Willerman et al. (1991) collected a sample of 40 right-handed Anglo introductory psychology students at a large southwestern university. Subjects took foursubtests (Vocabulary, Similarities, Block Design, and Picture Completion) of the Wechsler Adult Intelligence Scale-Revised (1981). The researchers usedMagnetic Resonance Imaging (MRI) to determine the brain size of the subjects. The research hypothesis was that brain size is correlated with IQ. The data arebelow:Participant12346178Which statistical test would you select: Pearson CorrelationHo: p = 0H₁: p = 0◆◆Enter the data above into and JAMOVI and use the correct statistical procedure to test the null hypothesis - answer the following questions:What is the coefficient that indicates the strength of the relationship? r ◆ = 0What is the standard error?What is the appropriate 95% CI? [What is the d-effect size?Brain Size11090100929498104111What is the t-observed?What is the actual p-value?Complete the APA Report.THREE decimal placesBased on the p-value (a = .05) should you reject the Ho (yes/no)?Wechsler IQ112821108774105147125◆Results

Given information: Participant Brain Size Wechsler IQ 1 110 112 2 90 82 3 100 110 4 92 87 5 94 74 6 98 105 7 104 147 8 111 125a) We should use t -test for correlation significance. Null Hypothesis: H0: The population correlation coefficient is not significantly different from 0(ρ=0). Alternate Hypothesis: HA: The population correlation coefficient is significantly different from 0(ρ≠0). b) Use the given data to form the table: Participant X Y X*Y X*X Y*Y 1 110 112 12320 12100 12544 2 90 82 7380 8100 6724 3 100 110 11000 10000 12100 4 92 87 8004 8464 7569 5 94 74 6956 8836 5476 6 98 105 10290 9604 11025 7 104 147 15288 10816 21609 8 111 125 13875 12321 15625 Sum 799 842 85113 80241 92672 Determine the correlation coefficient using the formula: r=n(∑xy)-∑x·∑yn∑x2-(∑x)2n∑y2-(∑y)2∑X=47,∑Y=51,∑XY=458,∑X2=429,∑Y2=493r=8(85113)-(799)(842)8(80241)-(799)2(8)(92672)-(842)2=0.7619c) S.E=1n-3=18-3=15=0.447213595≈0.4472 The standard error is 0.4472. d) Perform the fisher transformation and determine the log lower and log upper bound: zr=ln1+r1-r2=ln1+0.76191-0.76192=ln1.76190.23812=1.00728649L=zr-zα/2n-3=1.00728649-1.968-3=1.00728649-0.876538647=0.130747843U=zr+zα/2n-3=1.00728649+1.968-3=1.00728649+0.876538647=1.883825137 Determine the 95% confidence interval: Confidence interval=e2L-1e2L+1,e2U-1e2U+1=e2·0.13075-1e2·0.13075+1,e2·1.88383-1e2·1.88383+1=e0.261496-1e0.261496+1,e3.767650274-1e3.767650274+1=1.298871746-11.298871746+1,43.27825323-143.27825323+1=0.2988717462.298871746,42.2782532344.27825323=0.1301, 0.9548 95% confidence interval is C.I = (0.1301, 0.9548). e) r=d2d2+40.7619=d2d2+4d2d2+4=0.58049161d2=0.58049161d2+2.321966440.41950839d2=2.32196644d2=5.534970206d=2.352651739≈2.353 The d-effect size is 2.353.f) Significance level = 0.05 Degree of freedoms:- D.F = n - 2 D.F = 8 - 2 Test statistics:- t=r×n-21-r2 =0.7619×8-21-(0.7619)2=0.7619×3.781858711=2.881398152≈2.88 g) P-value = P(t < - 2.88) + P(t > 2.88) Use the excel formula, "=T.DIST.2T(2.88,6)" to determine the P-value. P-value = 0.028 h) Since p value(0.028) is less than the significance value(0.05), hence we have to reject the null hypothesis H0.