Motor Programs Lab Introduction. This lab will simulate an important experiment performed by Henry and Rogers (1960). The task involved the subject responding to an external signal then executing a simple, two-response movement, or four-response movement as fast as possible. The movement of the least complexity is a simple lifting of the reaction time key. The other two responses require two and four subsequent actions. The two and four subsequent action conditions involve flipping two or four switches in a row as fast as possible. Procedure: The subject will be seated in front of the task with their non-dominate finger on the reaction time key. The experimenter before each trial will pre-cue the subject about which movement conditions the will need to complete. It is important to emphasize that the subject is to react as quickly as possible to the stimulus and to move as quickly as possible. The subject should begin each trial by holding down the reaction time key. A warning signal should precede the stimulus to move. Experimenter has randomized the order of the tasks. A or B but the experimenter must present the order in a randomized fashion. For example in trial 1, you have the subject complete TASK B, then A. For trial 2, you have the subject complete task A, then B. Randomize the order that you want the subject to perform the tasks so they are required to change their motor program prior to each attempt. Each subject will perform 20 trials on each of the two movement tasks. Subject s performing Task A will lift of their index finger of their non-dominate hand and turn over 2 keys in order pre-cued by the experimenter as fast as possible and depress the second key to complete the trial. Task B will be switching 4 keys. The subject will be pre-cued prior to their performance on every trial about what key(s) and order they will be turning them over. Results 1. Record your reaction and movement time in ms for each trial on the individual data Table 1 below. Table I: Individual Data RT Trials Task A Order TaskA RT TaskA MT Task B Order Task B RT Task B RT 1 ks 1 & 3 keys 1,3,4,2 2 ks 1 & 4 keys 1,4,3,2 3 ks 1 & 2 keys 2,1,4,3 4 ks 3 & 2 keys 3,4,1,2 5 keys 2 & 3 keys 2,4,1,3
6 keys 3 & 4 keys 4,2,3,1 7 keys 1 & 2 keys 1,3,2,4 8 keys 2 & 4 keys 4,1,3,2 9 keys 4 & 1 keys 3,1,2,4 10 keys 4 & 2 keys 1,2,4,3 11 keys 1 & 3 keys 2,3,4,1 12 keys 1 & 2 keys 4,2,1,3 13 keys 2 & 4 keys 2,3,1,4 14 keys 4 & 1 keys 4,2,3,1 15 keys 2 & 3 keys 1,3,4,2 16 keys 1 & 2 keys 4,2,1,3 17 keys 4 & 3 keys 2,1,4,3 18 keys 3 & 1 keys 4,1,3,2 19 keys 1 & 4 keys 1,4,2,3 20 keys 3 & 2 keys 2,3,4,1 Mean xxxxxxxxxx xxxxxxxxxx 2. From your individual data sheet, calculate the mean RT and MT for each task. Record your means on Table 1. Graphing. Develop one bar graphs for RT and MT. Task A and B needed to be labeled on the x axis and ms are on the Y-axis. Entitle the graph, "RT and MT s between Movement tasks." On top of every bar the mean score should be indicated. Perform a Ttest. Conduct a T-test to determine if there was a between Task A and Task B reaction time means. In order to perform the Ttest, you will need to complete the following table and use the several formulas provided. See the example below. Trials RT Task A RT Task B Difference (A-B) taking in the sign (- /+) 1 2 3 4 5 6 7 8 9 Standard Deviation of the Deviation of (Squared_
10 11 12 13 14 15 16 17 18 19 20 Sum( ) Task1= Task2= D = xxxxxxxxx DS= Mean of the = D/N Number of trials = N Mean of Task 1 = Task1/Number of trials = Mean of Task 2 = Task1/Number of trials = Mean of = D/Number of trials = Standard deviation of s = SD = ( DS/N-1) = Deviation from mean = Standard deviation of s - D for each trial Standard error of the = SE = SD/ N = t (mean of /SE) = (this is called the t ratio) Probability = P < or > (circle one) = (e.g., P <.01). Use the below Table of values of t-test of significant to determine probability df 1% 5% 18 2.093 2.861 19 2.086 2.845 20 2.080 2.831 Discussion Questions: 1. Write a summary of the RT results that includes the results of the Ttest and your means.
2. Write a summary of your MT results based on your bar graph. 3. After reading the Classic Experiment by Henry and Rogers found on page 177 of your text, did the trend of your reaction times support or not support Henry & Rogers results as task difficulty increased? Do explain using your data! 4.How did this experiments attempt to demonstrate the existence of a motor program? Ttest example. The t is the ratio of the between means and an indication of variability of the sample distribution (your scores of two tasks). It tells us if there was a real between the mean scores of two variables (task A and B) that its occurrence should not be attributed to pure chance. Ttests are used in hypothesis testing. By taking into consideration the ratio of s between the means of the two sets of scores or groups and their standard deviations, a can be given quantitative value through the use of a t test. The relationship obtained through use of such a formula is called a critical ratio and can be used to determine if the between the scores or groups is large enough to show real (not chance) s. Once these critical ratios or t values are known they can, by reference to appropriate tables, be interpreted to show how much confidence can be placed in them. If a t is found to be significant at the.01 or one per cent level of confidence, for example, then the obtained would be expected to occur to chance only once in hundred times. If is found that an event or circumstance would be expected to occur five times out of a hundred, then the investigator accepts, with lesser degree of confidence (.05 or five per cent), that the results is not due to pure chance. Here is simple example of the computation required in a t test: Trials RT Task A RT Task B Difference (A- B) taking in the sign (-/+) Standard Deviation of the Deviation of mean (Squared) 1 23 20-3 -1 +1 2 20 18-2 0 0 3 19 15-4 -2 +4 4 18 14-4 -2 +4 5 16 14-2 0 0 6 16 13-3 -1 +1 7 15 15 0-2 +4 8 15 13-2 0 0 9 13 10-3 -1 +1 10 11 14 +3-5 +25 11 10 8-2 0 0 Sum( ) Task1= 176 Task2= 154 D = -22 xxxxxxxxx DS= +40 Mean of the =22/11 = 2.0
Number of trials is N = 11 Mean of Task 1 = Task1/N = 176/11 Mean of Task 2 = Task1/N = 154/11 Mean of = D/N = -22/11 = -2 Standard deviation of s is SD = ( DS/N-1) = ( 40/10) = 2.0 Deviation from mean = D for each trial - Standard deviation of s (e.g, trial 1 is -3-2 = -1; Trial 2 is -2-2 = 0; etc ) Standard error of the is SE = SD/ N = (2// 11) =.6 To determine the T ratio use the following formula: t(ratio) = mean of /SE t (mean of /SE) = (-2/.6) = -3.3 (ratio) Next step is to go to a table of values for T test of significance. In the table you will find a degrees of freedom column and levels of confidence. There is critical t value that is associated with each df and level of confidence (1% or 5%). Degrees of freedom (df) is usually N-1 of the paired scores and in this example N was 11 therefore df was N-1 or 10. If your t ratio is equal to or less than the critical t value then it is significant at that level of confidence. In our example the t ratio was -3.3 was well below the critical value of 2.22 for 1% and 3.17 for 5% confidence level. df 1% 5% 10 2.22 3.17 11 2.20 3.10 12 2.18 3.05 Then one might say that the probability that these two tasks or groups were different was at the.01 level of confidence, which indicated that they were significantly different. If the T ratio is above the critical value then one can only say there was no significant between the two tasks or groups Probability = P <.01