Introduction to the Practice of Statistics
Introduction to the Practice of Statistics
9th Edition
ISBN: 9781319013387
Author: David S. Moore, George P. McCabe, Bruce A. Craig
Publisher: W. H. Freeman
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Concept explainers

Correlation
Correlation defines a relationship between two independent variables. It tells the degree to which variables move in relation to each other. When two sets of data are related to each other, there is a correlation between them.
Linear Correlation
A correlation is used to determine the relationships between numerical and categorical variables. In other words, it is an indicator of how things are connected to one another. The correlation analysis is the study of how variables are related.
Regression Analysis
Regression analysis is a statistical method in which it estimates the relationship between a dependent variable and one or more independent variable. In simple terms dependent variable is called as outcome variable and independent variable is called as p…
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Chapter 2.4, Problem 73E

(a)

To determine

To find: The predicted count values.

(a)

Expert Solution
Check Mark

Answer to Problem 73E

Solution: The predicted count values are: 528.1, 378.7, 229.3, and 79.9.

Explanation of Solution

Calculation: The provided least-squares regression equation is:

count = 602.8−(74.7×time)

The time values are provided in the data as 1, 3, 5 and 7.

Substituting the values of time in the above linear regression equation, the following results are obtained:

For time(t1)=1, the predicted count (c^1) is calculated as:

count = 602.8−(74.7×time)=602.8−(74.7×t1)=602.8−(74.7×1)c^1=528.1

For time(t2)=3, the predicted count (c^2) is calculated as:

count = 602.8−(74.7×time)=602.8−(74.7×t2)=602.8−(74.7×3)c^2=378.7

For time(t3)=5, the predicted count (c^3) is calculated as:

count = 602.8−(74.7×time)=602.8−(74.7×t3)=602.8−(74.7×5)c^3=229.3

For time(t4)=7, the predicted count (c^4) is calculated as:

count = 602.8−(74.7×time)=602.8−(74.7×t4)=602.8−(74.7×7)c^4=79.9

Hence, the predicted count values obtained are 528.1, 378.7, 229.3, and 79.9.

(b)

Section 1

To determine

To find: The difference between the observed and the predicted counts.

(b)

Section 1

Expert Solution
Check Mark

Answer to Problem 73E

Solution: The differences obtained are 49.9, −61.7, −26.3, and 38.1.

Explanation of Solution

Calculation: The respective observed counts are provided in the exercise as:

c1=578

c2=317

c3=203

c4=118

for the times

t1=1

t2=3

t3=5

t4=7

The respective predicted counts for the above times are obtained from part (a) as:

c^1=528.1

c^2=378.7

c^3=229.3

c^4=79.9

The differences (di) where i=1, 2, 3, and 4 between the observed and the predicted counts are calculated as follows:

For time(t1) as 1, the difference (d1) is calculated as:

d1=c1−c^1=578−528.1=49.9

For time(t2) as 3, the difference (d2) is calculated as:

d2=c2−c^2=317−378.7=−61.7

For time(t3) as 5, the difference (d3) is calculated as:

d3=c3−c^3=203−229.3=−26.3

For time(t4) as 7, the difference (d4) is calculated as:

d4=c4−c^4=118−79.9=38.1

Hence, the differences obtained are 49.9, −61.7, −26.3, and 38.1.

Section 2

To determine

The number of positive differences between the observed and the predicted counts.

Section 2

Expert Solution
Check Mark

Answer to Problem 73E

Solution: The two positive differences are 49.9 and 38.1.

Explanation of Solution

The differences between the observed and the predicted counts are obtained in the Section 1 of part (b) above as:

d1=49.9

d2=−61.7

d3=−26.3

d4=38.1

Clearly, the positive difference values are d1, d4 as 49.9 and 38.1. Hence, 2 differences d1 and d4 are positive.

To determine

The number of negative differences between the observed and the predicted counts.

Expert Solution
Check Mark

Answer to Problem 73E

Solution: The two negative differences are −61.7 and −26.3.

Explanation of Solution

The differences between the observed and the predicted counts are obtained in the Section 1 of part (b) above as

d1=49.9

d2=−61.7

d3=−26.3

d4=38.1

Clearly, the negative difference values are d2, d3 −61.7 and −26.3. Hence, 2 differences d2 and d3 are negative.

(c)

Section 1

To determine

To find: Squares of the differences obtained in part(b).

