PRACTICE OF STATISTICS F/AP EXAM
PRACTICE OF STATISTICS F/AP EXAM
6th Edition
ISBN: 9781319113339
Author: Starnes
Publisher: MAC HIGHER
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Chapter 12, Problem AP4.45CPT

(a)

To determine

To describe the association between the seed count and weight shown in the scatterplot.

(a)

Expert Solution
Check Mark

Explanation of Solution

It is given the data and the scatterplot on the mean number of seeds produced in a year by several common tree species and the mean weight of the seeds produced. By looking at the scatterplot we can say that,

The direction of the scatterplot is negative because the pattern in the scatterplot slopes downward. And the form of the scatterplot is curved because there is a strong curvature present in the scatterplot. Also the strength of the scatterplot is strong because the points in the scatterplot do not deviate much from the general pattern in the points. The unusual features in the scatterplot: There appear to be one outlier because the right most point in the scatterplot lies far from the other points in the scatterplot.

(b)

To determine

To find out which model A or B is more appropriate for predicting seed weight from the seed count.

(b)

Expert Solution
Check Mark

Answer to Problem AP4.45CPT

Model B is appropriate.

Explanation of Solution

It is given the data and the scatterplot on the mean number of seeds produced in a year by several common tree species and the mean weight of the seeds produced. The two alternative models are proposed to predict the seed weight from the seed count. Thus, the scatterplot of model A contains strong curvature and the residual plot of the model A contains strong curvature as well which indicates that model A is not appropriate. Whereas, the scatterplot B contains no strong curvature and the residual plot of model B contains no strong curvature as well. Moreover, the residuals in the residual plot appear to be randomly scattered about the horizontal line at zero and thus model B appears to be appropriate for predicting seed weight from the seed count.

(c)

To determine

To predict the seed weight if the seed count is 3700 , using model you chose in part (b).

(c)

Expert Solution
Check Mark

Answer to Problem AP4.45CPT

The predicted seed weight is 19.7766 mg.

Explanation of Solution

It is given the data and the scatterplot on the mean number of seeds produced in a year by several common tree species and the mean weight of the seeds produced. The two alternative models are proposed to predict the seed weight from the seed count. In part (b), we find out the model B is more appropriate. Then we will use the model B. Thus, the general equation of the least square regression line is:

  y^=b0+b1x

Thus, form the computer output, we have that the estimate of the constant is given in the row “Constant” and in the column “Coef” as:

  b0=15.491

The slope b1 is given in the row “Mentos” and in the column “Coef” of the given computer output as:

  b1=1.5222

Now replacing the values in the equation we have,

  y^=b0+b1x=15.4911.5222x

Now, take the logarithm in the equation and solve it as:

  lny^=15.4911.5222x

Replace xby 3700 ,

  lny^=15.4911.5222x=15.4911.5222(3700)=2.9845

Now taking exponential on both sides we have,

  y^=elny^=e2.9845=19.7766

Thus, the predicted seed weight is 19.7766 mg.

