chapter 8

Chapter 8

T-tests – comparison of Means

How to find the critical value when alpha, α, is given

• Use the t distribution table. If α = .05 and two-tailed test, and the df = 18, what is the critical value?

So divide α by 2.05 /2 = .025Look at the t distribution table at df = 18And under α = .025.t critical, t crit = 2.101

Ho: There are no differences in the mean scores i.e. µ1 = µ2.H1: µ1 ≠ µ2 ( a 2-tailed test) meaning µ1 > µ2 or µ1 < µ2

p = .025p = .025

t = + 2.101t = - 2.101

Reject Null hypothesis


Do not rejectNull hypothesis

If your calculated t value or t obtained = 3.20, it is in the rejection zone, and so youwill reject the null hypothesis and retain the alternative hypothesis of H1

If your calculated t value is = 1.89, it is in the retain (do not reject) zone, so you will not reject the null hypothesis

• Use the t distribution table. If α = .05 and one-tailed test, and the df = 18, what is the critical value?

Ho: There are no differences in the mean scores i.e. µ1 = µ2.H1: µ1 < µ2 ( a 1-tailed test)

p = ,05

t = + 1.734



If your calculated t value or t obtained = 3.20, it is in the rejection zone, and so youwill reject the null hypothesis and retain the alternative hypothesis of H1

If your calculated t value is = 1.20, it is in the retain (do not reject) zone, so you will not reject the null hypothesis

Type I and Type II Error

• Type I error is committed if a true null hypothesis is rejected.

• Type II error is committed when a false null hypothesis is retained.

TRUE STATE OF AFFAIRS Ho is True Ho is False

YOUR DECISION Retain Ho Correct Type II error

Reject Ho Type I error Correct

Testing the Significance of the differences between Means

µ1 µ2

σ1 σ2

N1 N2

X1

S1

n1

X2

S2

n2

Sample 1

Sample 2

Population 1 Population 2

|u1 - µ2|x1 - x2

Mean of sampling Distribution of x1 - x2

25

µ1 - µ2 = 25Samplingdistribution µ1 = 25 µ1 = 27 µ1 = 25 µ1 = 35

The samplingDistribution of the difference of means

|u1 - µ2| = 0|u1 - µ2| = 2

|u1 - µ2| = 10

• The mean of the sampling distribution of the x1-x2 is equal to the difference between the population means u1 - µ2.

• If the population is normally distributed, the sampling distribution will be normal.

• If sample sizes are sufficiently large, the sampling distribution will be normal whether or not the populations are normally distributed

Assumptions of t-tests

• 1) Data must be interval or ratio

• 2) Data must be obtained via random sampling from population

• 3) Data must be normally distributed

• 4) The variance of each group must be similar

Step 1: Define the null and alternative hypotheses:

Ho: µ1 = µ2

H1: µ1 ≠ µ2

Step 2: Set level of significance, alpha. Alpha is set at .05

2121

2221

21 11

2

1121 nnnn

nsnsxx

s

21 xxs Variance Formula for

21

21

xx

obt s

xxt

Step 3: To calculate t:

2121

2

2

222

1

2

121

11

2

_

21 nnnn

n

xx

n

xx

xxs

Computational Formula for 21 xx

s

Example: Test whether there is any significant differences between theMaths marks obtained by Male and Female students

Male Female84 8878 9767 7487 8080 8778 9078 9079 8682 8481 78

X1 = 79.40S1 = 5.25Σx1 = 794Σx1

2 = 63292n1 = 10

X2 = 85.40S2 = 6.69Σx2 = 854Σx2

2 = 73334n2 = 10

79.40 - 85.40t = 63292 – (794) 2 73334 – (854) 2 10 10

10 + 10 - 2

1 + 110 10

t = -2.23

Using the computational formula for

Step 4: Locate the critical value of t from the t distribution table

The degree of freedom is n1 + n2 - 2 10 + 10 – 2 = 18

For the 2-tailed test, alpha for each tail

Enter the left column of the t table at 18 and locate the number in the column alpha = .025, the critical value of t is 2.10

025.2

05.

Step 5: Compare t obtained of (-2.23) with the critical value of 2.10. Since the absolute value of t obtained is greater than the critical value of 2.10, the null hypothesis is rejected and the alternative hypothesis is retained.

p = .025p = .025

t = + 2.10t = - 2.10




-2.23 is in the reject zon

Step 6: Interpret and report your findings There are significant differences in the Maths marks between male and female students. The means indicate that female students (x = 85.40) obtain significantly higher Maths marks than male students (x = 79.40).

4 Assumptions of t-test comparisons

• 1) Random sampling: it is assumed that the samples are representative of the populations from which they are drawn.

• 2) The populations from which the samples are taken are normally distributed. This assures that the sampling distributions for x1 and x2 are normally distributed. As such the sampling distribution of the differences of the means are also normally distributed. If one of the sample is not normally distributed, the assumption is violated unless the total sample size is sufficiently large (more than 30). The differences of means sampling distribution approach normal distribution as the sample size increases

• 3) Homogeneity of variance – the variances of the populations area equal

• 4) Independent observations – the scores within each sample are independent of one another – meaning each subject or respondent supplies only one score.

Effect Size for t-test

• What is Effect size for a t-test analysis?

• It is the standardized measure of the effect of one variable (say Gender) on another variable (say Maths marks).

