4. T-tests Pt2

0.0(0)
studied byStudied by 0 people
full-widthCall with Kai
GameKnowt Play
learnLearn
examPractice Test
spaced repetitionSpaced Repetition
heart puzzleMatch
flashcardsFlashcards
Card Sorting

1/38

encourage image

There's no tags or description

Looks like no tags are added yet.

Study Analytics
Name
Mastery
Learn
Test
Matching
Spaced

No study sessions yet.

39 Terms

1
New cards

What defines an independent-samples design, and how is it used in psychology?

  • Subjects in each group are not matched.

  • Groups are independent (e.g., control vs experimental).

  • Typical research process:

    1. Randomly sample subjects.

    2. Randomly assign to control/experimental group.

    3. Compare group means on the dependent variable.

  • Used widely in between-subjects experiments.

2
New cards

How else can independent samples be formed besides random assignment?

  • Randomly sample from pre-existing groups.

    • Example: only-children vs first-borns (with younger siblings).

  • Steps:

    1. Select participants from each group.

    2. Administer test of interest.

    3. Compare means between groups.

  • Rationale: compare natural categories when manipulation isn’t possible.

  • Think-pair-share question: Why only “oldest child” vs “only child”? → To isolate sibling-effect without mixing birth-order confounds.

3
New cards

What are the three main types of t-tests and when are they used?

  • One-sample t-test: Compare sample mean X-bar to known population mean μ

  • Dependent (paired) t-test: Compare mean of differences D-bar between matched or repeated measures.

  • Independent t-test: Compare means of two independent groups X1 - X2

  • All use:

    • Numerator = observed difference.

    • Denominator = standard error of that difference.

    • Logic: signal (effect) ÷ noise (variation).

<ul><li><p><strong>One-sample t-test</strong>: Compare sample mean X-bar to known population mean μ</p></li><li><p><strong>Dependent (paired) t-test</strong>: Compare mean of differences D-bar between matched or repeated measures.</p></li><li><p><strong>Independent t-test</strong>: Compare means of two independent groups X1 - X2</p></li><li><p>All use:</p><ul><li><p><strong>Numerator</strong> = observed difference.</p></li><li><p><strong>Denominator</strong> = standard error of that difference.</p></li><li><p><strong>Logic</strong>: signal (effect) ÷ noise (variation).</p></li></ul></li></ul><p></p>
4
New cards

What does the denominator represent in an independent-sample t-test?

  • It’s the variance of the difference in sample means.

  • Equivalent to: Standard Error of the Difference (SED).

  • This is the SD of the sampling distribution of differences between two independent group means.

  • Formula (general):

5
New cards

What is the denominator in the independent t-test called, and why is it important?

  • Standard error of the difference between means (SED).

  • Captures how much difference we expect by chance.

  • Smaller SE → easier to detect real group differences.

  • Denominator is the key to making inference valid.

<ul><li><p><strong>Standard error of the difference between means (SED)</strong>.</p></li><li><p>Captures how much difference we expect <strong>by chance</strong>.</p></li><li><p>Smaller SE → easier to detect real group differences.</p></li><li><p>Denominator is the key to making inference valid.</p></li></ul><p></p>
6
New cards

How does the variance sum law apply to the independent-samples t-test?

  • Law: Var(X ± Y) = Var(X) + Var(Y), if X & Y are independent.

  • Applied to means:

  • Taking square root gives standard error of difference.

  • Explains why both group variances contribute to denominator.

<ul><li><p>Law: Var(X ± Y) = Var(X) + Var(Y), if X &amp; Y are independent.</p></li><li><p>Applied to means:</p><img src="https://knowt-user-attachments.s3.amazonaws.com/2d391f07-1b43-416e-9eca-f48f9264135e.png" data-width="75%" data-align="center"></li></ul><ul><li><p>Taking square root gives <strong>standard error of difference</strong>.</p></li><li><p>Explains why both group variances contribute to denominator.</p></li></ul><p></p>
7
New cards

How do we compute the variance of the difference in sample means?

8
New cards

What is the formula for the independent t-statistic using group variances?

  • Numerator = mean difference.

  • Denominator = standard error of difference.

  • Shows difference in relation to variability.

9
New cards

How do we set up an independent t-test with two groups (School A vs School B)?

