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John LohmanMichael Mattocks

Aubrey Urwick

Chapter : 10Independent Samples t Test

Key Terms: Don’t Forget Notecards

Hypothesis Test (p. 233) Null Hypothesis (p. 236) Alternative Hypothesis (p. 236) Alpha Level (level of significance) (pp. 238 & 245) Critical Region (p. 238) Estimated Standard Error (p. 286) t statistic (p. 286) Degrees of Freedom (p. 287) t distribution (p. 287) Confidence Interval (p. 300) Directional (one-tailed) Hypothesis Test (p. 304) Independent-measures Research Design (p.318)

Formulas Estimated Standard Error: Estimated Standard Error: Pooled Variance: t-Score Formula: Degrees of Freedom:

n1 = n2

n1 ≠ n2

More Formulas Cohen’s d:

Confidence Interval: Hartley’s F-max Test:

Identifying the Independent-Measures Design

Question 1: What is the defining characteristic of an independent-measures research study?

Identifying the Independent-Measures Design

Question 1 Answer: An independent-measures study uses a separate group of

participants to represent each of the populations or treatment conditions being compared.

Pooled Variance and Estimated Standard Error

Question 2: One sample from an independent-measures study has n = 4 with SS = 100. The other sample has n = 8 and SS = 140.

a) Compute the pooled variance for the sample.b) Compute the estimated standard error for the mean difference.

Pooled Variance and Estimated Standard Error

Question 2 Answer:

a)

t Test for Two Independent Samples -- Two-tailed Example

Question 3: A researcher would like to determine whether access to computers has an effect on grades for high school students. One group of n = 16 students has home room each day in a computer classroom in which each student has a computer. A comparison group of n = 16 students has home room in a traditional classroom. At the end of the school year, the average grade is recorded for each student The data are as follows:

Computer TraditionalM = 86 M = 82.5

SS = 1005 SS = 1155

t Test for Two Independent Samples -- Two-tailed Example

Question 3:a) Is there a significant difference between the two groups? Use

a two-tailed test with α = 0.05.b) Compute Cohen’s d to measure the size of the difference.c) Compute the 90% confidence interval for the population mean

difference between a computer classroom and a regular classroom.

t Test for Two Independent Samples -- Two-tailed Example

Question 3a Answer: Step 1: State hypotheses

H0: Treatment has no effect. ( = 0) H1: Treatment has an effect. ( ≠ 0)

Step 2: Set Criteria for Decision (α = 0.05) Critical t = ± 2.042

df = 16 + 16 – 2 = 30

If -2.042 ≤ tsample ≤ 2.042, fail to reject H0

If tsample < -2.042 or tsample > 2.042, reject H0

t Test for Two Independent Samples -- Two-tailed Example

t = - 2.042 t = + 2.042

df = 30

t Distribution with α = 0.05 Critical regionCritical region

t Test for Two Independent Samples -- Two-tailed Example

Question 3a Answer: Step 3: Compute sample statistic

a) b)

Two-Tailed Hypothesis Test Using the t Statistic

Critical regionCritical region

t = - 2.042 t = + 2.042

df = 30

t Distribution with α = 0.05

t = 1.17

t Test for Two Independent Samples -- Two-tailed Example

Question 3a Answer: Step 4: Make a decision

For a Two-tailed Test:

tsample (1.17) < tcritical (2.042) Thus, we fail to reject the null and cannot conclude that access to

computers has an effect on grades.

If -2.042 ≤ tsample ≤ 2.042, fail to reject H0

If tsample < -2.042 or tsample > 2.042, reject H0

t Test for Two Independent Samples -- Two-tailed Example

Question 3b Answer:a) Cohen’s d: b) This is a small, or small-to-medium, effect.

