ? ?2 or 1 ? 2

Notice the subscript numbers next to the symbols for means: this is simply referring to each

group. Each symbol with a ?1? subscript requires the descriptive data from Group 1 and each

symbol with a ?2? subscript requires the descriptive data from Group 2. It does not necessarily

matter which group you refer to as ?1? and ?2?; what is important is that you are consistent

throughout the hypothesis test processes*. If you mix the groups throughout the steps, you will

end up with incorrect and invalid results. Now that we have stated our hypothesis, we must list

the descriptive statistics needed for the hypothesis test.

Group 1, as we stated, is the male prisoners. Since we have used this data throughout our

Worked Examples, we already have all the information needed.

N1 = 10 1 = 4 s1
2
= 3.4

Recall N is our sample size. is the mean. S2 is the variance. This is very important; the standard
deviation is not required in this test, we need to use the variance (standard deviation ?squared?).

Group 2, as we stated, is the female prisoners. For our purposes, I will simply supply the

descriptive statistics needed for the hypothesis test.

N2 = 12 2 = 2 s2
2
= 2

Now that we have the required information we can begin to test the hypothesis that the mean

number of months incarcerated is equal between male and female prisoners. From our two small

samples we see that males have an average of 4 months and females have an average of 2

months. Remember, these are very small samples so we must use that information in

combination with the variability to determine if we have enough information to conclude the

difference is significant.

Step 1: Compute the standard error.

This step is tedious and requires a lot of information.

S 1- 2= v(

) (

) S 1- 2= v(

) (

)

S 1- 2 = .73

Step 2: Compute the test statistic (t-value). We simply subtract the mean of group 2 from the

mean of group 1 and divide by the standard error, the value we just calculated.

T=

T=

T= 2.74

Step 3: Determine the critical value. This step requires knowledge of what alpha level you will

be using and a T-distribution table of values. As is common in criminal justice research we will

look at alpha levels of .05 and .01.

The critical value for our hypothesis test with an alpha level of .05 is 2.086

The critical value for our hypothesis test with an alpha level of .01 is 2.845

Step 4: Compare test statistic (t-value) and critical value. Interpret.

We computed a test statistic of 2.74 which is larger than the first critical value of 2.086; we reject

the null hypothesis that the mean number of months incarcerated is equal between males and

females. We are stating that based on the information provided to us we can say at the .05 alpha

level (or with 95% confidence) the means are different.

When examining the test statistic of 2.74 at the .01 alpha level we fail to reject the null

hypothesis that the mean number of months incarcerated is equal between males and females.

This is due to the test statistic being lower than the critical value of 2.845. We are stating that

based on the information provided to us we can?t say at the .01 alpha level the means are

N1

N1 N1

N1

different. Essentially, we can be 95% confident that the observed difference is true or not due to

sampling error/chance but we cannot be 99% confident.

*As a note, depending on the values of the means and the order in which you subtract one from

the other you may end up with a negative test statistic (t-value). That is fine. When this happens

simply compare the numerical value itself to the critical value just as you would if the value was

positive. The negative only implies directionality, a component we are not focusing on. The
interpretation of actual difference is what is important.

Steps to test the null hypothesis that the mean number of months incarcerated for

males is equal to the mean number of months incarcerated for females:

Find the standard error of the difference between means using:

S 1- 2= v(

) (

N
)

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)

McCormack

CRIM.3950.031

Tests of Equality II

Thus far, we have extended our description of data; most recently, through Chapter 5, we learned about

measures of variability and dispersion. However, usually, the main objective in data gathering is not

description. The most common research initiatives revolve around hypothesis testing.

In hypothesis testing, we are estimating a population value (of a specific variable) based on sample data.

Hypothesis testing, in part, allows us to test the difference between means; frequencies, proportions, etc.

We can examine different subgroups with the same variable: number of crimes by adults, juveniles, etc.

We can examine different components of the same variables: number of assaults, murders, etc.

