Independent-Samples t Test

Size: px

Start display at page:

Download "Independent-Samples t Test"

Tracy Hall
5 years ago
Views:

1 Chapter 14 Aplia week 8 (Two independent samples) Testing hypotheses about means of two populations naturally occurring populations introverts vs. extroverts neuroticism experimentally defined (random assignment) neutral vs. aggressive model child s aggression level 1

2 Testing hypotheses about means of two independent populations Hypotheses about population means H 0 : µ 1 = µ H 1 : µ 1 µ (H 1 : µ 1 > µ H 1 : µ 1 < µ ) random sampling from two populations real or hypothetical Draw a sample from each population and compare the means of the two samples Build a model of all possible outcomes, assuming the null hypothesis is true

3 Sampling from each population is described by a distribution of sample means Population 1 Population H 0 is true Distributions of raw scores X µ 1 = µ Distributions of sample means M Distribution of differences between sample means M 1 M All random samples of size n 1 or n All pairs of sample means σ M1 M µ M1 = µ M σ M1 = σ 1 n 1 µ M1 M = 0 σ M = σ 1 + σ n 1 n = σ n 3

4 Use the distribution of differences between means to determine whether observed difference between sample means implies rejection of H 0 Distribuiton of differences between means Need to know σ M1 M.05 0 M 1 M.05 Must estimate from s 1 and assumption of equal population variance ( = ) s 1 treat and as independent estimates and average them to get pooled estimate, s p s s σ 1 σ 4

5 Recall the variance of the distribution of differences between means So the standard deviation is σ M1 M = σ 1 + σ n 1 σ M1 M = σ 1 + σ n 1 Replace σ 1 and σ with the estimate s p n n yielding an estimate of σ M1 M Standard deviation of difference between means s M1 M = s p + s p n 1 n 5

6 Implications of using estimated standard error of difference between means Experimental Control M n s M 1 M = = 10! " # = %&'(%'' # s M1 M = s p + s p n 1 n = 30 = = 4 What is the model of possible outcomes? 6

7 Distribution of differences between means normal, with mean = 0 if H 0 is true But where is a difference of 10 points in this distribution? M1 M Standardize the values in the distribution of differences by dividing each one by its own s M 1 M This produces a t distribution with df = n 1 + n 7

8 Experimental Control M n s Independent-Samples t Test M 1 M = = 10 s M1 M = 4 Standardize the observed difference between means t = M 1 M s M1 M = 10/4 =.50 Relevant t distribution:

9 What if unequal n? recall how to compute sample variance s = Σ( X M ) N 1 = SS df Pooling two sample estimates of variance combine SS from each sample and divide by combined df s p = SS 1 + SS s p = ( n 1 1)s 1 +( n 1)s df 1 + df n 1 + n weighted average 9

10 Full numerical example of t test two random samples: introverts and extroverts scores on a neuroticism self-report scale Introverts X X-M M = 7.0 Σ(X M) = 16.0 s = 4.0 Extroverts X X-M M = 6.4 Σ(X M) = 17. s = 4.3 s p = SS 1 + SS df 1 + df s p = = =

11 Compute s M1 M for the previous example Introverts X X-M M = 7.0 Σ(X M) = 16.0 s = 4.0 Extroverts X X-M M = 6.4 Σ(X M) = 17. s = 4.3 s M1 M = s p + s p = 4.15 n 1 n s p = = = 4.15 = =

12 Continuing the example to compute a t test Introverts X X-M M = 7.0 Σ(X M) = 16.0 s = 4.0 t = M 1 M s M1 M = Extroverts X X-M s p = s M = 6.4 M1 = 1.9 M Σ(X M) = 17. s = 4.3 = = = 4.15 df = 8 α =.05 critical t =?? (nondirectional) 1