Chapter 12 Solutions

PRACTICE OF STATISTICS F/AP EXAM

Ch. 12.1 - Prob. 11ECh. 12.1 - Prob. 12ECh. 12.1 - Prob. 13ECh. 12.1 - Prob. 14ECh. 12.1 - Prob. 15ECh. 12.1 - Prob. 16ECh. 12.1 - Prob. 17ECh. 12.1 - Prob. 18ECh. 12.1 - Prob. 19ECh. 12.1 - Prob. 20ECh. 12.1 - Prob. 21ECh. 12.1 - Prob. 22ECh. 12.1 - Prob. 23ECh. 12.1 - Prob. 24ECh. 12.1 - Prob. 25ECh. 12.1 - Prob. 26ECh. 12.1 - Prob. 27ECh. 12.1 - Prob. 28ECh. 12.1 - Prob. 29ECh. 12.1 - Prob. 30ECh. 12.1 - Prob. 31ECh. 12.1 - Prob. 32ECh. 12.2 - Prob. 33ECh. 12.2 - Prob. 34ECh. 12.2 - Prob. 35ECh. 12.2 - Prob. 36ECh. 12.2 - Prob. 37ECh. 12.2 - Prob. 38ECh. 12.2 - Prob. 39ECh. 12.2 - Prob. 40ECh. 12.2 - Prob. 41ECh. 12.2 - Prob. 42ECh. 12.2 - Prob. 43ECh. 12.2 - Prob. 44ECh. 12.2 - Prob. 45ECh. 12.2 - Prob. 46ECh. 12.2 - Prob. 47ECh. 12.2 - Prob. 48ECh. 12.2 - Prob. 49ECh. 12.2 - Prob. 50ECh. 12.2 - Prob. 51ECh. 12.2 - Prob. 52ECh. 12.2 - Prob. 53ECh. 12.2 - Prob. 54ECh. 12.2 - Prob. 55ECh. 12.2 - Prob. 56ECh. 12.2 - Prob. 57ECh. 12.2 - Prob. 58ECh. 12 - Prob. R12.1RECh. 12 - Prob. R12.2RECh. 12 - Prob. R12.3RECh. 12 - Prob. R12.4RECh. 12 - Prob. R12.5RECh. 12 - Prob. R12.6RECh. 12 - Prob. T12.1SPTCh. 12 - Prob. T12.2SPTCh. 12 - Prob. T12.3SPTCh. 12 - Prob. T12.4SPTCh. 12 - Prob. T12.5SPTCh. 12 - Prob. T12.6SPTCh. 12 - Prob. T12.7SPTCh. 12 - Prob. T12.8SPTCh. 12 - Prob. T12.9SPTCh. 12 - Prob. T12.10SPTCh. 12 - Prob. T12.11SPTCh. 12 - Prob. T12.12SPTCh. 12 - Prob. AP4.1CPTCh. 12 - Prob. AP4.2CPTCh. 12 - Prob. AP4.3CPTCh. 12 - Prob. AP4.4CPTCh. 12 - Prob. AP4.5CPTCh. 12 - Prob. AP4.6CPTCh. 12 - Prob. AP4.7CPTCh. 12 - Prob. AP4.8CPTCh. 12 - Prob. AP4.9CPTCh. 12 - Prob. AP4.10CPTCh. 12 - Prob. AP4.11CPTCh. 12 - Prob. AP4.12CPTCh. 12 - Prob. AP4.13CPTCh. 12 - Prob. AP4.14CPTCh. 12 - Prob. AP4.15CPTCh. 12 - Prob. AP4.16CPTCh. 12 - Prob. AP4.17CPTCh. 12 - Prob. AP4.18CPTCh. 12 - Prob. AP4.19CPTCh. 12 - Prob. AP4.20CPTCh. 12 - Prob. AP4.21CPTCh. 12 - Prob. AP4.22CPTCh. 12 - Prob. AP4.23CPTCh. 12 - Prob. AP4.24CPTCh. 12 - Prob. AP4.25CPTCh. 12 - Prob. AP4.26CPTCh. 12 - Prob. AP4.27CPTCh. 12 - Prob. AP4.28CPTCh. 12 - Prob. AP4.29CPTCh. 12 - Prob. AP4.30CPTCh. 12 - Prob. AP4.31CPTCh. 12 - Prob. AP4.32CPTCh. 12 - Prob. AP4.33CPTCh. 12 - Prob. AP4.34CPTCh. 12 - Prob. AP4.35CPTCh. 12 - Prob. AP4.36CPTCh. 12 - Prob. AP4.37CPTCh. 12 - Prob. AP4.38CPTCh. 12 - Prob. AP4.39CPTCh. 12 - Prob. AP4.40CPTCh. 12 - Prob. AP4.41CPTCh. 12 - Prob. AP4.42CPTCh. 12 - Prob. AP4.43CPTCh. 12 - Prob. AP4.44CPTCh. 12 - Prob. AP4.45CPTCh. 12 - Prob. AP4.46CPT
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