The Effect Of Flavor On The Decision Making Process

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The histograms created for the correct number of flavors guessed with prior knowledge displays a skewed left distribution, which directly correlates to the p-value generated, as the subjects being cognizant of the type of flavors produced higher proportions. Moreover, the sample mean was 4.5476 beverages correctly guessed, out ofthe five beverages provided, illuminating the influence of flavor knowledge in the decision making process. Since x is relatively high, most of the data is clustered on the right side of the graph, or the higher value proportions. In the without prior knowledge histogram, the distribution is approximately normal, displaying the variability in correct answers. The sample mean generated was 3.3182, which is…show more content…
This experiment yielded a x of 4.5476, while the non-dyed drinks produced a x of 4.7837. Moreover, the sample standard deviation for the undyed experiment, .5341, is expectantly lower than the dyed drinks experiment, 1.4069, exemplifying a smaller range of data and consequently suggesting the presence of synaesthesia. A larger standard deviation would symbolize more variability in the answers collected, as there were two noticeable outliers in the correctly guessed dyed beverages with prior knowledge. The outliers of one out of five correctly guessed beverages are more extreme than the outliers in the correctly guessed non-dyed beverages without prior knowledge. In the non-dyed experiment, the only deviation from the typical pattern of correctly guessing all flavors were the proportions of three out of five. In regard to clustering, the vast majority of the data in both histograms were centered around the higher proportions (four to five correct) or the right side of the graph. However, the no prior knowledge with non-dyed beverages experiment is less heavily skewed than its counterpart, due to higher variability. With such a significantly low p-value of 2.1909 x 10-9, the histograms for the two different treatments support the alternative hypothesis. 5. Conclusion The p-value obtained from the data collected did not agree with the original hypothesis conjured, as 2.633 x 10-6 is less

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