HFS 3283 independent t test

HFS3283INDEPENDENT T-TEST

DR. SHARIFAH WAJIHAH WAFA BTE SST WAFASchool of Nutrition and Dietetics

Faculty of Health [email protected]

KNOWLEDGE FOR THE BENEFIT OF HUMANITY

SCHOOL OF NUTRITION AND DIETETICS . FACULTY OF HEALTH SCIENCES

Topic Learning Outcomes

At the end of this lecture, the student should be able to:

1• Understand structure of research study appropriate

for independent-measures t hypothesis test

2• Test between two populations or two treatments

using independent-measures t statistics

3• Understand how to evaluate the assumptions

underlying this test


Independent-measures Design Introduction

• Most research studies compare two (or more) sets of data– Data from two completely different, independent

participant groups (an independent-measures or between-subjects design)

– Data from the same or related participant group(s) (a within-subjects or repeated-measures design)


Independent-Measures Design Introduction (continued)

• Computational procedures are considerably different for the two designs

• Each design has different strengths and weaknesses

• Consequently, only between-subjects designs are considered in this lecture; repeated-measures designs will be reserved for discussion in Next Lecture


When to use the independent samples t-test

• It is used to compare differences between separate groups.

• This test can be used to explore differences in naturally occurring groups.

• For example, we may be interested in differences of IQ between males and females.


When to use the independent samples t-test (cont.)

• Any differences between groups can be explored with the independent t-test, as long as the tested members of each group are reasonably representative of the population. [1]

[1] There are some technical requirements as well. Principally, each variable must come from a

normal (or nearly normal) distribution.


Independent-Measures Design t Statistic

• Null hypothesis for independent-measures test

• Alternative hypothesis for the independent- measures test

0: 210 H

0: 211 H


Example 1

• Suppose we put people on 2 diets: the nasi lemak diet and the roti canai

diet.

• Participants are randomly assigned to either 1-week of breakfast eating exclusively nasi (NL) lemak or 1-week of exclusively roti canai (RC).


Example 1-con’t

• At the end of the week, we measureweight gain by each participant.

• Which diet causes more weight gain?

• In other words, the null hypothesis is:

Ho: wt. gain NL diet =wt. gain RC diet.


Example 1-con’t

• Why? • The null hypothesis is the opposite of what we

hope to find. • In this case, our research hypothesis is that

there ARE differences between the 2 diets. • Therefore, our null hypothesis is that there are

NO differences between these 2 diets.


Example 1-con’t Column 3 Column 4

X1 : NL X2 : RC1 3 1 12 4 0 02 4 0 02 4 0 03 5 1 12 4

0.4 0.4

211 )( 2

22 )(

1 2

n

sx2

2 )(


Example 1-con’t

• The first step in calculating the independent samples t-test is to calculate the variance and mean in each condition.

• In the previous example, there are a total of 10 people, with 5 in each condition.

• Since there are different people in each condition, these “samples” are “independent” of one another;

giving rise to the name of the test.


Example 1-con’t

• The variances and means are calculated separately for each condition (NL and RC).

• A streamlined calculation of the variance for each condition was illustrated previously. (See Slide 11.)

• In short, we take each observed weight gain for the NL condition, subtract it from the mean gain of the NL dieters ( 2) and square the result (see column 3).

1


Example 1-con’t

• Next, add up column 3 and divide by the number of participants in that condition (n1 = 5) to get the sample variance,

• The same calculations are repeated for the “RC” condition.

4.02 xs


Formula

The formula for the independent samples t-test is:

, df = (n1-1) + (n2-1)

11 2

2

1

221

21

nS

nS

txx


Example 1 (cont.)

• Where = Mean of first sample = Mean of second samplen1 = Sample size (i.e., number of observations) of first samplen2 = Sample size (i.e., number of observations) of second sample = sample variance of first sample = sample variance of second samplesp = Pooled standard deviation

21xs22xs

1

2


Example 1 (cont.)

• From the calculations previously, we have everything that is needed to find the “t.”

47.4

44.

44.42

t , df = (5-1) + (5-1) = 8

• After calculating the “t” value, we need to know if it is large enough to reject the null hypothesis.


