experimental design data analysis

Experimental Design Data analysisGROUP 5

PAIRED T-TESTPRESENTER: TRAN THI NGAN GIANG

Introduction A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample.

For example:

A diagnostic test was made before studying a particular module and then again after completing the module. We want to find out if, in general, our teaching leads to improvements in students’ knowledge/skills.

First, we see the descriptive statistics for both variables.

The post-test mean scores are higher.

Next, we see the correlation between the two variables.

There is a strong positive correlation. People who did well on the pre-test also did well on the post-test.

Finally, we see the T, degrees of freedom, and significance.

Our significance is .053

If the significance value is less than .05, there is a significant difference.If the significance value is greater than. 05, there is no significant difference.

Here, we see that the significance value is approaching significance, but it is not a significant difference. There is no difference between pre- and post-test scores. Our test preparation course did not help!

INDEPENDENT SAMPLES T-TESTSPRESENTER: DINH QUOC MINH DANG

Outline

1. Introduction

2. Hypothesis for the independent t-test

3. What do you need to run an independent t-test?

4. Formula

5. Example (Calculating + Reporting)

Introduction The independent t-test, also called the two sample t-test or student's t-test is an inferential statistical test that determines whether there is a statistically significant difference between the means in two unrelated groups.

Hypothesis for the independent t-test

The null hypothesis for the independent t-test is that the population means from the two unrelated groups are equal:

H0: u1 = u2

In most cases, we are looking to see if we can show that we can reject the null hypothesis and accept the alternative hypothesis, which is that the population means are not equal:

HA: u1 ≠ u2

To do this we need to set a significance level (alpha) that allows us to either reject or accept the alternative hypothesis. Most commonly, this value is set at 0.05.

What do you need to run an independent t-test?

In order to run an independent t-test you need the following:

1. One independent, categorical variable that has two levels.

2. One dependent variable

Formula

M: mean (the average score of the group)

SD: Standard Deviation

N: number of scores in each group

Exp: Experimental Group

Con: Control Group

Formula

Example

Example

Effect Size Cohen’s d measures difference in means in standard deviation units.

Cohen’s d = Interpretation:

◦ Small effect: d = .20 to .50 ◦ Medium effect: d = .50 to .80◦ Large effect: d = .80 and higher

Reporting the Result of an Independent T-Test When reporting the result of an independent t-test, you need to include the t-statistic value, the degrees of freedom (df) and the significance value of the test (P-value). The format of the test result is: t(df) = t-statistic, P = significance value.

Example result (APA Style) An independent samples T-test is presented the same as the one-sample t-test:

t(75) = 2.11, p = .02 (one –tailed), d = .48

Example: Survey respondents who were employed by the federal, state, or local government had significantly higher socioeconomic indices (M = 55.42, SD = 19.25) than survey respondents who were employed by a private employer (M = 47.54, SD = 18.94) , t(255) = 2.363, p = .01 (one-tailed).

Degrees of freedom Value of

statisticSignificance of statistic

Include if test is one-tailed

Effect size if available

Analysis of Variance (ANOVA)PRESENTER : MINH SANG

Introduction We already learned about the chi square test for independence, which is useful for data that is measured at the nominal or ordinal level of analysis.

If we have data measured at the interval level, we can compare two or more population groups in terms of their population means using a technique called analysis of variance, or ANOVA.

Completely randomized designPopulation 1 Population 2….. Population k

Mean = 1 Mean = 2 …. Mean = k

Variance=12 Variance=2

2 … Variance = k2

We want to know something about how the populations compare. Do they have the same mean? We can collect random samples from each population, which gives us the following data.

Completely randomized designMean = M1 Mean = M2 ..… Mean = Mk

Variance=s12 Variance=s2

2 …. Variance = sk2

N1 cases N2 cases …. Nk cases

Suppose we want to compare 3 college majors in a business school by the average annual income people make 2 years after graduation. We collect the following data (in $1000s) based on random surveys.

