The two tables are data related to ratings 5 testers gave on a scale of 1-100 fo
ID: 3132471 • Letter: T
Question
The two tables are data related to ratings 5 testers gave on a scale of 1-100 for 2 energy drinks. For the first question, relating to the table on the left, we were asked to assess whether there was a significant difference in preferences based on an alpha of .05. For the second question, relating to the table on the right, we were asked to come up with an interval with a C of .95 for the same data. I don't understand why the first table subtracts column B from column A, whereas the second table subtracts column A from column B. Since subtraction is not commutitive the two tables result in different sample means which influences the resulting interval. In summary why does the table on the left take A-B and the table on the right take B-A?
To assess whether these is a difference in preferences, we test We compute the mean and standard deviation for the 5 differences Here is the true mean difference in rates between the two drinks. The null hypothesis says that there is no difference between the rates for the two drinks, and the alternative hypothesis that the rates are significantly different. Subject Drink A Drink B Difference B-A We compute the mean and standard deviation for the 5 differences 48 Subject Drink A Drink B Difference 83 85 A-B 67 61 -6 48 4 76 83 85 73 70 67 61 6 4 76 71 73 70 3Explanation / Answer
We USe Tabel A not Table B
H0: No difference between two means of drinks
H1: There is a difference between two means of drinks
alpha:5%
test statistics:
we use paired t test
ence from Excel
Conclusion: t cal=0.412
P value=0.7017
P > alpha we accept the null hypothesis
t-Test: Paired Two Sample for Means Variable 1 Variable 2 Mean 69.4 68.4 Variance 176.3 129.8 Observations 5 5 Pearson Correlation 0.914237 Hypothesized Mean Difference 0 df 4 t Stat 0.411693 P(T<=t) one-tail 0.350834 t Critical one-tail 2.131847 P(T<=t) two-tail 0.701669 t Critical two-tail 2.776445Related Questions
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