When we are hypothesis testing, we do so by analyses that result in inferential statistics. An important

facet of inferential statistics is the probability of empirical outcomes ? how likely we are to see that result

in the population. When a sample statistic is not equal to a population parameter, this can be

conceptualized in two ways.

? It may be a product of random fluctuations in sample statistics called sampling error. The disparity

could represent a legitimate statistical effect.

? The sample may not be appropriate to generalize to the population. A genuine discrepancy exists

between the sample statistic and population parameter.

The overarching purpose of hypothesis testing is to determine which of the above explanations is more

valid.

Perhaps the most commonly used inferential test is the t-test. This test is used to compare means across

groups, e.g. do males, on average, commit more crimes than females? A t-test uses estimation to compare

ONE value across TWO groups. We are asking, are those values significantly different?

What is the average age of our class? Let?s use hypothetical data.

Mean of 32 years overall: 35 males, 29 females (even gender split)

McCormack

CRIM.3950.031

In this case, we could use a t-test to answer the question, is there a significant difference between the

average age of males and females? Using our common levels of confidence, we can be 95/99% confident

in our answer, that there is a significant difference or there is not a significant difference. On the surface,

we see there is a difference: 35 ? 29

Why do we need to statistically test for a difference? What if we had just 2 folks in the class, 1 male (35

years old) and 1 female (29 years old)? Should two people be representative of a population? Doubtful!

We need to take into account some of the groups? descriptive information to assess if the 35 value is

significantly different from the 29 value based on the their variability and how many cases were

used to calculate them (e.g., is our sample big enough?). Here is what we need to know for each group.

? What are the averages?

o (???1) and (???2)

? What are the variations?

o (?1
2) and (?2

2)

? How many cases?

o (n1) and (n2)

That information is either always readily available or simple to obtain. We already know how to calculate

those values! What is next?

McCormack

CRIM.3950.031

Essentially, we end up calculating (or the software does it for us), a t-value that represents the amount of

difference between the two values (mean of group 1 and mean of group 2). Then, based on our level of

confidence (95% or 99%) and the number of cases, we compare it to a critical value. If/when the t-value

is greater than the critical value, we have enough information (the observed difference is larger enough)

to say there is a significant difference between groups. How do we get a difference that is enough?

? A large numerical difference and/or

? A large amount of cases

If the t-value (we calculated) is greater than the critical value (determined via a statistical table)

There is a significant difference in the average _____ between groups.

If the t-value (we calculated) is less than the critical value

There is a not significant difference in the average _____ between groups.

The t-value we calculated represents the difference between the groups. The critical value that is pre-

determined represents how different the groups? values must be to be significantly different. There are

only a few short formulae needed if we were to compute this by hand?.

Let?s see an example.

1) Calculate/obtain descriptives

Number of months: Is there a significant difference in sentence length between male and female offenders?

Males: mean = 4 variance = 3.4 N = 10

Females mean = 2 variance = 2 N = 12

2) Calculate the standard error

3) Calculate the test statistic (t-value)

McCormack

CRIM.3950.031

4) Compare to critical t-value

2) Calculate the standard error

3) Calculate the test statistic (t-value)

4) Compare to critical t-value

McCormack

CRIM.3950.031

Our t- value is 2.74. To be 95% confident we have a significant difference in sentence length between

male and female inmates, our t-value needs to be greater than 2.086. Achieved.

We are 95% confidence the sentence length is significantly different between males and females.

Our t-value is 2.74. To be 99% confident we have a significant difference in sentence length between male

and female inmates, our t-value needs to be greater than 2.845. Not achieved.

We are not 99% confident the sentence length is significantly different between males and females.

If the 99% level of confidence was our threshold, we would accept the null hypothesis that the sentence

lengths are equal between the two groups (male and female).

In this example, we only have enough cases/the difference is large enough to be 95% confident it

exists in the population, but not 99% confident. This can happen.

If you achieve significance at the 99% level, you will achieve it at the 95% level. If you don?t

achieve significant at the 95% level, you won?t achieve it at the 99% level.

