Statistics for engineering and the sciences 5th edition mendenhall solutions manual

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CHAPTER 10 …………………………………………………………..

Simple Linear Regression 10.2

a.

b.

y = 2 + 2x

y=4+x

c.

d.

y = −2 + 4x 10.4

a.

y = −4 − x

Using MINITAB, the least squares line is: Fitted Line P lot y = 18.13 + 6.210 x S R-Sq R-Sq(adj)

140

0 100.0% 100.0%

120 100

y

80 60 40 20 0 0

5

10 x

15

20

The slope of the line is positive – as x increases, y also increases.

Simple Linear Regression

215

10.6

b.

βˆ0 = 18.13:

Since we have no information as to whether x = 0 is within the observed range, we will assume that it is. Then, the estimate of the mean magnitude when redshift level is 0 is 18.13.

c.

β̂1 = 6.21:

For each unit increase in redshift level, the mean value of magnitude is estimated to increase by 6.21.

a.

b.

Some preliminary calculations are:

∑ x = 120

∑ x = 1240

∑ y = .37

∑ y = .0127

2

∑ xy = 3.86

2

( ∑ x ) = 1240 − (120) = 280 x − 2

2

SS xx =

∑

SS xy =

∑ xy − ∑ n∑ = 3.86 −

x=

2

15

n

x

y

∑ x = 120 = 8

βˆ1 =

n

SSxy SS xx

15

=

y=

120(.37) = .9 15

∑ y = .37 = .0247 n

15

.9 = .00321 280

βˆ0 = y − βˆ1 x = .0247 − (.00321)8 = −.00105 yˆ = −.00105 + .00321x c.

216

The least squares line is plotted on the graph above.

Chapter 10

d.

10.8

a.

βˆ0 = −.00105 :

We estimate the mean congestion time to be −.00105 minutes when the number of vehicles is 0.

βˆ1 = .00321 :

We estimate the mean congestion time will increase .00321 minutes for every additional vehicle.

The scattergram appears below: 7 6

Mass

5 4 3 2 1 0 0

10

20

30

40

50

60

70

Time

b.

Excel was used to fit the simple linear regression model to the data and the results appear here:

Regression Statistics Multiple R 0.92376344 R Square 0.853338893 Adjusted R Square 0.846355031 Standard Error 0.857257302 Observations 23 ANOVA df 1 21 22

SS 89.79419524 15.43269171 105.226887

MS 89.79419524 0.734890082

F 122.1872461

Coefficients 5.220695364 -0.114022801

Standard Error 0.295977687 0.010315226

t Stat 17.63881399 -11.053834

P-value 4.55174E-14 3.25999E-10

Regression Residual Total

Intercept TIME

Simple Linear Regression

Significance F 3.25999E-10

217

d.

10.10

a.

βˆ0 = 5.22:

When the time is 0 minutes, we estimate the mass of the spill to be 5.22 pounds.

βˆ1 = −.114:

For each additional one minute in spill time, we estimate the mass of spill to decrease by .114 pounds.

Some preliminary calculations are:

∑ x = 6167

∑ y = 135.8

∑ x = 1,641,115

∑ x y = 34765

i

i

2 i

SSxy =

∑x y −

SSxx =

∑

i i

n = 24

i i

( ∑ x )( ∑ y ) i

i

n (6167)(135.8) = −129.94167 = 34765 − 24 xi2 −

(∑ x )

2

i

n (6167) 2 = 56452.958 = 1,641,115 − 24

βˆ1 =

SSxy SSxx

=

−129.94167 = −.002301769 ≈ −.0023 56452.958

βˆ0 = y − βˆ1 x =

135.8 ⎛ 6167 ⎞ = 6.249792065 ≈ 6.25 − (−.002301769) ⎜ ⎟ 24 ⎝ 24 ⎠

The least squares line is ŷ = 6.25 − .0023x b.

c.

218

βˆ0 = 6.25:

Since x = 0 is not in the observed range, β̂ 0 has no interpretation other than being the y-intercept.

βˆ1 = −.0023:

For each additional increase of 1 part per million of pectin, the mean sweetness index is estimated to decrease by .0023.

ŷ = 6.25 − .0023(300) = 5.56

Chapter 10

10.12

a.