Magnitude of d Evaluation of Effect Sized = 0.2 Small effect (mean difference around 0.2 standard deviations)

d = 0.5 Medium effect (mean difference around 0.5 standard deviations)

d = 0.8 Large effect (mean difference around 0.8 standard deviations)

t Test for Two Independent Samples -- Two-tailed Example

Question 3c Answer: df = 30 Critical t = ± 1.697

Thus, the population mean difference is estimated to be between – 1.591 and 8.591. The fact that zero is an acceptable value (inside the interval) is consistent with the decision that there is no significant difference between the two population means. (Fail to reject the null)

Our α = 0.10 because our confidence intervalleaves 10% split between the 2-tails.

t Test for Two Independent Samples -- One-tailed Example

A researcher is using an independent-measures design to evaluate the difference between two treatment conditions with n = 8 in each treatment. The first treatment produces M = 63 with a variance of s2 = 18, and the second treatment has M = 58 with s2 = 14.

a) Use a one-tailed test with α = 0.05 to determine whether the scores in the first treatment are significantly greater than scores in the second.

b) Measure the effect size with r2.

t Test for Two Independent Samples -- One-tailed Example

Question 4 Answer: Step 1: State hypotheses

H0: No difference between treatments. () H1: Treatment 1 has a greater effect. (> 0)

Step 2: Set Criteria for Decision (α = 0.05)

Critical t = 1.761

df = 8 + 8 – 2 = 14

If tsample ≤ 1.761, fail to reject H0

If tsample > 1.761, reject H0

t Test for Two Independent Samples -- One-tailed Example

t Distribution with α = 0.05

df = 14

t = + 1.761

Critical region

Because this is a one-tailed test‚there is only one critical region.

t Test for Two Independent Samples -- One-tailed Example Question 4 Answer:

Step 3: Compute sample statistica) b)

t Test for Two Independent Samples -- One-tailed Example

t Distribution with α = 0.05

t = + 1.761

Critical region

df = 14

t = 2.50

Because this is a one-tailed test‚there is only one critical region.

t Test for Two Independent Samples -- One-tailed Example Question 4 Answer:

Step 4: Make a decision For a One-tailed Test:

tsample (2.50) > tcritical (1.761) Thus, we reject the null and conclude that the treatment 1 has a

significantly greater effect.

If tsample ≤ 1.761, fail to reject H0

If tsample > 1.761, reject H0

t Test for Two Independent Samples -- One-tailed Example

Question 4 Answer:

b) This is a large effect.

Percent of Variance Explained as Measured by r2 Evaluation of Effect Sizer2 = 0.01 (0.01*100 = 1%) Small effect

r2 = 0.09 (0.09*100 = 9%) Medium effect

r2 = 0.25 (0.25*100 = 25%) Large effect

Assumptions Underlying the Independent-Measures t Test

Question 5: What three assumptions must be satisfied before you use the independent-measures t formula for hypothesis testing?

Assumptions Underlying the Independent-Measures t Test

Question 5 Answer:1) The observations within each sample must be independent.2) The two populations from which the samples are selected must

be normal.3) The two populations from which the samples are selected must

have equal variances (homogeneity of variance).

Testing for Homogeneity of Variance

Question 6: Suppose that two independent samples each have n = 10 with sample variances of 12.34 and 9.15. Do these samples violate the homogeneity of variance assumption? (Use Hartley’s F-Max Test with α = 0.05)

Testing for Homogeneity of Variance

Question 6 Answer: F-max = df = 10 – 1 = 9 k = 2 Critical levels for α = 0.05 are in regular type;

critical levels for α = 0.01 are in bold type.

Critical F-max = 4.03

Testing for Homogeneity of Variance

Question 6 Answer: Because the obtained F-max value (1.35) is smaller than the

critical value (4.03), we can conclude that the data do not provide evidence that the homogeneity of variance assumption has been violated.

Frequently Asked Questions FAQs

What’s the difference between and ? There seems to be some confusion as to how to work the

formula for pooled variance. In the first example (the correct one), the numerator is the sum of SS1 and SS2. The denominator is the sum of df1 and df2.

In the second example (the incorrect method), we’re dividing, then adding. This yields a completely different, and ultimately wrong, result.

SS = 10; df = 15SS = 5; df = 20

𝑠𝑝2=

𝑆𝑆1+𝑆𝑆2𝑑𝑓 1+𝑑𝑓 2

= 10+515+20

=1535

=0.429

𝑠𝑝2=

𝑆𝑆1

𝑑𝑓 1+𝑆𝑆2

𝑑𝑓 2=1015

+ 520

=0.67+0.25=0.92