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Linear Regression Rg

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(i) Saturation (X1) and Transisomers (X2) hypothesis testing:












The p-value for a two-tailed t-score of 9.308 is 0.000.
The p-value for a two-tailed t-score of 11.505 is 0.000.
We find that the p-values for both Saturation and Transisomer variables is less than
0.05, and so we have enough data to reject the null hypothesis. We conclude that both regressors
display a significant linear relationship to the given Y.
(ii) The value around which the predicted interval will be constructed:
     


  



  
(iii) The data points move randomly about the line y=0 (a completely horizontal line from the
residual y axis). Thus we appear to be correct in assuming that the relationship between X1,
X2 and Y is linear. There is the tell-tale “horizontal band” that forms around the y=0 line,
which tells us that the variances of the error terms are about equal. There is no discernable
pattern to the distribution. Finally, the point at ~ (60, 3) does not appear to be in line with
the other points, suggesting that the original data set contains an outlier.
Without more data, it is difficult to gauge whether or not the dataset has an outlier, and it is
even more difficult to determine whether or not there is a pattern—or an undetectable
pattern hidden by a lack of data points—in the residual plot. As is always the case, my
suggestion is to get more data. That is always the solution, but in this particular case, there
is truly a need for more than 7 data points.

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(i) Saturation (X1) and Transisomers (X2) hypothesis testing: 𝐻0 : 𝛽1 = 𝛽2 = 0 𝐻𝐴 : 𝐴𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝛽 ≠ 0 𝐷𝑓 = 7 − 2 = 5 𝑏1 2.802 𝑡𝑠𝑎𝑡 = = = 9.308 𝑆𝐸 0.3 𝑏2 1.073 𝑡𝑡𝑟𝑎𝑛𝑠 = = = 11.505 𝑆𝐸 0.093 The ...
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