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Unformatted text preview: Elementary Research Methods
Chapter 9 Research Hypothesis States a predicted relationship between the independent variable and the dependent variable Examples – Contextual knowledge increases comprehension – Warm breakfasts increases test scores Form Equivalent Groups Random assignment to different levels of the independent variable – A. each subject has equal chance of being assigned to your experimental groups, A1 or A2 – B. Assignment of subjects is independent Notation Dealing with two levels of IV Form two equivalent groups: Group A1 One group assigned to A1, another to A2 Group A2 nA1 = No of subs nA2 = no of subs in A1 subs in A2 XA1 = mean A1 XA2 = mean A2 s2A1 = variance A1 s2A2 = variance of A2 Value of Equivalent Groups Eliminates subject characteristics or differences from creating confounding sources of extraneous variables. Extraneous Variables:
– Anything other than the IV that might explain differences between groups A1 and A2 on the dependent variable Confounding: – A difference between groups A1 and A2 that is due to an extraneous variable (something other than the IV) Simple Experiment After two equivalent groups are formed The IV is manipulated The DV is measured Statistical Analysis of Data – 1. to describe central tendency and variance of data – 2. to determine if XA1 is different from XA2 DID THE IV HAVE AN EFFECT ON THE DV? Example Experiment 1. Research hypothesis: – Contextual knowledge has an effect on language comprehension – IV: The presence (A1) and absence (A2) of contextual knowledge – Factor: Contextual knowledge Levels: A1 presence A2 absence Example Experiment 2) Form Equivalent groups What’s ideal? Convenience sample – Random selection – Random assignment I’ve got this fabulous new technique that I Example Experiment: Why do we need equivalent groups? believe that should help improve people’s ability to shoot free throws. To test it out, I give team 1 instruction on the superduper new technique and I don’t give team 2 instruction Then, I make each person on each team shoot 10 free throws to see if there are any differences. Team 1 Team 2 Example Experiment 3) Possible Confounds Something other than the levels of the IV differ between the two groups, and might cause differences in DV Ss in the two groups are not equivalent
– – – – Profession athletes vs. professional dancers Team 1 is consistently taller than team 2 Guys/girls Team 1 has more prior experience playing basketball than team 2… Example Experiment 4) Manipulate the IV 5) Efforts to control Extraneous variables – Administer levels A1 & A2 of the context factor – Onefactor, betweensubjects design – A. random assignment – B. Placebo condition/single blind experiment – C. Blind E to which condition subjects are in (double blind) Contextual effect experiment: Procedure If the balloons popped, the sound wouldn’t be able to carry since everything would be too far away from the correct floor. A closed window would also prevent the sound from carrying, since most building tend to be well insulated. Since the whole operation depends on a steady flow of electricity, a break in the middle of the wire would also cause problems. Of course, the fellow could shout, but the human voice is not loud enough to carry that far. An additional problem is that a string could break on the instrument. Then there could be no accompaniment to the message. It is clear that the best situation would involve less distance. Measure the DV
How well did understand the passage? Record your answer, using the following scale… Comprehension Scale 1 2 3 4 5 6 7 8 9 very low moderate very high
comprehension comprehension comprehension Measure the DV
Comprehension Scale 1 2 3 4 5 6 7 8 9 very low moderate very high
comprehension comprehension comprehension A1(presence) A2( absence) nA1 = 8 nA2 = 12 XA1 = 8.6 XA2 = 3.6 s2A1 = 1.5 s2A2 = 4.3 Testing the differences between 2 groups When you collect data from 2 separate groups, statistical differences are tested using the t test for two independent groups (also called the two sample t test for independent groups) tind = (XA1 –XA2) – (μA1 μA2) SXA1XA2 How to find What if μA1, μA2 , σA1 , and σA2 are unknown?