Completely randomized designAccounting Marketing Finance

27 23 48

22 36 35

33 27 46

25 44 36

38 39 28

29 32 29

Completely randomized designCan the dean conclude that there are differences among the major’s incomes?

Ho: 1 = 2 = 3

HA: 1 2 3

In this problem we must take into account:

1) The variance between samples, or the actual differences by major. This is called the sum of squares for treatment (SST).

Completely randomized design2) The variance within samples, or the variance of incomes within a single major. This is called the sum of squares for error (SSE).

Recall that when we sample, there will always be a chance of getting something different than the population. We account for this through #2, or the SSE.

F-StatisticFor this test, we will calculate a F statistic, which is used to compare variances.

F = SST/(k-1)

SSE/(n-k)

SST=sum of squares for treatment

SSE=sum of squares for error

k = the number of populations

N = total sample size

F-statisticIntuitively, the F statistic is:

F = explained variance

unexplained variance

Explained variance is the difference between majors

Unexplained variance is the difference based on random sampling for each group (see Figure 10-1, page 327)

Calculating SSTSST = ni(Mi - )2

= grand mean or = Mi/k or the sum of all values for all groups divided by total sample size

Mi = mean for each sample

k= the number of populations

Calculating SSTBy major

Accounting M1=29, n1=6

Marketing M2=33.5, n2=6

Finance M3=37, n3=6

= (29+33.5+37)/3 = 33.17

SST = (6)(29-33.17)2 + (6)(33.5-33.17)2 + (6)(37-33.17)2 = 193

Calculating SSTNote that when M1 = M2 = M3, then SST=0 which would support the null hypothesis.

In this example, the samples are of equal size, but we can also run this analysis with samples of varying size also.

Calculating SSESSE = (Xit – Mi)2

In other words, it is just the variance for each sample added together.

SSE = (X1t – M1)2 + (X2t – M2)2 +

(X3t – M3)2

SSE = [(27-29)2 + (22-29)2 +…+ (29-29)2]

+ [(23-33.5)2 + (36-33.5)2 +…]

+ [(48-37)2 + (35-37)2 +…+ (29-37)2]

SSE = 819.5

Statistical OutputWhen you estimate this information in a computer program, it will typically be presented in a table as follows:

Source of df Sum of Mean F-ratio

Variation squares squares

Treatment k-1 SST MST=SST/(k-1) F=MST

Error n-k SSE MSE=SSE/(n-k) MSE

Total n-1 SS=SST+SSE

Calculating F for our exampleF = 193/2

819.5/15

F = 1.77

Our calculated F is compared to the critical value using the F-distribution with

F, k-1, n-k degrees of freedom

k-1 (numerator df)

n-k (denominator df)

The ResultsFor 95% confidence (=.05), our critical F is 3.68 (averaging across the values at 14 and 16

In this case, 1.77 < 3.68 so we must accept the null hypothesis.

The dean is puzzled by these results because just by eyeballing the data, it looks like finance majors make more money.

The ResultsMany other factors may determine the salary level, such as GPA. The dean decides to collect new data selecting one student randomly from each major with the following average grades.

New dataAverage Accounting Marketing Finance M(b)

A+ 41 45 51 M(b1)=45.67

A 36 38 45 M(b2)=39.67

B+ 27 33 31 M(b3)=30.83

B 32 29 35 M(b4)=32

C+ 26 31 32 M(b5)=29.67

C 23 25 27 M(b6)=25

M(t)1=30.83 M(t)2=33.5 M(t)3=36.83

= 33.72

Randomized Block DesignNow the data in the 3 samples are not independent, they are matched by GPA levels. Just like before, matched samples are superior to unmatched samples because they provide more information. In this case, we have added a factor that may account for some of the SSE.

Two way ANOVANow SS(total) = SST + SSB + SSE

Where SSB = the variability among blocks, where a block is a matched group of observations from each of the populations

We can calculate a two-way ANOVA to test our null hypothesis. We will talk about this next week.