  • School A: X-bar=45, s=6, n=25

  • School B: X-bar=41, s=5, n=25

  • Steps:

    1. State hypotheses: H0:μ1=μ and H1:μ1≠μ2

    2. Compute mean difference = 4.

    3. Compute SE using formula.

    4. Calculate t.

    5. Compare with critical value or p-value.

<ul><li><p>School A: X-bar=45, s=6, n=25</p></li><li><p>School B:&nbsp;X-bar=41, s=5, n=25</p></li><li><p>Steps:</p><ol><li><p>State hypotheses: H0:μ1=μ and H1:μ1≠μ2</p></li><li><p>Compute mean difference = 4.</p></li><li><p>Compute SE using formula.</p></li><li><p>Calculate t.</p></li><li><p>Compare with critical value or p-value.</p></li></ol></li></ul><p></p>
10
New cards

What does the result of the happiness study tell us?

  • Obtained statistic: t(48)=2.56

  • p = 0.013 (two-tailed).

  • Interpretation: Reject H0

  • Students’ happiness differs significantly between schools.

<ul><li><p>Obtained statistic: t(48)=2.56</p></li><li><p>p = 0.013 (two-tailed).</p></li><li><p>Interpretation: Reject H0</p></li><li><p>Students’ happiness <strong>differs significantly</strong> between schools.</p></li></ul><p></p>
11
New cards

What is the distribution of differences between means?

  • It’s the sampling distribution of X1-X2

  • Center = μ1−μ2

  • Shape ≈ normal (by CLT).

  • Spread = SE of difference.

  • Important: this is the theoretical distribution against which our observed difference is tested.

<ul><li><p>It’s the sampling distribution of X1-X2</p></li><li><p>Center = μ1−μ2</p></li><li><p>Shape ≈ normal (by CLT).</p></li><li><p>Spread = SE of difference.</p></li><li><p>Important: this is the <strong>theoretical distribution</strong> against which our observed difference is tested.</p></li></ul><p></p>
12
New cards

How do we get from populations to the distribution of mean differences?

  • Process:

    1. Sample repeatedly from population 1 → distribution of means.

    2. Sample repeatedly from population 2 → distribution of means.

    3. Subtract group means for each pair of samples.

    4. Result = distribution of differences between sample means.

  • That’s the reference distribution for the independent t-test.

<ul><li><p>Process:</p><ol><li><p>Sample repeatedly from <strong>population 1</strong> → distribution of means.</p></li><li><p>Sample repeatedly from <strong>population 2</strong> → distribution of means.</p></li><li><p>Subtract group means for each pair of samples.</p></li><li><p>Result = distribution of differences between sample means.</p></li></ol></li><li><p>That’s the reference distribution for the independent t-test.</p></li></ul><p></p>
13
New cards

What defines the distribution of differences between means in t-tests (independent samples t-test)?

  • Center: observed mean difference X1-X2

  • Width: standard error sX1-X2

  • Test statistic:

<ul><li><p>Center: observed mean difference X1-X2</p></li><li><p>Width: standard error s<sub>X1-X2</sub></p></li><li><p>Test statistic:</p><img src="https://knowt-user-attachments.s3.amazonaws.com/1f3418d3-7ad1-45e2-9ccf-08fb25775f36.png" data-width="25%" data-align="center"></li></ul><p></p>
14
New cards

What is the general form of the independent t-test?

<img src="https://knowt-user-attachments.s3.amazonaws.com/75fac574-3c56-4854-9778-9e939082e1d3.png" data-width="100%" data-align="center"><p></p>
15
New cards

Why do we use pooled variance when sample sizes are unequal?

  • Unequal n → SE formula can overweight one group’s variance.

  • Solution: use pooled variance (weighted average of group variances).

  • Weights each variance by its df, giving fairer estimate of true variance.

Formula:

<ul><li><p>Unequal n → SE formula can overweight one group’s variance.</p></li><li><p>Solution: use <strong>pooled variance</strong> (weighted average of group variances).</p></li><li><p>Weights each variance by its df, giving fairer estimate of true variance.</p></li></ul><p>Formula:</p><img src="https://knowt-user-attachments.s3.amazonaws.com/950ba203-dd65-45c4-a962-16cbd70c7447.png" data-width="50%" data-align="center"><p></p>
16
New cards

When should we use pooled variance?

  • Use if:

    • S2≈ S2(similar variances, less than 4× different).

    • Especially when n1≠n2

  • Don’t pool if:

    • Variances differ a lot (≥4×).