(c)

Section 1

Expert Solution
Check Mark

Answer to Problem 73E

Solution: The squares of the differences are:

d12=2490.01

d22=3806.89

d32=691.69

d42=1451.61

Explanation of Solution

Calculation: The differences obtained in the part (b) above are:

d1=49.9

d2=−61.7

d3=−26.3

d4=38.1

The squares of the differences obtained are calculated as follows:

The square of the difference (d1) is calculated as:

d12=d1×d1=49.9×49.9=2490.01

The square of the difference (d2) is calculated as:

d22=d2×d2=(−61.7)×(−61.7)=3806.89

The square of the difference (d3) is calculated as:

d32=d3×d3=(−26.3)×(−26.3)=691.69

The square of the difference (d4) is calculated as:

d42=d4×d4=38.1×38.1=1451.61

Hence, the squares of the differences obtained are 2490.01, 3806.89, 691.69, and 1451.61.

Section 2

To determine

To find: The sum of the squares of the differences obtained in Section 1 above.

Section 2

Expert Solution
Check Mark

Answer to Problem 73E

Solution: The sum of the differences obtained is 8440.2_.

Explanation of Solution

Calculation: The squares of the differences are obtained in Section 1 above as:

d12=2490.01

d22=3806.89

d32=691.69

d42=1451.61

The sum of the differences (Σdn2) where n is 1, 2, 3, and 4 is calculated as follows:

Σdn2n=14=d12+d22+d32+d42=(2490.01)+(3806.89)+(691.69)+(1451.61)=8440.2

Hence, the sum of the differences obtained is 8440.2_.

(d)

Section 1

To determine

To find: The predicted count values.

(d)

Section 1

Expert Solution
Check Mark

Answer to Problem 73E

Solution: The predicted count values obtained are:

c^5=400

c^6=200

c^7=0

c^8=−200

Explanation of Solution

Calculation: The provided least-squares regression equation is:

count = 500−(100×time)

The time values are provided in the data as 1, 3, 5 and 7.

Substituting the values of time in the above linear regression equation, the following results are obtained:

For time(t1)=1, the predicted count (c^5) is calculated as:

count = 500−(100×time)=500−(100×t1)=500−(100×1)c^5=400

For time(t2)=3, the predicted count (c^6) is calculated as:

count = 500−(100×time)=500−(100×t2)=500−(100×3)c^6=200

For time(t3)=5, the predicted count (c^7) is calculated as:

count = 500−(100×time)=500−(100×t3)=500−(100×5)c^7=0

For time(t4)=7, the predicted count (c^8) is calculated as:

count = 500−(100×time)=500−(100×t4)=500−(100×7)c^8=−200

Hence, the predicted count values obtained are:

c^5=400

c^6=200

c^7=0

c^8=−200

For the times 1, 3, 5 and 7 respectively.

Section 2:

To determine

To find: The difference between the observed and the predicted counts.

Section 2:

Expert Solution
Check Mark

Answer to Problem 73E

Solution: The differences obtained are

d5=178

d6=117

d7=203

d8=318

Explanation of Solution

The observed counts are provided in the exercise as: 578, 317, 203 and 118 for the times as 1, 3, 5 and 7 respectively. The predicted counts obtained from section 1 above are: 400, 200, 0, and −200 for the times as 1, 3, 5 and 7 respectively.

The differences (di) where i is 5, 6, 7, and 8 between the observed and the predicted counts are calculated as follows:

For time(t1) as 1, the difference (d5) is calculated as:

d5=c1−c^5=578−400=178

For time(t2) as 3, the difference (d6) is calculated as:

d6=c2−c^6=317−(200)=117

For time(t3) as 5, the difference (d7) is calculated as:

d7=c3−c^7=203−0=203

For time(t4) as 7, the difference (d8) is calculated as:

d8=c4−c^8=118−(−200)=318

Hence, the differences obtained are 178, 117, 203, and 318.

Section 3:

To determine

The number of positive differences between the observed and the predicted counts.

Section 3:

Expert Solution
Check Mark

Answer to Problem 73E

Solution: The four positive differences are 178, 117, 203, and 318.

Explanation of Solution

The differences between the observed and the predicted counts are obtained in the Section 2 above as,

d5=178

d6=117

d7=203

d8=318

Clearly, all the difference values are positive. Hence, all the 4 differences d5, d6, d7, and d8 are positive.

To determine

The number of negative differences between the observed and the predicted counts.

Expert Solution
Check Mark

Answer to Problem 73E

Solution: There are no negative differences.

Explanation of Solution

The differences between the observed and the predicted counts are obtained in the Section 1 above as,

d5=178

d6=117

d7=203

d8=318

Clearly, none of the values are negative. Hence, 0 out of 4 differences are negative.

Section 4:

To determine

To find: Squares of the differences obtained in section 2 of part (d).

