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- How to write a primary research paper

- How to write a primary research paper

- How to keep a research notebook

- Types of variables and scales

- How to write a primary research paper

- How to keep a research notebook

- Types of variables and scales

- Types of distributions

- How to write a primary research paper

- How to keep a research notebook

- Types of variables and scales

- Types of distributions

- Attributes of the normal distribution

- testing single values – “z” test

(one sample t-test)

- How to write a primary research paper

- How to keep a research notebook

- Types of variables and scales

- Types of distributions

- Attributes of the normal distribution

- Comparing means of two independent groups

- t-test and MWU

- How to write a primary research paper

- How to keep a research notebook

- Types of variables and scales

- Types of distributions

- Attributes of the normal distribution

- Comparing means of two independent groups

- Experimental designs

- How to write a primary research paper

- How to keep a research notebook

- Types of variables and scales

- Types of distributions

- Attributes of the normal distribution

- Comparing means of two independent groups

- Experimental designs

- Comparing > 2 groups, and multiple effects

- ANOVA

- Differences Between Means
- A. 1 Sample
- one sample t-test (z test)
- B. 2 samples
- 1. Independent Groups
- - t-test (parametric)
- - MWU (non-parametric)
- 2. Related Samples
- - Paired t-test

- Paired t-test

- Paired t-test

Suppose we wanted to assess the effect of a muscle-building supplement, and randomly assign people to two groups – placebo and experimental. People differ in many characteristics (ethnicity, sex, weight, diet, etc.), and so these have been randomized across groups. Effects due to these variables are part of the “within group variance” in the denominator.

- Paired t-test

But if we give everyone the drug, and assess their performance before and after, then there is no within group variance between sample points – they are the same individual.

- Paired t-test

We can look at the distribution of differences between before and after weights, and do a “z-test” asking the question: is “0” an unusual value for this sample? Or, how likely is it that “0” (no difference) is a part of this population of differences? If “0” is unlikely, then our population is different from zero; the difference between “before” and “after” is NOT zero – there IS an effect.

Person wt. Before wt. After Difference

1 150 156.5 1.5

2 155 156.3 1.3

3 158 159.6 1.6

4 160 161.4 1.4

5 163 164.5 1.5

6 167 166.8 -0.2

7 175 176.3 1.3

8 180 181.5 1.5

9 185 186.1 1.1

10 191 192.6 1.6

mean = 1.26

sd = 0.536

1.26 – 0

0.536 / 10

t =

= 7.46

In a two-tailed test, we are asking if a value (or sample) IS DIFFERENT FROM a sample… (it can differ because it is LARGER or SMALLER.)

In a one-tailed test, we have a preconceived hypothesis about the direction of the effect. In our experiment here, “0” should be LOWER than the mean of our differences. So, our type one error can be “pooled” into one tail of the distribution.

So, our type one error can be “pooled” into one tail of the distribution. This means we use the t value in the table A.2 corresponding to p = 0.1 to test at the p = 0.05 level.

- Differences Between Means in SPSS
- A. 1 Sample
- one sample t-test (z test)
- B. 2 samples
- 1. Independent Groups
- - t-test (parametric)
- - MWU (non-parametric)
- 2. Related Samples
- - Paired t-test
- - Sign Test (non-parametric)

2. Related Samples in SPSS

- Paired t-test

- Sign Test (non-parametric)

For the matched pairs, you simply record whether one partner is greater (+) or less than (-) the other:

Person wt. Before wt. After Sign in SPSS

1 150 156.5 +

2 155 156.3 +

3 158 159.6 +

4 160 161.4 +

5 163 164.5 +

6 167 166.8 -

7 175 176.3 +

8 180 181.5 +

9 185 186.1 +

10 191 192.6 +

Person wt. Before wt. After Sign in SPSS

1 150 156.5 +

2 155 156.3 +

3 158 159.6 +

4 160 161.4 +

5 163 164.5 +

6 167 166.8 -

7 175 176.3 +

8 180 181.5 +

9 185 186.1 +

10 191 192.6 +

Now, if we are testing the hypothesis of NO effect, then we would expect the “after” to be greater 1/2 the time (p = 0.5), and less than ½ the time (q = 0.5).