  • If n1=n2, pooled variance = simple average of group variances.

<ul><li><p>Use if:</p><ul><li><p>S<sup>2</sup><sub>1&nbsp;</sub>≈ S<sup>2</sup><sub>2&nbsp;</sub>(similar variances, less than 4× different).</p></li><li><p>Especially when n1≠n2</p></li></ul></li><li><p>Don’t pool if:</p><ul><li><p>Variances differ a lot (≥4×).</p></li></ul></li><li><p>If n1=n2, pooled variance = simple average of group variances.</p></li></ul><p></p>
17
New cards

How does the independent t-test formula change with pooled variance?

Replace each variance with pooled variance:

  • If variances are similar and n1≠n2​ → use pooled.

  • If variances are too different → don’t pool.

  • If variances similar & n1=n2​ → either method works (same result).

<p>Replace each variance with pooled variance:</p><img src="https://knowt-user-attachments.s3.amazonaws.com/c057c46c-98ee-46ef-9479-4507f692e551.png" data-width="25%" data-align="center"><ul><li><p>If variances are similar and n1≠n2​ → use pooled.</p></li><li><p>If variances are too different → don’t pool.</p></li><li><p>If variances similar &amp; n1=n2​ → either method works (same result).</p></li></ul><p></p>
18
New cards

What happens when sample sizes are equal?

Pooled variance simplifies to:

  • Just the average variance.

  • With equal n, pooled and non-pooled give identical t.

<p>Pooled variance simplifies to:</p><img src="https://knowt-user-attachments.s3.amazonaws.com/a8f2a782-b959-4367-8c9c-517c08cd6f88.png" data-width="25%" data-align="center"><ul><li><p>Just the <strong>average variance</strong>.</p></li><li><p>With equal n, pooled and non-pooled give identical t.</p></li></ul><p></p>
19
New cards

How is the DrugX experiment set up for an independent t-test?

  • IV: Drug condition (DrugX vs Placebo).

  • DV: Number of maze errors (lower = better).

  • Group 1 (DrugX): X-bar = 14, n=9

  • Group 2 (Placebo): X-bar=16, n=10

  • H0:μ1=μ2

  • H1:μ1≠μ2

<ul><li><p><strong>IV</strong>: Drug condition (DrugX vs Placebo).</p></li><li><p><strong>DV</strong>: Number of maze errors (lower = better).</p></li><li><p>Group 1 (DrugX): X-bar = 14, n=9</p></li><li><p>Group 2 (Placebo): X-bar=16, n=10</p></li><li><p>H0:μ1=μ2</p></li><li><p>H1:μ1≠μ2</p></li></ul><p></p>
20
New cards

What are the null and alternative hypotheses for DrugX?

  • H0​: μ1=μ2​ → Drug has no effect.

  • H1​: μ1≠μ2​ → Drug does affect learning.

  • Inference: Reject or fail to reject based on t-test outcome.

21
New cards

What formula is used for the DrugX study’s t-test?

22
New cards

What are the key descriptive stats for the DrugX groups?

  • Group 1 (DrugX): n=9, X-bar=14, s2=23.25

  • Group 2 (Placebo): n=10, X-bar=16, s2=15.78

  • Hypotheses:

    • H0: No difference in errors.

    • H1: Difference exists.

  • Data collected: number of maze errors (lower = better).

<ul><li><p>Group 1 (DrugX): n=9, X-bar=14, s<sup>2</sup>=23.25</p></li><li><p>Group 2 (Placebo): n=10, X-bar=16, s<sup>2</sup>=15.78</p></li><li><p>Hypotheses:</p><ul><li><p>H0: No difference in errors.</p></li><li><p>H1: Difference exists.</p></li></ul></li><li><p>Data collected: number of maze errors (lower = better).</p></li></ul><p></p>
23
New cards

Why must pooled variance be used in the DrugX example?

  • Sample sizes unequal (n=9 vs n=10).

  • Variances similar (ratio < 4).

  • Rule: with unequal n and similar variances → must pool.

<ul><li><p>Sample sizes unequal (n=9 vs n=10).</p></li><li><p>Variances similar (ratio &lt; 4).</p></li><li><p>Rule: with unequal n and similar variances → <strong>must pool</strong>.</p></li></ul><p></p>
24
New cards

What are the calculated statistics for the DrugX example?