Section 4:

Expert Solution
Check Mark

Answer to Problem 73E

Solution: The squares of the differences are:

d52=31,684

d62=13,689

d72=41,209

d82=101,124

Explanation of Solution

Calculation: The differences obtained in the section 2 of part (d) above are:

d5=178

d6=117

d7=203

d8=318

The squares of the differences obtained are calculated as follows:

The square of the difference (d5) is calculated as:

d52=d5×d5=178×178=31,684

The square of the difference (d6) is calculated as:

d62=d6×d6=117×117=13,689

The square of the difference (d7) is calculated as:

d72=d7×d7=203×203=41,209

The square of the difference (d8) is calculated as:

d82=d8×d8=318×318=101,124

Hence, the squares of the differences obtained are 31,684, 13,689, 41,209 , and 101,124.

Section 5:

To determine

To find: The sum of the squares of the differences obtained in Section 3 above.

Section 5:

Expert Solution
Check Mark

Answer to Problem 73E

Solution: The sum of the differences is 187,706_.

Explanation of Solution

The squares of the differences are obtained in Section 4 above as:

d52=31,684

d62=13,689

d72=41,209

d82=101,124

The sum of the differences (Σdn2) where n=5,6,7, and 8 is calculated as follows:

Σdn2n=58=d52+d62+d72+d82=(31,684)+(13,689)+(41,209)+(101,124)=187,706

Hence, the sum of the differences is 187,706_.

(e)

To determine

To explain: The least-squares inference based on the calculations performed.

(e)

Expert Solution
Check Mark

Answer to Problem 73E

Solution: The following least-square regression is a better measure of the relationship between the count and the time:

count = 602.8−(74.7×time)

Explanation of Solution

In a linear least-square equation, the graph of the residuals must have an unbiased pattern, that is, the scatter plot should be scattered above and below the x-axis. The residual plots must be both positive and negative. In the calculations performed above, it is observed that for the regression line,

count = 602.8−(74.7×time)

The differences (residuals) are both positive and negative values, whereas, for the regression line,

count = 500−(100×time)

all the differences (residuals) are high and positive values. Thus, the residual plots for the first regression line will lie both below and above the x-axis. Also, its predicted regression line will lie much near to the observed regression line. Whereas, the residual plots for the second regression line will lie only above the x-axis. Also, with such high and positive differences, the predicted regression line will lie far to the observed regression line. Hence, the first line better depicts the relationship between the count and the time because it gives a better approximate value of the response variable.