•

Effect Size

221

___

21

__

ssXX

EffectSize

X1 = 15.08 s1 = 4.05 X2 = 14.36 s2 = 3.63

Example:

Result: Effect Size (Cohen’s d) = .1875 (Small effect size)

Note: Effect size ~ .3 small; ~ .5 (medium); ~ .8 (high)

1875.84.3

72.0

263.305.436.1408.15

EffectSize

Effect Size measured by Cohen’s d

Cohen’ d Interpretation ~ .2 Small~ .5 Moderate~ .8 Large

Presentation of t-test results using the APA format

Table 2

T-tests comparisons of CRA scores by gender

Father Mother

Mean

SD

15.06 14.36

4.05 3.63

t-value p < .05

5.38 NS

(n =13) (n =12) EffectSize

.18

Power of a test

• Power of a statistical test is the probability of observing a treatment effect when it occurs.

• It is the probability that it will correctly lead to the rejection of a false null hypothesis (Green, 2000)

• The statistical power is the ability of the test to detect an effect if it actually exists (High, 2000)

• The statistical power is denoted by 1 – β, where β is the Type II error, the probability of failing to reject the null hypothesis when it is false.

• Conventionally, a test with a power greater than .8 level (or β = < .2) is considered statistically powerful.

α = is the probability of rejecting the true null hypothesis (Type I error)

β = is the probability of not rejecting the false null hypothesis (Type II error)

There are four components that influence the power of a test:

• 1) Sample size, or the number of units (e.g., people) accessible to the study

• 2) Effect size, the difference between the means, divided by the standard deviation (i.e. 'sensitivity')

• 3) Alpha level (significance level), or the probability that the observed result is due to chance

• 4) Power, or the probability that you will observe a treatment effect when it occurs

Usually, experimenters can only change the sample size (population) of the study and/or the alpha value

Exercise 1: The following are the Statistics marks obtained for 2 levels of anxiety. High Anxiety Low Anxiety x1 = 4.2 x2 = 2.2

s12 = .5 s2

2 = .7

n1 = 10 n2 = 10

1) Specify the null and alternative hypothesis2) Test the null hypothesis with α = .05. What is the tcrit for α = .05, two-tailed

test?3) Perform the t test4) Interpret the findings5) Suppose the variance is increased so that s1

2=5.2 and s22 = 5.4

a) perform a t test with α = .05 b) Interpret the findings. c) What effect has increasing the variability had on tobt and the final conclusion about the null hypothesis?6) Write your report in the APA format together with the APA table.

Exercise 2: Test whether there is any significant differencesbetween the Motivation levels of Form 5A and 5F students at α = .05.Write your report in the APA format together with the APA table.

Form 5A Form 5F27 2439 2825 2333 2521 2435 2230 2926 2625 2931 2835 1930 29 28

X1 = S1 = Σx1 = Σx1

2 = n1 =

X2 = S2 = Σx2 = Σx2

2 = n2 =

How to test the significance of the differences of means for dependent Samples – Use Paired t-test

• Example: You wish to test whether there are significant differences in the Maths scores of a class before and after the treatment or intervention (eg special math tuition) at α = .05

• The same students are involved.

• So this comparison is called comparison of the means of dependent samples

Paired t-test

• Assumptions 1) Normality of the population difference of

scores – this is ascertained by ensuring the normality of each variable separately.

2) the other assumptions similar to group t – test a) Data must be interval or ratio b) Data must be obtained via random sampling from population c) Data must be normally distributed

Example: Maths scores Maths scores D D2 Before tuition after tuition 5 3 2 4 6 4 2 4 7 7 0 0 6 7 -1 1

ΣD = 3 ΣD2 = 9

Step 2: Compute SD

s D = ΣD2 - (ΣD)2 / np

np - 1

sD = 9 - (3)2 / 4 = 6.75 = 1.50 4 – 1 3

Step 1: State the null hypothesis and alternate hypothesis

Step 3: Substitute sD in the following formula:

sD = sD

np

= 1.50 = .75

4

Step 4: Substitute into the following formula to calculate t

t = x – y with df = np - 1 sD

t = 1.45

Example: t-test comparison for paired data

Maths Maths D D2

Pretest Posttest

8 4 4 16 12 8 4 16 6 2 4 6 11 8 15 9 8 5 7 4

x = 8.88 y = 5.75 ΣD = 25 ΣD2= 115

SD = SD

np

= 2.30 8

= .81

sD == ΣD2 - [(ΣD)2/ np] np – 1 = 2.30

tobt = 8.88 – 5.75 .81t obt = 3.86 df = np – 1T obt = 2.365Since 3.86 > 2.365, Reject Ho: µx = µy

Retain H1: µx ≠ µy

tcrit = +2.36

.025

tobt =+3.36

.025

0

tcrit = -2.36

Interpretation:

Table 1 shows that there is a significant difference between the Maths pretest and Maths posttest mean scores, t (7) = 3.86, p < .05. The means indicate that MathsPosttest mean score are significantly lower (y = 5.75) than the Maths PrestestMean score (x = 8.88). This indicates that that the treatment (tuition) has a significant negative effect on the students’ performance in Mathematics!

Insert your APA Table 1 here

Exercise 3:

The following are the Motivation scores before and after a MotivationProgram. Analyse the data and ascertain whether the Motivation programIs effective in enhancing the level of motivation of the students.

Pretest Posttest95 8088 7069 8879 89 77 8767 5645 6757 8187 6759 67

(a) State the null and the alternative hypothesis(b) What is the tobt?(c) What is the tcrit?(d) Interprete the findings.

chapter 8

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