  • Pooled variance: sp2=19.3

  • Standard error: SE=2.02

  • Observed t: t=−0.99

  • df = 17.

  • Group 1: X-bar=14.

  • Group 2: X-bar=16

<ul><li><p>Pooled variance: sp2=19.3</p></li><li><p>Standard error: SE=2.02</p></li><li><p>Observed t: t=−0.99</p></li><li><p>df = 17.</p></li><li><p>Group 1: X-bar=14. </p></li><li><p>Group 2: X-bar=16</p></li></ul><p></p>
25
New cards

What does the t-test reveal about DrugX’s effect on learning?

  • tobs=−0.99

  • Critical t (df=17, α=0.05, 2-tailed) = ±2.11.

  • ∣tobs∣<tcrit→ fail to reject H0.

  • Conclusion: No significant evidence DrugX affects maze learning errors.

  • p-value confirms (p > .05).

<ul><li><p>tobs=−0.99</p></li><li><p>Critical t (df=17, α=0.05, 2-tailed) = ±2.11.</p></li><li><p>∣tobs∣&lt;tcrit→ <strong>fail to reject H0</strong>.</p></li><li><p>Conclusion: No significant evidence DrugX affects maze learning errors.</p></li><li><p>p-value confirms (p &gt; .05).</p></li></ul><p></p>
26
New cards

How do we compute a 95% confidence interval for an independent t-test?

<img src="https://knowt-user-attachments.s3.amazonaws.com/6097a1f9-4633-4c3b-8849-6dc57f3f8b43.png" data-width="75%" data-align="center"><p></p>
27
New cards

What did the 95% CI for the DrugX study show, and what does it mean?

  • Calculation: 

  • CI includes 0 → difference could plausibly be 0.

  • Interpretation: Fail to reject H0H_0H0​.

  • Conclusion: No evidence DrugX significantly changed maze errors.

<ul><li><p>Calculation:&nbsp;</p></li></ul><img src="https://knowt-user-attachments.s3.amazonaws.com/c56a8edc-e0bf-441d-93cd-b4ccc754b482.png" data-width="100%" data-align="center"><ul><li><p>CI <strong>includes 0</strong> → difference could plausibly be 0.</p></li><li><p>Interpretation: Fail to reject H0H_0H0​.</p></li><li><p>Conclusion: No evidence DrugX significantly changed maze errors.</p></li></ul><p></p>
28
New cards

What is the “homogeneity of variance” assumption in t-tests?

  • Assumption: population variances equal (σ2122)

  • If violated → “heterogeneity of variance.”

  • Rule of thumb: difference ≥ 4× → too different → don’t pool variances.

  • If heterogeneous: use separate variance formula instead of pooled.

<ul><li><p>Assumption: population variances equal (σ<sup>2</sup><sub>1</sub>=σ<sup>2</sup><sub>2</sub>)</p></li><li><p>If violated → “heterogeneity of variance.”</p></li><li><p>Rule of thumb: difference ≥ 4× → too different → don’t pool variances.</p></li><li><p>If heterogeneous: use separate variance formula instead of pooled.</p></li></ul><p></p>
29
New cards

What are the three main assumptions of an independent-sample t-test?

  • Equal variances (homogeneity).

    • Even if means differ, assume variances are the same.

  • Normality.

    • Group means should come from normal populations.

    • CLT helps if n ≥ 30 per group; issue if very small n.

  • Independence.

    • Individuals’ scores not related (no pairing)

<ul><li><p><strong>Equal variances</strong> (homogeneity).</p><ul><li><p>Even if means differ, assume variances are the same.</p></li></ul></li><li><p><strong>Normality</strong>.</p><ul><li><p>Group means should come from normal populations.</p></li><li><p>CLT helps if n ≥ 30 per group; issue if very small n.</p></li></ul></li><li><p><strong>Independence</strong>.</p><ul><li><p>Individuals’ scores not related (no pairing)</p></li></ul></li></ul><p></p>
30
New cards

Why use effect size with t-tests, and how is it calculated?

  • Problem: p-values depend on sample size.

  • Effect size gives a standardized difference, less sensitive to n.

  • Formula:

  • Important: denominator = standard deviation of differences, not standard error.

  • Uses variance sum law for calculation.

