A researcher is interested in whether they can manipulate the rate at which a participantrefreshes (thinks about) something they are trying to remember. The participant is shown twoletters and asked to remember the letters over a 10-second delay. During the delay, theparticipant sees either letters or numbers on the screen. Whenever the participant rehearses(thinks about) the letters they are trying to remember, they press a button. We hypothesizethat the participant will change their rate of rehearsal when, during the delay, they seedistractors that are similar to what they are trying to remember (e.g., letters). The data arebelow:Participant11123415Ho:Which statistical test would you select:Similar Distractors◆◆1171019979107H₁:Enter the data above into and JAMOVI and use the correct statistical procedure to test thenull hypothesis - answer the following questions:What is the center of the distribution?What are the critical values? +Dissimilar Distractors76708857973 decimal placesReport all values (excluding critical values) to TWO decimal places.Report actual p-value to THREE decimal places. Report all values (excluding critical values) to TWO decimal places.Report actual p-value to THREE decimal places.MDActual p-valuenDdfDSED =tobs• What can we say about the tobs in relation to the p-value and significance?.05o p95% CI around mean difference =dz:o Our observation=• What can we say about the CI in relation to the center of the distribution and significance?o The 95% CIinclude zero.statistically significant. Weo Our observationstatistically significant. WeComplete the APA Report.Report all values to TWO decimal places.Results

Given information: Participant Similar Distractor Dissimilar Distractors 1 117 76 2 101 70 3 99 88 4 79 57 5 107 97JAMOVI Procedure: Step 1) Select Analyses > T-Tests > Paired Samples T-Test. Step 2) Drag and drop variables to Paired Variables Step 3) Click next to get the results. JAMOVI Output:a) We should use the Paired t test for the given scenario. Null hypothesis: HA: μDifference=0 Alternate hypothesis: HA: μDifference≠0 This is two tailed test and significance level is 0.05. b) We will be using the t-distribution for the given scenario, the centre of the t-distribution is 0. Use the excel formula, "=T.INV.2T(0.05,4)" to determine the critical value: The critical values are + 2.776. c) Use the sample data to calculate the standard deviation of the differences(s), the standard error (S.E) of the mean difference, the degrees of freedom(DF), and the t statistic test statistic(t). Participant Similar Distractor Dissimilar Distractors d d-d¯ d-d¯2 1 117 76 41 18 324 2 101 70 31 8 64 3 99 88 11 -12 144 4 79 57 22 -1 1 5 107 97 10 -13 169 Sum 503 388 115 0 702 Mean 100.6 77.6 23 0 140.4 MD=∑dn=1155=23 Actual P-value = 0.018 nD = 5 dfD = nD - 1 dfD = 5 - 1 dfD = 4c) s=∑(d-d¯)2n-1=7025-1s=13.24764130≈13.2476S.ED=sn=13.24765=5.924525297≈5.925 d) t=[(x1-x2)-D]S.E=23-05.9225=3.882167574≈3.88 e) The P-value for this test is the probability that a t statistic having 4 degrees of freedom is less than - 3.88 or greater than 3.88. P-value = P(t < - 3.88) + P(t > 3.88) Use the excel formula, "=T.DIST.2T(3.88,4)" to determine the P-value. P-value = 0.018 Decision: Since the P-value (0.018) is less than the significance level (0.05), hence we have to reject the null hypothesis. P-value < 0.05 Our observation are statistically significant. We have to reject the null hypothesis.f) Use the excel formula, "=T.INV.2T(0.05,4)" for determining the critical value:tCritical = 2.776 C.I=x¯d±tα2*S.E=23±2.776*5.925=23±16.4478C.I=(6.55, 39.45) 95% confidence interval around mean difference is (6.55, 39.45). The 95% confidence interval does not include zero. Our observation are statistically significant. We have to reject the null hypothesis. g) dz=|x1-x2|SD=2313.2476=1.736163532≈1.75