• The “t” is calculated under the assumption, called the null hypothesis,

“that there are no differences between the NL and RC diet”.

• If this were true, when we repeatedly sample 10 people from the population and put them in our 2 diets, most often we would calculate a “t” of “0.”

Some theory


• Look again at the formula for the “t”. • Most often the numerator (X1-X2) will be “0,”

because the mean of the two conditions should be the same under the null hypothesis.

• That is, weight gain is the same under both the NL and RC diet.

Some theory - Why?


• Sometimes the weight gain might be a bit higher under the NL diet, leading to a positive “t” value.

• In other samples of 10 people, weight gain might be a little higher under the RC diet, leading to a negative “t” value.

• The important point, however, is that under the null hypothesis we should expect that most “t” values that we compute are close to “0.”

Some theory - Why (cont.)


• Our computed t-value is not “0,” but it is in fact negative (t(8) = -4.47).

• Although the t-value is negative, this should not bother us.

• Remember that the t-value is only - 4.47 because we named the NL diet X1 and the RC diet X2.

– This is, of course, completely arbitrary. • If we had reversed our order of calculation, with the

NL diet as X2 and the RC diet as X1, then our calculated t-value would be positive 4.47.

Some theory - Why (cont.)


• The calculated t-value is 4.47 (notice, I’ve eliminated the unnecessary “-“ sign), and the degrees of freedom are 8.

• In the research question we did not specify which diet should cause more weight gain, therefore this t-test is a so-called “2-tailed t.”

Example 1 (again) Calculations


Degrees of freedom

• Degrees of freedom (df) for t statistic isdf for first sample + df for second sample

)1()1( 2121 nndfdfdf

Note: this term is the same as the denominator of the pooled variance


• In the last step, we need to find the critical value for a 2-tailed “t” with 8 degrees of freedom.

• This is available from tables that are in the back of any Statistics textbook.

• Look in the back for “Critical Values of the t-distribution,” or something similar.

• The value you should find is: C.V. t(8), 2-tailed = 2.31.



• The calculated t-value of 4.47 is larger in magnitude than the C.V. of 2.31, therefore we can reject the null hypothesis.

• Even for a results section of journal article, this language is a bit too formal and general. It is more important to state the research result, namely:

Participants on the RC diet (M = 4.00) gained significantly more weight than those on the NL diet (M = 2.00), t(8) = 4.47, p < .05 (two-tailed).



Repeat from previous slide:Participants on the RC diet (M = 4.00) gained significantly more weight than those on the NL diet (M = 2.00), t(8) = 4.47, p < .05 (two-tailed).

• Making this conclusion requires inspection of the mean scores for each condition (NL and RC).

Example 1 (concluding comment)


Example 1 Using SPSS

• First, the variables must be setup in the SPSS data editor.

• We need to include both the independent and dependent variables.

• Although it is not strictly necessary, it is good practice to give each person a unique code

(e.g., personid):


Example 1 Using SPSS (cont.)

• In the previous example:– Dependent Variable

= wtgain (or weight gain)– Independent Variable = diet

• Why? • The independent variable (diet) causes

changes in the dependent variable (weight gain).



• Next, we need to provide “codes” that distinguish between the 2 types of diets.

• By clicking in the grey box of the “values” field in the row containing the “diet” variable, we get a pop-up dialog that allows us to code the diet variable.

• Arbitrarily, the NL diet is coded as diet “1” and the RC diet is diet “2.”

• Any other 2 codes would work, but these suffice

See next slide.



.



• Moving to the data view tab of the SPSS editor, the data is entered.

• Each participant is entered on a separate line; a code is entered for the diet they were on (1 = NL, 2 = RC); and the weight gain of each is entered, as follows











• Click Analyze > Compare Means > Independent-Samples T Test... on the top menu, as shown below:



• You will be presented with the Independent-Samples T Test dialogue box.

• Transfer the dependent variable, wtgain, into the Test Variable(s): box, and transfer the independent variable, diet, into the Grouping Variable: box, by highlighting the relevant variables and pressing the SPSS Right Arrow Button buttons. You will end up with the following screen in the next slide



You then need to define the groups (diet). Click on the button. You will be presented with the Define Groups dialogue box, as shown above:



• Enter 1 into the Group 1: box and enter 2 into the Group 2: box. Remember that we labelled the Nasi Lemak group as 1 and the Roti Canai group as 2.