Assignment 2 McCormack
CRIM.3950.01

Open the Excel data file in the Week 8 folder.

Your spreadsheet should like this:

This week we will complete the calculation and explanation of descriptive statistics with the help

of Excel. Below are the commands needed to compute the respective values.

Measures of Variability/Dispersion1:

Min. =min(array of values)

Max. =max(array of values)

Range =(Max-Min)

Variance =VAR.P(array of values)2

Standard Deviation =STDEV.P(array of values)

Quartiles:

Q1: =QUARTILE.INC(array of values, 1)3

Q3: =QUARTILE.INC(array of values,3)

1 In order to calculate the Range in Excel, we need to calculate the High and Low value(s) in the data set(s).
2 In Excel, like all statistics programs, you have the option of calculating some measures for a ?sample? or for a
?population?. We are going to use the Population form. The alternative form of the function in Excel for variance
and standard deviation would have ?.S? instead of a ?.P?.
3 In Excel, there is the option to INClude the median the calculation of the quartiles or EXClude the median. Including,
as we have done, will affect the Q1 and Q3 calculations and provide more symmetrical quartiles. The disadvantage
to this is that it will make it more difficult to identify outliers. Thus, we assume more normality in our distribution.

Assignment 2 McCormack
CRIM.3950.01

IQR4: =(Q1-Q3)

Normality:

Skewness: =SKEW.P(array of values)

Kurtosis: =KURT(array of values)

The measures of skewness and kurtosis may be unfamiliar and/or new to you. These values can

also be used to describe how normally distributed a set of data is. Each also has a threshold – or

rule of thumb – value, which one can use to determine how normal (or not) a distribution is.

Skewness measures the symmetry of the distribution. The closer the value is to 0 the more

symmetrical the distribution is, e.g. the less skewed. The value measures the relative size of the

two tails. Data that has skewness measured greater than | 1 | are highly skewed (positive or negative

direction indicates the direction of the skew). Kurtosis measures the ?peakedness? of the

distribution. Excel calculates this value using the ?minus 3? rule – a correction that actually reflects

a normal distribution with a value of 0. Thus, in Excel, the closer the value is to 0, the more

normally distributed the distribution is. Both values, like many of our descriptive measures, are

heavily influenced by our sample size.

Like before, add these measures in the Excel spreadsheet. Begin below ?mean?, in cell A49. Your

spreadsheet should look like this (open your previous assignment if needed):

Using the commands above, let?s calculate the values. Ultimately, you should come up with the

following values ? formatted in tabular form:

4 The Interquartile Range (IQR) describes where the middle 50% of the data is located.

Assignment 2 McCormack
CRIM.3950.01

All Reported Crimes

Annual Number in Boston 1985-2014

Mode N/A (multimodal)

Median 35,788

Mean 43,069

Min. 22,018

Max. 70,003

Range 47,985

Variance 258,567,142.6

Standard Deviation 16,080.02

Q1 31,718.75

Q3 56,188

IQR 24,452.25

Skewness 0.47

Kurtosis -1.27

How would we explain these results? First, we see a fairly large range. Annually, we have seen a

near consistent decrease in the overall number of reported crime in Boston. Its peak was in 1989,

with just over 70,000 crimes reported, and a low in 2014, with just over 22,000 crimes reported.

With a large range usually comes a large standard deviation. A value of just over 16,000 indicates

that the typical distance each annual total is away from the mean is about 16,000. So, the annual

values tend to differ. We know the IQR represents where the middle 50% of the data lie, so half

of all years had between about 31,000 and 56,000 reported crimes.

Variability and Dispersion

Calculate the measures of variability and dispersion for the same 3 offenses you have worked with

previously. Create a similar table for those same 3 offenses.

Copy/paste them into or create in a Word document (.doc or .docx) which will be

submitted.

Beneath each table (3 you picked and created), write a 100-word paragraph describing the

measures of variability and dispersion.

Ensure all of your tables and write-ups are submitted in one Word (.doc or .docx) file for

Assignment 3.