Review of Tests: Use this in conjunction with the hypothesis testing review. This page is just a review of the actual tests, it’s step #5 from the review.  These all go together!

 

  1. Z-test

-for comparing obtained sample means (x bars) to populations with known parameters (both m and s are known).

-questions like this will make reference to a known population: “All other years’ grades” or ”On average at all other colleges” or “In the population” (something like that).

Example:  A university professor believes that in the past couple of years,  the average age of students attending his university has changed. He takes a sample from this years’ class of 150 students. The mean age of these students is 23.5, with a standard deviation of 8.4 years.  He then collapses across all the census data taken every other years at the university for the last ten years. That mean age is 22.4, with a standard deviation of 7.6 years. 

 Steps:

a.         determine the parameters of the null hypothesis population. This is what the population                

would look like if the null hypothesis is true.  The population here is the “collapsed census data over the last ten years”. m = 22.4, s = 7.6.

 

b.      derive the sampling distribution of the mean for this population:

m= m=22.4. 

s= s/√N = 7.6/√150 = .6205

 

c.       do the test: (compare your obtained mean to what the mean would be if the null is true).

 = (23.5-22.4)/.6205 = 1.77

 

d. do step #6:  evaluate the results and state a conclusion.

 

  1. t-test for single sample

-for comparing obtained sample means to populations where m is known, but s is unknown.

-questions like this are just like questions for the Z-test, but without a population standard deviation. We are going to estimate it from the sample standard deviation.

*Note: if both a population standard deviation and a sample standard deviation are given, use the Z-test with the population standard deviation

Example: we are going to use the example from above, but assuming the population standard deviation is not given.

Steps:

determine the parameters of the null hypothesis population. This is what the population                

a.       would look like if the null hypothesis is true.  The population here is the “collapsed census data over the last ten years”. m = 22.4

b.      derive the sampling distribution of t for this population:

m= m=22.4. 

S = S/√N = 8.4/√150 = .686       (We use this value since the actual population standard deviation is not                                                  given)

 

c.       do the test: (compare your obtained mean to what the mean would be if the null is true).

tobt = = (23.5-22.4)/.686 = 1.603

 

d.      do step #6:  evaluate the results and state a conclusion.

 

  1. t-test for correlated groups

-for analyzing within-subjects designs (where a group of people = N) are in two conditions.

-questions like this will clearly state that there is one group of people who are measured on the DV twice. Usually it’s after some treatment or intervention.

-this test ‘reduces’ to a single sample t (like above)

 

We will do the entire process here so you can see another example of how it works: (we will do all the steps from the hypothesis testing review!)

 

Example: An advertising firm is considering a new campaign to increase people’s consumption of ‘yokel’ (a delicious egg yolk sandwich which has been on the market for several months already).  To test whether their ad campaign will work, they select a sample of 10 people and monitor their consumption of yokel for one week, then they see yokel commercials every day for a week, and then their consumption of yokel is measured for one week again after the ad campaign. The DV is number of yokel sandwiches eaten each week.

 

Step 1: read it carefully and decide what it being asked.

You know it’s correlated groups, so use a t-test for correlated groups (remember, you should always favor a t-test over the sign test when enough information is given).

The hypothesis seems to be directional, right? They want to know if the campaign increases consumption.

Think here about what an increase in consumption will mean. If they eat more after the campaign than before, then subtracting post from pre (pre-post) will yield a negative number.  If someone ate 10 sandwiches before the campaign and 12 after the campaign (this is the predicted direction), then (10-12) = -2.  If everyone goes in this direction (the predicted direction), then the obtained mean of the difference scores will be negative. Negative numbers are the predicted direction! You need to use this to make your hypotheses:

 

      Step 2: state the hypotheses:

                        Null:                 the campaign does not increase consumption.

                                                differences are due to chance

                        (because if the campaign doesn’t work, people will have the same or higher scores              after the campaign)

 

                        Alternative:       the campaign does increase consumption

                                                differences are due to the campaign

                                                (because negative numbers are the predicted direction)

 

Step 3: set α = .051tail

Step 3a. Decide on the cut-offs: tcrit for α = .051tail,, (df = N-1 = 9)  is 1.833 (from table D.)

 

Step 4: draw the distribution:  this is important to think about because we have a one-tailed test with a predicted direction of negative t-values. Your cut-off score (1.833) is negative. It goes on the left side of the distribution, less than the hypothesized mean value of 0.

 

Step 5: Do the test: Remember, this ends up being a single sample t-test. We will use the same steps as above:

Here is the data:

 

                                               

person#

pre-ad

post-ad

1

6

4

2

5

3

3

8

6

4

7

7

5

9

5

6

4

1

7

8

5

8

14

12

9

3

1

10

2

2

 

            a. determine the parameters of the null hypothesis population. This is what the population                

            would look like if the null hypothesis is true.

 (the actual difference in the population would be 0)

s is not given, so we will estimate it by using the standard deviation of the differences.                                  

                       

 

b.         Derive the sampling distribution of t for this population:

 

s is not known, so we will estimate it with the standard deviation of the sample.

 S = SD/√N.

To do this, we need to get our sample (recall that this is a one-sample test, and that the sample is actually the sample of difference scores)

 

 

D

D^2

 

2

4

 

2

4

 

2

4

 

0

0

 

4

16

 

3

9

 

3

9

 

2

4

 

2

4

 

0

0

SUMS

20

54

 

           

SD is nothing more than the standard deviation of the difference scores.  

 

            Where SSD =       OR       SSD =

In this case:  SSD =  = 54  - ((202)/N) = 54-40=14. Don’t stop, that’s just the sum of squares of D.

Now, the standard deviation of D (SD)   is  √ (14/9) = 1.247

 

But, we are still not done. 1.247 is the standard deviation of the difference scores, but the SE of the sampling distribution (S) is

SD/√N = 1.247/√10 = .395

 

Here’s the other way to do it:  this way skips the step of computing the standard deviation of the difference scores (SD ) and directly computes the standard error of the sampling distribution (S).  You still have to compute SSD, though. 

 

                        S =   = = .395

 

 

 

We are done with b: deriving the sampling distribution of t for this problem. Whew.

 

c.       Do the test!  We are evaluating the reasonableness of assuming that our obtained mean (of differences) comes from a null hypothesis population where the true mean of differences is 0. We therefore need the actual mean of our differences. It’s  ∑D /N = (20/10) = 2.

 

  = (2 – 0)/.395 = 5.06

 

d. evaluate the results.  Can you reject the Null hypothesis that the ad campaign does not increase                                    consumption?