13 13

14 Continuing the example to compute a t test Introverts X X-M M = 7.0 Σ(X M) = 16.0 s = 4.0 t = M 1 M s M1 M = Extroverts X X-M s p = s M = 6.4 M1 = 1.9 M Σ(X M) = 17. s = 4.3 = = = 4.15 df = 8 α =.05 critical t =.306 (nondirectional) 14

15 Summary of steps for computing an independentsamples t test 1. Obtain SS for each group from given information Introverts Extroverts M = 7.0 s = SS M = 6.4 s = SS n = 5 df n = 5 df s = 4.0 SS = s (df ) s = 4.3 SS = s (df ) SS = 4.0(4) = 16.0 SS = 4.3(4) = 17. s Pool SS and df from each p = group to obtain pooled estimate of σ = =

16 3. Use pooled variance estimate to obtain estimated standard deviation of difference between means s M1 M = s p + s p = 4.15 n 1 n = = Form the t ratio by dividing the difference between means by the estimated standard deviation of difference between means (also called the standard error of difference) t = M 1 M s M1 M = = 0.47 t (df = 8)

17 5. Compare the obtained t ratio with the critical t ratio and reject the null hypothesis if the obtained value is more extreme α =.05, nondirectional df = 8 critical t =.306 (observed t = 0.47) Do not reject H 0 t (df = 8)

18 Another example: attitude toward water conservation Treatment Control M n 17 5 s 10 1 s p = ( n 1 1)s 1 +( n 1)s n 1 + n s M1 M = s p + s p n 1 n 18

19 Assumptions Independent-Samples t Test random (independent) sampling normal distribution of differences between sample means usually by assuming normal distributions of raw scores equal variance in populations of raw scores (homogeneity of variance) violation of this assumption (heterogeneity of variance) if sample variances differ by approx. a ratio of 4:1 or more if unequal variance: compute t ratio without pooling variance estimates; test t ratio using df from the smaller of the two samples 19

20 Example with heterogeneity of variance drug therapy improves concentration ability of children with ADHD Placebo group Drug group M =.5 M = 6.6 n = 1 n = 8 s = 8.8 s =.1 H 0 : µ 1 = µ H 1 : µ 1 < µ s M1 M = s 1 + s = n 1 n t = M 1 M s M1 M = = 1.55 = =.65 df = 7, α =.05 critical t = Do not reject H 0 0

21 Application example Independent-Samples t Test certain type of amnesic patients (e.g., HM) suffer impaired conscious recollection free recall of list of words hypothesize more words recalled in control group versus amnesic group 1

22 Hypothetical data for free recall test Control Amnesic X X M (X M) X X M (X M) M=10 Σ(X M) =40 M=5 Σ(X M) =44 s p = SS 1 + SS df 1 + df s p = = 4.67 s s p M1 M = + s p n 1 n = = 0.97

23 Hypothetical data for free recall test H 0 : µ 1 = µ H 1 : µ 1 > µ α =.05 df = = 18 critical t = s M1 M = 0.97 t = = 5.15 Reporting results of the t test: t(18) = 5.15, p <.05 H 0 is rejected, conclude that amnesic subjects are impaired on a test of conscious recollection 3

24 t.test function in R Independent-Samples Design > data= read.table(file.choose(new=f),header=t) > describe(data) > t.test(data$control,data$amnesic, var.equal=t) Two Sample t-test data: dat$control and dat$amnesic t = , df = 18, p-value = 6.36e-05 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: sample estimates: mean of x mean of y

25 Effect size Independent-Samples t Test standard method of quantifying the size of an effect particularly useful when considering statistical power d = M 1 M s p From example with amnesia and recall: s p = 4.67 s p = 4.67 =.16 d = =.31 Try plotting two normal population distributions of raw scores representing this effect size 5

Chapter 11: Inference for Distributions Inference for Means of a Population 11.2 Comparing Two Means

Chapter 11: Inference for Distributions 11.1 Inference for Means of a Population 11.2 Comparing Two Means 1 Population Standard Deviation In the previous chapter, we computed confidence intervals and performed