Question

Data on alcohol content and wine quality was collected from variants of a particular wine. From a sample of 45 ​wines, a model was created using the percentages of alcohol to predict wine quality. The data are modeled by Yi=−0.329+0.5449Xi​, where Xi is the alcohol content​ (%) and Yi is the rated quality of the wine. For these​ data, SYX=0.9792​, X=10.24, and hi=0.096892 when X=8.

A. Construct a 95​% confidence interval estimate of the mean wine quality rating for all wines that have 8​% alcohol.

B. Construct a 95​% prediction interval of the wine quality rating of an individual wine that has 8% alcohol.

C. Explain the differences in the results from​ (a) and​ (b).

Answer 1

Data  
X Value = 8
Confidence Level   95%
  
Intermediate Calculations  
Sample Size ,n= 45
Degrees of Freedom   43
t Value = 2.0167


Standard Error of the Estimate =Se = 0.9792
h Statistic = 0.096892
Predicted Y (YHat) = −0.329+0.5449*8 = 4.0302
  

a)

For Average Y  
Interval Half Width ,E= t*Se*√(Hi) = 0.6147
Confidence Interval Lower Limit =Y-E= 3.4155
Confidence Interval Upper Limit =Y^+E = 4.6449
  

b)

For Individual Response Y  
Interval Half Width ,E =t*Se*√(1+Hi) =   2.0682
Prediction Interval Lower Limit = 1.9620
Prediction Interval Upper Limit = 6.0984

c)

Prediction interval is for individual value while confidence interval is for mean value.prediction interval is wider than confidence interval