EBK PRACTICE OF STAT.F/AP EXAM,UPDATED
EBK PRACTICE OF STAT.F/AP EXAM,UPDATED
6th Edition
ISBN: 9781319287573
Author: Starnes
Publisher: MPS PUB
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Question
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Chapter 3, Problem R3.4RE

(a)

To determine

To make a scatterplot that is suitable for predicting when the cherry trees will blossom from the temperature and which variable did you choose as the explanatory variable.

(a)

Expert Solution
Check Mark

Explanation of Solution

Since we expect the temperature to influence the days in April to first blossom temperature is the explanatory variable and the days in April to first blossom is the response variable. Thus, the scatterplot is as:

  EBK PRACTICE OF STAT.F/AP EXAM,UPDATED, Chapter 3, Problem R3.4RE , additional homework tip  1

(b)

To determine

To use the technology to calculate the correlation and the equation of the regression line and interpret the slope and y intercept of the line in the setting.

(b)

Expert Solution
Check Mark

Answer to Problem R3.4RE

  y^=33.12034.6855x .

Explanation of Solution

Using calculator, press on STAT and then select 1 : Edit . and then enter the data of sugar in the list L1 and enter the data of calories in the list L2 .

Next, press on STAT select CALC and then select Linreg(a+bx) . Next we need to finish the command by entering L1L2 .

  Linreg(a+bx)L1L2

Finally, pressing on ENTER then gives us the following result:

  y=a+bxa=33.1203b=4.6855r=0.8511

This then implies the regression line as:

  y^=a+bxy^=33.12034.6855x

This then implies that on average, the days in April to first blossom decreases by 4.6855 days per degree Celsius. And the days in April to first blossom is 33.1203 days when the temperature is zero degree Celsius.

(c)

To determine

To explain would you be willing to use the equation in part (b) to predict the date of the first blossom.

(c)

Expert Solution
Check Mark

Answer to Problem R3.4RE

No.

Explanation of Solution

The regression line in part (b) is:

  y^=33.12034.6855x

To predict the date to first blossom at 4.20C is then,

  y^=33.12034.6855x=33.12034.6855(4.2)=5.3052

We then note that the predicted day to first blossom in April is 5.3052 . However this does not make dense as a day is always a positive integer and thus we are not willing to use the equation in part (b).

(d)

To determine

To calculate and interpret the residual for the year when the average March temperature was 4.50C .

(d)

Expert Solution
Check Mark

Answer to Problem R3.4RE

Residual is 2.033 .

Explanation of Solution

The regression line in part (b) is:

  y^=33.12034.6855x

Now, the days to first blossom when average March temperature was 4.50C is:

  y^=33.12034.6855x=33.12034.6855(4.5)=12.033

And the actual value is 10 from the table given.

Thus, the residual is as:

  Residual=yy^=1012.033=2.033

This then implies that we overestimated the number of days in April to first blossom by 2.033 days when using the regression line with the temperature as the explanatory variable.

(e)

To determine

To use the technology to help construct the residual plot and describe what you see.

(e)

Expert Solution
Check Mark

Explanation of Solution

The residual plot is as:

  EBK PRACTICE OF STAT.F/AP EXAM,UPDATED, Chapter 3, Problem R3.4RE , additional homework tip  2

Thus, there is no obvious pattern in the residual plot and thus the linear regression line seems to be a good fit.

Chapter 3 Solutions

EBK PRACTICE OF STAT.F/AP EXAM,UPDATED

Ch. 3.1 - Prob. 11ECh. 3.1 - Prob. 12ECh. 3.1 - Prob. 13ECh. 3.1 - Prob. 14ECh. 3.1 - Prob. 15ECh. 3.1 - Prob. 16ECh. 3.1 - Prob. 17ECh. 3.1 - Prob. 18ECh. 3.1 - Prob. 19ECh. 3.1 - Prob. 20ECh. 3.1 - Prob. 21ECh. 3.1 - Prob. 22ECh. 3.1 - Prob. 23ECh. 3.1 - Prob. 24ECh. 3.1 - Prob. 25ECh. 3.1 - Prob. 26ECh. 3.1 - Prob. 27ECh. 3.1 - Prob. 28ECh. 3.1 - Prob. 29ECh. 3.1 - Prob. 30ECh. 3.1 - Prob. 31ECh. 3.1 - Prob. 32ECh. 3.1 - Prob. 33ECh. 3.1 - Prob. 34ECh. 3.1 - Prob. 35ECh. 3.1 - Prob. 36ECh. 3.2 - Prob. 37ECh. 3.2 - Prob. 38ECh. 3.2 - Prob. 39ECh. 3.2 - Prob. 40ECh. 3.2 - Prob. 41ECh. 3.2 - Prob. 42ECh. 3.2 - Prob. 43ECh. 3.2 - Prob. 44ECh. 3.2 - Prob. 45ECh. 3.2 - Prob. 46ECh. 3.2 - Prob. 47ECh. 3.2 - Prob. 48ECh. 3.2 - Prob. 49ECh. 3.2 - Prob. 50ECh. 3.2 - Prob. 51ECh. 3.2 - Prob. 52ECh. 3.2 - Prob. 53ECh. 3.2 - Prob. 54ECh. 3.2 - Prob. 55ECh. 3.2 - Prob. 56ECh. 3.2 - Prob. 57ECh. 3.2 - Prob. 58ECh. 3.2 - Prob. 59ECh. 3.2 - Prob. 60ECh. 3.2 - Prob. 61ECh. 3.2 - Prob. 62ECh. 3.2 - Prob. 63ECh. 3.2 - Prob. 64ECh. 3.2 - Prob. 65ECh. 3.2 - Prob. 66ECh. 3.2 - Prob. 67ECh. 3.2 - Prob. 68ECh. 3.2 - Prob. 69ECh. 3.2 - Prob. 70ECh. 3.2 - Prob. 71ECh. 3.2 - Prob. 72ECh. 3.2 - Prob. 73ECh. 3.2 - Prob. 74ECh. 3.2 - Prob. 75ECh. 3.2 - Prob. 76ECh. 3.2 - Prob. 77ECh. 3.2 - Prob. 78ECh. 3.2 - Prob. 79ECh. 3.2 - Prob. 80ECh. 3 - Prob. R3.1RECh. 3 - Prob. R3.2RECh. 3 - Prob. R3.3RECh. 3 - Prob. R3.4RECh. 3 - Prob. R3.5RECh. 3 - Prob. R3.6RECh. 3 - Prob. T3.1SPTCh. 3 - Prob. T3.2SPTCh. 3 - Prob. T3.3SPTCh. 3 - Prob. T3.4SPTCh. 3 - Prob. T3.5SPTCh. 3 - Prob. T3.6SPTCh. 3 - Prob. T3.7SPTCh. 3 - Prob. T3.8SPTCh. 3 - Prob. T3.9SPTCh. 3 - Prob. T3.10SPTCh. 3 - Prob. T3.11SPTCh. 3 - Prob. T3.12SPTCh. 3 - Prob. T3.13SPT
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