Some preliminary calculations are:

∑ x = 45.12

∑ y = 114.6

∑ x = 88.7788

∑ y = 575.02

2

( ∑ x ) = 88.7788 − (45.12) = 3.9532 2

2

SS xx =

∑

SS xy =

∑ xy − ∑ n∑ = 225.04 −

x=

xi2 −

i

24

n

x

y

∑ x = 45.12 = 1.88

βˆ1 =

24

n

SSxy SSxx

=

∑ xy = 225.04

2

(45.12)(114.6) = 9.592 24 y=

∑ y = 114.6 = 4.775 n

24

9.592 = 2.4264 3.9532

βˆ0 = y − βˆ1 x = 4.775 − (2.4264)(1.88) = .2134 The least squares line is yˆ = −.2134 + 2.4264 x b.

c.

βˆ0 = .2134: βˆ1 = 2.4264:

Simple Linear Regression

When the unflooded area ratio is 0, we estimate the heat transfer enhancement to be .2134. Note: this value falls outside the range of our data, and the interpretation is not useful.

For every 1-unit increase in the unflooded area ratio, we estimate the heat transfer enhancement to increase by 2.4264 units.

219

10.14

a.

b.

c.

10.16

Radiata pine:

For each additional one unit increase in the natural logarithm of number of cycles, we estimate that the mean stress value will decrease 2.50 MPa.

Hoop pine:

For each additional one unit increase in the natural logarithm of number of cycles, we estimate that the mean stress value will decrease 2.36 MPa.

Radiata pine:

When the natural logarithm of number of cycles is 0, we estimate that the mean stress value will be 97.37 MPa.

Hoop pine:

When the natural logarithm of number of cycles is 0, we estimate that the mean stress value will be 122.03 MPa.

It appears that the hoop pines timber has a higher stress level to start with and that the stress level decreases at a slower rate than does the radiata pine timber.

The two linear equations are:

∑x =∑y βˆ ∑ x + βˆ ∑ x = ∑ x y

nβˆ0 + βˆ1

i

i

0

i

2 i

1

i i

From the first equation, divide every term by n.

βˆ0 + βˆ1 x = y ⇒ βˆ0 = y − βˆ1 x Now substitute the value of βˆ0 into the second equation.

( y − βˆ x ) ∑ x + βˆ ∑ x =∑ x y i

1

⎛ ⇒⎜ ⎜ ⎝

2 i

1

i i

∑ y − βˆ ∑ x ⎞⎟ ∑ x + βˆ ∑ x =∑ x y i

i

n ⎟⎠

1

n

i

1

2 i

i i

∑ x ∑ y − βˆ ( ∑ x ) + βˆ ∑ x = ∑ x y ⇒ 2

i

i

i

1

n

⎛ ⇒ βˆ1 ⎜ ⎜⎜ ⎝

n

( ∑ x ) ⎞⎟ = 2

∑

xi2 −

i

n

1

2 i

i i

∑ xy − ⎟⎟ ∑ n

xi yi

i i

⎠

⇒ βˆ1SS xx = SSxy

⇒ βˆ1 =

220

SSxy SSxx

Chapter 10

10.18

( )

Show that E βˆ0 = β 0 First, the model is yi = β 0 + β1 xi + ε i . E ( yi ) = E ( β 0 + β1 xi + ε i ) = β 0 + β1 xi

( ) (

)

E βˆ0 = E y − βˆ1 x = E ( y ) − β1 x

=

1 E n

(from Exercise 10.17)

( ∑ y ) − β ∑n = 1n ∑ ( β + β x ) − β ∑n = 1n ( nβ + β ∑ x ) − β ∑n xi

xi

1

i

0

1 i

1

0

1

i

xi

1

= β0 10.20

a.

From Exercise 10.6,

∑ y = .37, ∑ y = .0127, βˆ = .00321, and SS = .9 2

1

xy

( ∑ y ) = .0127 − .37 = .0035733 y − 2

SS yy =

∑

2

15

n

SSE = SS yy − βˆ1 SS xy = .0035733 − (.00321)(.9) = .0006843

s2 =

SSE .0006843 = = .0000526 15 − 2 n−2

s = s 2 = .0000526 = .0072 We expect most of the observed congestion times to fall within +2s = .0144 minutes of their least squares prediction equation. b.