<ul><li><p>Problem: p-values depend on sample size.</p></li><li><p>Effect size gives a <strong>standardized difference</strong>, less sensitive to n.</p></li><li><p>Formula:</p></li></ul><img src="https://knowt-user-attachments.s3.amazonaws.com/2d27240e-3442-4a91-9068-8d91bca77e9e.png" data-width="25%" data-align="center"><ul><li><p><strong>Important</strong>: denominator = <strong>standard deviation</strong> of differences, not standard error.</p></li><li><p>Uses variance sum law for calculation.</p></li></ul><p></p>
31
New cards

How do we compute Cohen’s d for independent samples?

<img src="https://knowt-user-attachments.s3.amazonaws.com/1fca0f3a-8f18-48eb-a7a7-0b7a9638090c.png" data-width="100%" data-align="center"><p></p>
32
New cards

How was the mindfulness/anxiety study designed?

  • n=12 per group (low anxiety vs high anxiety).

  • IV: Anxiety group.

  • DV: Mindfulness (FFMQ scores).

  • Hypotheses:

    • H0​: No difference in mindfulness.

    • H1: Groups differ in mindfulness.

<ul><li><p>n=12 per group (low anxiety vs high anxiety).</p></li><li><p><strong>IV</strong>: Anxiety group.</p></li><li><p><strong>DV</strong>: Mindfulness (FFMQ scores).</p></li><li><p>Hypotheses:</p><ul><li><p>H0​: No difference in mindfulness.</p></li><li><p>H1: Groups differ in mindfulness.</p></li></ul></li></ul><p></p>
33
New cards

What are the 5 steps of hypothesis testing (applied to mindfulness/anxiety)?

  1. State hypotheses.

  2. Choose test (independent t-test).

  3. Gather what’s needed (means, variances, n, SE).

  4. Calculate t and make decision.

  5. Check effect size.

34
New cards

How do we choose the test and decide 1- vs 2-tailed for mindfulness/anxiety?

  • Test: Independent t-test.

  • Method:

    • Option 1: Compute tobs

    • Option 2: Compute CI.

  • Tail: TWO-tailed (question is whether mindfulness differs, not directional).

<ul><li><p>Test: Independent t-test.</p></li><li><p>Method:</p><ul><li><p>Option 1: Compute tobs</p></li><li><p>Option 2: Compute CI.</p></li></ul></li><li><p>Tail: TWO-tailed (question is whether mindfulness differs, not directional).</p></li></ul><p></p>
35
New cards

What info do we need to compute tobs​ in mindfulness/anxiety?

  • Group means (X1, X2)

  • Group sizes (n1, n2)

  • Variances (s21, s22)

  • Decide: pool or not?

    • If n1=n2​ and variances ≈ equal → pool.

    • If variances ≥4× different → don’t pool.

<ul><li><p>Group means (X1, X2)</p></li><li><p>Group sizes (n1, n2)</p></li><li><p>Variances (s<sup>2</sup><sub>1</sub>, s<sup>2</sup><sub>2</sub>)</p></li><li><p>Decide: pool or not?</p><ul><li><p>If n1=n2​ and variances ≈ equal → pool.</p></li><li><p>If variances ≥4× different → don’t pool.</p></li></ul></li></ul><p></p>
36
New cards

How do we compute pooled variance for mindfulness/anxiety?

<img src="https://knowt-user-attachments.s3.amazonaws.com/63f18aa2-aadb-4fc6-a953-438b198fe3bb.png" data-width="50%" data-align="center"><p></p>
37
New cards

What is the observed t-value for mindfulness/anxiety?

<img src="https://knowt-user-attachments.s3.amazonaws.com/bd515204-40e2-48a4-bf61-712534a1ab80.png" data-width="50%" data-align="center"><p></p>
38
New cards

What is the conclusion from the mindfulness/anxiety test?

  • tobs=2.73​=2.73, p = 0.012 (< 0.05).

  • Reject H0

  • Conclusion: Low- and high-anxiety groups differ in mindfulness scores.

<ul><li><p>tobs=2.73​=2.73, p = 0.012 (&lt; 0.05).</p></li><li><p>Reject H0</p></li><li><p>Conclusion: Low- and high-anxiety groups differ in mindfulness scores.</p></li></ul><p></p>
39
New cards

What is the effect size for mindfulness/anxiety, and how do we interpret it?

<p></p><img src="https://knowt-user-attachments.s3.amazonaws.com/9d1efcb2-8c23-4032-964c-4d0a8a847942.png" data-width="75%" data-align="center"><p></p>