A study was conducted to examine the quality of fish after seven days in ice storage. For this study: Y = measurement of fish quality (on a 10 point scale with 10 = BEST.); X = # of hours after being caught that the fish were packed in ice. The regression line is: y-hat = 8.5−.5x. From this we can say that:  A one hour delay in packing the fish in ice increases the estimated quality by .5  A one hour delay in packing the fish in ice decreases the estimated quality by .5  If the estimated quality increases by 1 the fish have been packed in ice two hours later.  If the estimated quality increases by 1 then the fish have been packed in ice one hour sooner.  Can’t really say until we see a plot of the data.

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Asked Dec 3, 2019
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A study was conducted to examine the quality of fish after seven days in ice storage. For this study: Y = measurement of fish quality (on a 10 point scale with 10 = BEST.); X = # of hours after being caught that the fish were packed in ice. The regression line is: 
y-hat = 8.5−.5x. From this we can say that:

   

A one hour delay in packing the fish in ice increases the estimated quality by .5

   

A one hour delay in packing the fish in ice decreases the estimated quality by .5

   

If the estimated quality increases by 1 the fish have been packed in ice two hours later.

   

If the estimated quality increases by 1 then the fish have been packed in ice one hour sooner.

   

Can’t really say until we see a plot of the data.

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Expert Answer

Step 1

Given equation of linear regression line is Y = 8.5−.5(X).

Here,

Y is dependent variable and X is independent variable.

So, the slope of regression line is -0.5.

Step 2

The slope is negative. So, for each unites increases in independe...

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