Pairs that don’t differ are dropped from the analysis, with reduction in n.

Person wt. Before wt. After Sign in SPSS

1 150 156.5 +

2 155 156.3 +

3 158 159.6 +

4 160 161.4 +

5 163 164.5 +

6 167 166.8 -

7 175 176.3 +

8 180 181.5 +

9 185 186.1 +

10 191 192.6 +

Now, if we are testing the hypothesis of NO effect, then we would expect the “after” to be greater 1/2 the time (p = 0.5), and less than ½ the time (q = 0.5). So, this reduces to calculating the probability of a particular BINOMIAL OUTCOME, OR SOMETHING MORE EXTREME.

Pairs that don’t differ are dropped from the analysis, with reduction in n.

So, if p = 0.5 and q = 0.5 (no effect), what would be the probability of having at least 9/10 individuals show a weight gain just by chance?

p(9) = (p9) (q1) = 0.009760

p(10) = (p10) (q0) = 0.0009760

10!

(9!) (1!)

10!

(10!) (0!)

0.010736

This is the one-tailed probability of seeing at least 9/10 showing a weight gain (directional).

2. Related Samples probability of having at least 9/10 individuals show a

- Paired t-test

- Sign Test (non-parametric)

- Wilcoxon Signed-ranks test

2. Related Samples probability of having at least 9/10 individuals show a

- Paired t-test

- Sign Test (non-parametric)

- Wilcoxon Signed-ranks test

In the sign test, the magnitude of the difference doesn’t matter. You could have 5 big positive diffs and 4 very small diffs, and it would still be 5 + and 4 - .

2. Related Samples probability of having at least 9/10 individuals show a

- Paired t-test

- Sign Test (non-parametric)

- Wilcoxon Signed-ranks test

In the sign test, the magnitude of the difference doesn’t matter. You could have 5 big positive diffs and 4 very small diffs, and it would still be 5 + and 4 - .

The signed-ranks test takes the magnitude of the difference into account.

Example 9.5 probability of having at least 9/10 individuals show a

AGGRESSION

Female w/o kittens w kittens Diff. Rank

1 3 7 4 4

2 2 8 6 6

3 5 4 -1 1.5(-)

4 6 9 3 3

5 5 10 5 5

6 1 9 8 7

7 8 9 1 1.5

Example 9.5 probability of having at least 9/10 individuals show a

AGGRESSION

Female w/o kittens w kittens Diff. Rank

1 3 7 4 4

2 2 8 6 6

3 5 4 -1 1.5(-)

4 6 9 3 3

5 5 10 5 5

6 1 9 8 7

7 8 9 1 1.5

Sum ranks with positive and with negative values:

Negative = 1.5

Positive = 26.5

T = lower value = 1.5

Example 9.5 probability of having at least 9/10 individuals show a

AGGRESSION

Female w/o kittens w kittens Diff. Rank

1 3 7 4 4

2 2 8 6 6

3 5 4 -1 1.5(-)

4 6 9 3 3

5 5 10 5 5

6 1 9 8 7

7 8 9 1 1.5

Sum ranks with positive and with negative values:

Negative = 1.5

Positive = 26.5

T = lower value = 1.5

N 0.05 probability of having at least 9/10 individuals show a

6 0

7 2

8 4

9 6

10 8

11 11

Compare SMALLER value to critical value, at n for number of paired samples. Reject Ho if calculated value is SMALLER THAN critical value, as our is here (1.5 < 2).

Fig. 15.1 probability of having at least 9/10 individuals show a

- Differences Between Means probability of having at least 9/10 individuals show a
- A. 1 Sample
- B. 2 Samples
- C. >2 Samples
- 1. 1 factor
- - One way ANOVA
- - Kruskal-Wallis

Consider 4 groups that are not normally distributed, with 10 values each.

Rank all values across categories.

Sum ranks for categories:

1 2 3 4

162.5 208.5 316.5 132.5

H = 14.273

Use Chi-square distribution,k-1 df.

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