2. You are interested in investigating how brain injury (BI) in different areas

ID: 3316315 • Letter: 2

Question

2. You are interested in investigating how brain injury (BI) in different areas affects performance on an IQ test ( = 0.052 tail). You collect IQ test data from 28 frontal lobe BI patients, 34 occipital lobe BI patients, 29 parietal lobe Bl patients, and 37 temporal lobe Bl patients. a. What test should you use? b. H1 in terms of the IV: c. Ho in terms of the IV: d. You discover the sum of squares between groups is 487, and the sum of squares within groups is 2013. Source of variance Between Groups Within Groups Total Complete the ANOVA table below: df MS obt e. Find the critical value: f. Make decision about the null hypothesis, and draw a conclusion: g. Calculate effect size: h. State whether effect size is small, medium, or large: i. Calculate the experiment-wise error rate if you were to conduct all possible pairs of t-tests instead:

Explanation / Answer

Part a

Here, we should use the single factor one way analysis of variance or ANOVA F test for checking the difference between population means. We are given more than two means for comparison and that’s why we should ANOVA test.

Part b

H1: There is a statistically significant difference in the average performance on an IQ test for the persons with brain injuries in different areas of brain.

H1: µ1 µ2 µ3 µ4 or at least two µi’s are not same.

Part c

H0: There is no statistically significant difference in the average performance on an IQ test for the persons with brain injuries in different areas of brain.

H0: µ1 = µ2 = µ3 = µ4

Part d

We are given

SSB = 487

SSW = 2013

SST = SSB + SSW = 487 + 2013 = 2500

Total number of groups = k = 4

Total number of observations = n = 28 + 34 + 29 + 37 = 128

Degrees of freedoms are given as below:

Total df = n – 1 = 128 – 1 = 127

Between df = k – 1 = 4 – 1 = 3

Within df = Total df – between df = 127 – 3 = 124

MSB = SSB / between df = 487/3 = 162.3333333

MSW = SSW / within df = 2013 / 124 = 16.23387097

Fobt = MSB / MSW = 162.3333333/16.23387097 = 9.999668816

Completed ANOVA table is given as below:

P-value

Between groups

487

162.3333

9.999669

0.00000596

Within groups

2013

124

16.23387

Total

2500

127