• Click the button.



• Two sections (boxes) appear in the output: Group Statistics– provides basic information about the group comparisons,

including the sample size (n), mean, standard deviation, and standard error for weight gain by group.



• SPSS gives the means for each of the conditions (NL Mean = 2 and RC Mean = 4).

• In addition, SPSS provides a t-value of -4.47 with 8 degrees of freedom.

• These are the same figures that were computed “by hand” previously.

• However, SPSS does not provide a critical value. • Instead, an exact probability is provided (p

= .002).



• The second section, Independent Samples Test, displays the results most relevant to the Independent Samples t Test. There are two parts that provide different pieces of information: (A) Levene’s Test for Equality of Variances and (B) t-test for Equality of Means.



Levene's Test for Equality of of Variances: This section has the test results for Levene's Test. From left to right:• F is the test statistic of Levene's test• Sig. is the p-value corresponding to this test statistic.The p-value of Levene's test is printed as “1.000“, so we ACCEPT the null of Levene's test and conclude that the variance in weight gain of NL diet is identical to the RC diet .This tells us that we should look at the "Equal variances assumed" row for the t-test results. (If this test result had been significant -- that is, if we had observed p < α -- then we would have used the "Equal variances not assumed" output.)

A



t-test for Equality of Means provides the results for the actual Independent Samples t Test. From left to right:• t is the computed test statistic• df is the degrees of freedom• Sig (2-tailed) is the p-value corresponding to the given test statistic and degrees

of freedom• Mean Difference is the difference between the sample means; it also

corresponds to the numerator of the test statistic• Std. Error Difference is the standard error; it also corresponds to the

denominator of the test statistic

B

B



• The mean difference is calculated by subtracting the mean of the second group from the mean of the first group. In this example, the mean weight gain for RC diet was subtracted from the mean weight gain for NL diet (2.00- 4.00 = -2.00). The sign of the mean difference corresponds to the sign of the t value.

• The negative t value in this example indicates that the mean weight gain for the first group, NL diet, is significantly lower than the mean for the second group, RC diet.

• The associated p value is printed as ".002".

B

B



• Confidence Interval of the Difference: This part of the t-test output complements the significance test results. Typically, if the CI for the mean difference contains 0, the results are not significant at the chosen significance level. In this example, the CI is [-3.03128, -0.96872], which does not contain zero; this agrees with the small p-value of the significance test.

C

C


In the Literature

• Report whether the difference between the two groups was significant or not

• Report descriptive statistics (M and SD) for each group

• Report t statistic and df• Report p-value• Report CI immediately after t


In the Literature

Participants on the RC diet (M = 4.00) gained significantly more weight than

those on the NL diet (M = 2.00), t(8) = 4.47, p < .01 (two-tailed).

Variables NL diet(n=5)

Mean (SD)

RC diet(n=5)

Mean (SD)

Mean diff (95% CI)

t statistics (df)

P-value

Weight gain 2.00 (0.71) 4.00 (0.71) -2.00(-3.03,-0.97) -4.47(8) <0.01

Table 1: Type of diet associated with weight gain



Repeat from previous slide:Participants on the RC diet (M = 4.00) gained significantly more weight than those on the NL diet (M = 2.00), t(8) = 4.47, p < .01 (two-tailed).

• In APA style we normally only display significance to 2 significant digits.

• Therefore, the probability is displayed as “p<.01,” which is the smallest probability within this range of accuracy.


Sample size??

• In each group, sample size is BIG (>30)• In case, the sample size is less than 30 in one group,

so that we need to check for normality assumption.

• How to check?


Normality assumptionA

BC


Normality assumption

C


recap

• FOUR assumptions 1. Random sample2. Observations are independent3. In each group, data are normally distributed or

sample size is big (>30)4. Population variances are not different between

2 groups Levene’s test


AnyQuestio

ns?

Concepts?

Data & Analytics

HFS 3283 independent t test