– Expected difference is always “0” if H0 is true! – H0: μA1= μA2 , or μA1 μA2 =0 1) Don’t need to know μA1, μA2 2) Need to estimate σ2A1 & σ2A2 from sample data – s2A1 & s2A1 – Use s2A1 & s2A2 = to compute SXA1XA2 Computing SXA1XA2
s2A1 = ∑(XXA1)2 s2A2 = ∑(XXA2)2 nA11 nA21 If nA1 = nA2 sXA1XA2 = s2A1 + s2A2 nA1 nA2 If nA1 ≠ nA2 sXA1XA2 = (nA11) s2A1 + (nA21) s2A2 1 1 + nA1+ nA22 nA1 nA2 tind
tind = (XA1 –XA2) SXA1XA2 Does contextual knowledge affect language comprehension? 5step procedure 1. H0: μA1= μA2 H1: μA1 ≠ μA2 2) α = .01 3) Set Decision Stage
– A. use twosample t for independent groups – B. reject – C. critical value df = N2 = nA1+ nA22 t(18) = +/2.878 – D. reject H0 if tobs ≥ 2.878 or ≤ 2.878 4) Collect Data and compute tind
A1(presence) A2( absence) nA1 = 8 nA2 = 12 XA1 = 8.6 XA2 = 3.6 s2A1 = 1.5 s2A2 = 4.3 tind = XA1 – XA2 + (nA11) s2A1 + (nA21) s2A2 1 1 nA1+ nA22 nA1 nA2 tind = 8.63.6 = 6.09 (7)1.5 +(11)4.3 1 1 + 8+122 8 12 5) Decision Reject H0, accept H1 Because t(18) = 6.09; is greater than critical limit of 2.878 Conclusion: Contextual knowledge has an effect on language comprehension Null Hypothesis for twosample tind H0: μA1= μA2 or μA1 μA2 = 0 Sampling Distribution of tind is a distribution of expected differences between XA1 & XA2 Expected difference between μA1& μA2 is 0 μA1 μA2 = 0, is the mean expected difference, given H0 Sampling Distribution of the expected differences between two groups is the basis for twosample ttest for independent groups tind = (XA1 –XA2) SXA1XA2 Where: SXA1XA2 = the estimated standard error of the distribution of differences between XA1 & XA2 Power The probability of rejecting H0 when it is Probability of finding an effect for an false and accepting H1 independent variable when one exists tind = (XA1 –XA2) SXA1XA2 Ways to increase power 1) maximize the effect of the IV 2) increase sample size – Increases the numerator of t formula – decreases the denominator of t formula – decreases the denominator of t formula 3) decrease variability of scores Strength of Effect
How 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 relationship eta squared – measure of strength of effect η2 = t 2obs t2obs+df Eta Squared Indicates the proportion of the variance in the DV that can be accounted for by the IV In psychological research η2 of .10 to .15 is considered a strong relationship Haase et al (1982) median eta = .083 – 0.0 indicates no relationship between IV an DV – 1.0 indicates perfect relationship: Knowledge of the IV condition allows for perfect prediction of DV Computing η 2 Context and comprehension experiment t(18) = 6.09 η2 = t 2obs = (6.09)2 t2obs+df (6.09)2 + 18 η2 = 37.1 = 37.1 = .67 37.1+18 55.5 .67 of the variance in comprehension can be predicted by the context condition A very, very strong effect!!!!!! Confidence Intervals on the Difference between two independent population means 95% (XA1 –XA2) – t.05 (SXA1XA2) to (XA1 –XA2) + t.05 (SXA1XA2) Where: (XA1 –XA2) = difference between two observed sample means (SXA1XA2) = standard error of the difference between means t.05 = twotailed tcrit at .05 significance with N2 df Confidence Intervals on the Difference between context and no context means on comprehension XA1 = 8.6 XA2 = 3.6 sXA1XA2 = .82 t(18)crit, α=.05 = 2.101 (XA1 –XA2) – t.05 (SXA1XA2) to (XA1 –XA2) + t.05 (SXA1XA2) 95% (8.63.6) 2.101(.82) to (8.63.6)+ 2.101(.82) 5.0 2.101(.82) to 5.0 + 2.101(.82) 5.0 – 1.72 to 5 + 1.72 3.28 to 6.72 95% confident that the difference between XA1 & XA2 is 3.28 to 6.72 units of comprehension ttest for Related Scores 1) Assumes scores from A1 pair up with scores from A2 2) Paired scores require that nA1 = nA2 3) Situations when used
– Use N pairs 4) Provides more power to reject H0 when scores are related in A1 & A2 – A. withinsubjects designs – B. Related or dependent score pairs in A1 & A2 ttest for Related Scores
trel = XA1 – XA2 SD SD = ∑D2(∑D)2 /Npairs Npairs (Npairs 1) D= XA1 XA2 = diff between an individual pair of scores in A1 & A2 Npairs = number of pairs of scores df = Npairs 1 Sample Experiment E wants to know if brothers and sisters receive different amounts of education 1. H0: μA1= μA2 H1: μA1 ≠ μA2 2) α = .05 3) Set the decision stage A. Use trel B.
reject C. t.05 with Npairs1 (4) df = 2.776 D. Reject H0 if tobs ≥ 2.776 or ≤ 2.776 Do Experiment and Compute trel A1 (brothers) A2(sisters) D D2 12 14 2 4 15 18 3 9 16 20 4 16 12 12 0 0 16 18 2 4 ∑ 71 82 11 33 X = 14.2 16.4 trel = XA1 – XA2 = 14.216.4 ∑D2(∑D)2 /Npairs 33(11)2/5 Npairs (Npairs 1) 5(4) trel = 2.2 = 2.2 = 3.33 3324.2 .66 20 5) Decision How strong is the effect? η2 = t 2obs = (3.33)___ = 11.1 t2obs+df (3.33)+4 11.4+4 η2 = 11.1 = .72 15.4 Thus: .72 of the variance in Education is accounted for by gender (within families) Reject H0, accept H1 because tobs(4) = 3.33 ≤ tcrit(4)α=.05 = 2.776 Confidence Intervals on the Difference between two independent population means 95% (XA1 –XA2) – t.05 (SD) to (XA1 –XA2) + t.05 (SD) Gender and education in families XA1 = 14.2 XA2 = 16.4 sD = .66 t.05 = 2.776 (14.216.4)2.776(.66) to (14.216.4)+2.776(.66) 2.21.83 to 2.2+1.83 4.03 to .37 The difference between the amount of education of brothers and sisters is between 4.03 to .37, with brothers getting less education ...
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 '07
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