T-test for two independent groupsChapter 9, part 2

What if A1, A2 , A1 , and A2 are unknown? 1) Dont need to know A1, A2 Expected difference is always 0 if H0 is true!H0: A1= A2 A1- A2 = 0

2) Need to estimate 2A1 & 2A2 from sample data s2A1 & s2A1 Use s2A1 & s2A2 = to compute SXA1-XA2

Computing SXA1-XA2s2A1 = (X-XA1)2 s2A2 = (X-XA2)2 nA1-1 nA2-1

sXA1-XA2 = (nA1-1) s2A1 + (nA2-1) s2A2 1 1 (nA1+ nA2 2) nA1 nA2

+

tindtind = (XA1 XA2) sXA1-XA2

Where:

sXA1-XA2 = (nA1-1) s2A1 + (nA2-1) s2A2 1 1 (nA1+ nA2 2) nA1 nA2

+

Does gratitude journal cause more happiness?

A1(presence) A2( absence) nA1 = 8 nA2 = 12 XA1 = 8.6XA2 = 3.6 s2A1 = 1.5 s2A2 = 4.3

5-step procedure

1)

2)

3) Set Decision StageA. B.

C. Critical value is found from table C.2 based upondf: N-2 Alpha .05 Thus, critical limit D. reject H0 if

A1(presence) A2( absence) nA1 = 8 nA2 = 12 XA1 = 8.6XA2 = 3.6 s2A1 = 1.5 s2A2 = 4.3

tobs = 8.6-3.6 = 6.11 (8-1)1.5 +(12-1)4.3 1 1 8+12-2 8 12

+

5) DecisionSince 6.11 is greater than 2.101, then it falls into the rejection region. Thus, I reject the Null Hypothesis and accept the Alternative Hypothesis. Thus, a daily gratitude journal (8.6) causes significantly more happiness than no gratitude journal (3.6).

PowerThe probability of rejecting H0 when it is false and accepting H1Probability of finding an effect for an independent variable when one exists

Ways to increase power1) maximize the effect of the IV

2) increase sample size

3) decrease variability of scores

tind = (XA1 XA2) SXA1-XA2

Strength of EffectHow strong is the causal relationship between the IV and the DV?Statistical significance rejecting and accepting tells us there is a relationship between IV & DV, but tells us nothing about the strength of the relationshipeta squared measure of strength of effect2 = t2obs t2obs+ df

Eta SquaredIndicates the proportion of the variance in the DV that can be accounted for by the IV

In psychological research 2 of .10 to .15 is considered a strong relationship

Computing 2In our experiment Remember that tobs = 6.11

2 = t2obs t2obs+df

.67 of the variance in the DV can be accounted for by the IV.67 of the variance in happiness can be predicted by the gratitude journal

Confidence Intervals on the Difference between two independent population means95% (XA1 XA2) t.05 (sXA1-XA2) to (XA1 XA2) + t.05 (sXA1-XA2)

Where: (XA1 XA2) = difference between two observed sample means (sXA1-XA2) = standard error of the difference between means t.05 = two-tailed critical value at .05 significance with N-2 df

95% C.I. (XA1 XA2) t.05 (SXA1-XA2) to (XA1 XA2) + t.05 (SXA1-XA2)

The interval 3.28 to 6.72 has a 95% probability of including the difference between population means

1. Null Hypothesis: H0: A1= A2 2. Since fell into rejection region, reject Null Hypothesis 3. So reject that A1= A2, so reject that A1 - A2 = 0

FYI - If 95% C.I does not contain zero, then reject Null Hypothesis