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Chapter 2 Solutions

Introduction to the Practice of Statistics

Ch. 2.1 - Prob. 1UYKCh. 2.1 - Prob. 2UYKCh. 2.1 - Prob. 3UYKCh. 2.1 - Prob. 4UYKCh. 2.1 - Prob. 5ECh. 2.1 - Prob. 6ECh. 2.1 - Prob. 7ECh. 2.1 - Prob. 8ECh. 2.1 - Prob. 9ECh. 2.2 - Prob. 10UYK
Ch. 2.2 - Prob. 11UYKCh. 2.2 - Prob. 12UYKCh. 2.2 - Prob. 13UYKCh. 2.2 - Prob. 14UYKCh. 2.2 - Prob. 15UYKCh. 2.2 - Prob. 16ECh. 2.2 - Prob. 17ECh. 2.2 - Prob. 18ECh. 2.2 - Prob. 19ECh. 2.2 - Prob. 20ECh. 2.2 - Prob. 21ECh. 2.2 - Prob. 22ECh. 2.2 - Prob. 23ECh. 2.2 - Prob. 24ECh. 2.2 - Prob. 25ECh. 2.2 - Prob. 26ECh. 2.2 - Prob. 27ECh. 2.2 - Prob. 28ECh. 2.2 - Prob. 29ECh. 2.2 - Prob. 30ECh. 2.2 - Prob. 31ECh. 2.2 - Prob. 32ECh. 2.2 - Prob. 33ECh. 2.2 - Prob. 34ECh. 2.2 - Prob. 35ECh. 2.2 - Prob. 36ECh. 2.2 - Prob. 37ECh. 2.3 - Prob. 38UYKCh. 2.3 - Prob. 39UYKCh. 2.3 - Prob. 40ECh. 2.3 - Prob. 41ECh. 2.3 - Prob. 42ECh. 2.3 - Prob. 43ECh. 2.3 - Prob. 44ECh. 2.3 - Prob. 45ECh. 2.3 - Prob. 46ECh. 2.3 - Prob. 47ECh. 2.3 - Prob. 48ECh. 2.3 - Prob. 49ECh. 2.3 - Prob. 50ECh. 2.3 - Prob. 51ECh. 2.3 - Prob. 52ECh. 2.3 - Prob. 53ECh. 2.3 - Prob. 54ECh. 2.3 - Prob. 55ECh. 2.3 - Prob. 56ECh. 2.3 - Prob. 57ECh. 2.3 - Prob. 58ECh. 2.3 - Prob. 59ECh. 2.3 - Prob. 60ECh. 2.4 - Prob. 61UYKCh. 2.4 - Prob. 62UYKCh. 2.4 - Prob. 63UYKCh. 2.4 - Prob. 64UYKCh. 2.4 - Prob. 65ECh. 2.4 - Prob. 66ECh. 2.4 - Prob. 67ECh. 2.4 - Prob. 68ECh. 2.4 - Prob. 69ECh. 2.4 - Prob. 70ECh. 2.4 - Prob. 71ECh. 2.4 - Prob. 72ECh. 2.4 - Prob. 73ECh. 2.4 - Prob. 74ECh. 2.4 - Prob. 75ECh. 2.4 - Prob. 76ECh. 2.4 - Prob. 77ECh. 2.4 - Prob. 78ECh. 2.4 - Prob. 79ECh. 2.4 - Prob. 80ECh. 2.4 - Prob. 81ECh. 2.4 - Prob. 82ECh. 2.4 - Prob. 83ECh. 2.4 - Prob. 84ECh. 2.4 - Prob. 85ECh. 2.4 - Prob. 86ECh. 2.4 - Prob. 87ECh. 2.4 - Prob. 88ECh. 2.4 - Prob. 89ECh. 2.4 - Prob. 90ECh. 2.4 - Prob. 91ECh. 2.5 - Prob. 92UYKCh. 2.5 - Prob. 93UYKCh. 2.5 - Prob. 94ECh. 2.5 - Prob. 95ECh. 2.5 - Prob. 96ECh. 2.5 - Prob. 97ECh. 2.5 - Prob. 98ECh. 2.5 - Prob. 99ECh. 2.5 - Prob. 100ECh. 2.5 - Prob. 101ECh. 2.5 - Prob. 102ECh. 2.5 - Prob. 103ECh. 2.5 - Prob. 104ECh. 2.5 - Prob. 105ECh. 2.5 - Prob. 106ECh. 2.5 - Prob. 107ECh. 2.5 - Prob. 108ECh. 2.5 - Prob. 110ECh. 2.5 - Prob. 111ECh. 2.5 - Prob. 112ECh. 2.6 - Prob. 113UYKCh. 2.6 - Prob. 114UYKCh. 2.6 - Prob. 115UYKCh. 2.6 - Prob. 116UYKCh. 2.6 - Prob. 117UYKCh. 2.6 - Prob. 118UYKCh. 2.6 - Prob. 119ECh. 2.6 - Prob. 120ECh. 2.6 - Prob. 121ECh. 2.6 - Prob. 122ECh. 2.6 - Prob. 123ECh. 2.6 - Prob. 124ECh. 2.6 - Prob. 125ECh. 2.6 - Prob. 126ECh. 2.6 - Prob. 127ECh. 2.6 - Prob. 128ECh. 2.6 - Prob. 129ECh. 2.6 - Prob. 130ECh. 2.7 - Prob. 131ECh. 2.7 - Prob. 132ECh. 2.7 - Prob. 133ECh. 2.7 - Prob. 134ECh. 2.7 - Prob. 135ECh. 2.7 - Prob. 136ECh. 2.7 - Prob. 137ECh. 2.7 - Prob. 138ECh. 2.7 - Prob. 139ECh. 2.7 - Prob. 140ECh. 2.7 - Prob. 141ECh. 2.7 - Prob. 142ECh. 2.7 - Prob. 143ECh. 2 - Prob. 144ECh. 2 - Prob. 145ECh. 2 - Prob. 146ECh. 2 - Prob. 147ECh. 2 - Prob. 148ECh. 2 - Prob. 149ECh. 2 - Prob. 150ECh. 2 - Prob. 151ECh. 2 - Prob. 152ECh. 2 - Prob. 153ECh. 2 - Prob. 154ECh. 2 - Prob. 155ECh. 2 - Prob. 156ECh. 2 - Prob. 157ECh. 2 - Prob. 158ECh. 2 - Prob. 159ECh. 2 - Prob. 160ECh. 2 - Prob. 161ECh. 2 - Prob. 162ECh. 2 - Prob. 163ECh. 2 - Prob. 164ECh. 2 - Prob. 165ECh. 2 - Prob. 166ECh. 2 - Prob. 167ECh. 2 - Prob. 168ECh. 2 - Prob. 169ECh. 2 - Prob. 170ECh. 2 - Prob. 171ECh. 2 - Prob. 172ECh. 2 - Prob. 173ECh. 2 - Prob. 174ECh. 2 - Prob. 175ECh. 2 - Prob. 176E
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