Using the calculations from Exercise 10.7:

∑ y = 1710.2, ∑ y = 97.8 2

( ∑ y ) = 1710.2 − ( 97.8) = 116.06 y − 2

SS yy =

∑

2

2

n

6

SSE = SS yy − βˆ1SS xy = 116.06 − 0.9665(77.18) = 41.46553 s2 =

SSE 41.46553 = = 10.366 s = s 2 = 10.366 = 3.220 n−2 6−2

We expect most of the apparent porosity percentages for bricks to fall within +2s = 6.44% of their least squares predicted values.

Simple Linear Regression

221

c.

Using the calculations from Exercise 10.8:

∑ y = 262.27, ∑ y = 60.1 2

( ∑ y ) = 262.27 − ( 60.1) = 105.226 y − 2

SS yy =

∑

2

2

23

n

SSE = SS yy − βˆ1SS xy = 105.226 − ( −.1140)(−787.511) = 15.450 s2 =

SSE 15.450 = = .7357 s = s 2 = .7357 = .857 n − 2 23 − 2

We expect most of the mass values to fall within +2s = 1.714 pounds of their least squares predicted values. d.

Using the calculations from Exercise 10.9:

∑ y = 214.41, ∑ y = 45.5 2

( ∑ y ) = 214.41 − ( 45.5) = 76.393 y − 2

SS yy =

∑

2

2

15

n

SSE = SS yy − βˆ1SS xy = 76.393 − (1.062)(54.5767) = 18.433 s2 =

SSE 18.433 = = 1.4161 s = s 2 = 1.4161 = 1.190 n − 2 15 − 2

We expect most of the rain gauge amounts to fall within +2s = 2.380 millimeters of their least squares predicted values. e.

Using the calculations from Exercise 10.10:

∑ y = 769.72, ∑ y = 135.8 2

( ∑ y ) = 769.72 − (135.8) = 1.318333 y − 2

SS yy =

∑

2

2

n

24

SSE = SS yy − βˆ1SS xy = 1.318333 − ( −.00230)( −129.94167) = 1.0192

s2 =

SSE 1.0192 = = .046225 s = s 2 = .046225 = .215 n − 2 24 − 2

We expect most of the sweetness index values to fall within +2s = 0.430 of their least squares predicted values.

222

Chapter 10

f.

From Exercise 10.11,

∑ y = 130.27, ∑ y = 1,081.3975, βˆ = .01439, and SS = 3640.46471 2

xy

1

( ∑ y ) = 1,081.3975 − (130.27) = 83.146153 y − 2

SS yy =

∑

2

2

17

n

SSE = SS yy − βˆ1 SSxy = 83.146153 − .01439(3640.46471) = 30.7679 s2 =

SSE 30.7679 = = 2.0512 n−2 17 − 2

s = s 2 = 2.0512 = 1.4322 We expect most of the observed times to drill to fall within + 2s = 2.8644 minutes of their least squares prediction equation.) g.

Using the calculations from Exercise 11.8:

∑ y = 575.02 2

( ∑ y ) = 575.02 − (114.6) = 27.805 y − 2

SS yy =

∑

2

2

24

n

SSE = SS yy − βˆ1 SS xy = 27.805 − 2.4264(9.592) = 4.531 s2 =

SSE 4.531 = = .206 n − 2 24 − 2

s = s 2 = .206 = .4538

We expect most of the observed heat transfer enhancement values to fall within ±2s = .9076 of their least squares predicted value. h.

Using the calculations from Exercise 10.13:

∑ y = 40.6022 ( ∑ y ) = 40.6022 − (16.3) = 2.6465 SS = ∑ y − n 7 2

2

2

yy

SSE = SS yy − βˆ1 SSxy = 2.6465 − (.987)(2.4267) = .2513 s2 =

SSE .2513 = = .0503 n−2 7−2

s = s 2 = .0503 = .2242

We expect most of the observed shear strength values to fall within ±2s = .4484 of their least squares predicted value.

Simple Linear Regression

223

10.22

a.

Some preliminary calculations are:

∑ x = 5.5075 ∑ y = 10,255 ∑ xy = 237.10

∑ x = 5.45 ∑ y = 239

2

2

( ∑ x ) = 5.5075 − (5.45) = 0.5570833 x − 2

SS xx =

∑

SS xy =

∑ xy −

x=

2

2

n

6

( ∑ x∑ y ) = 237.10 − (5.45)(239) = 20.008333 n

6

∑ x = 5.45 = 0.908333 n

SSxy

βˆ1 =

SS xx

y=

6

=

∑ y = 239 = 39.83333 n

6

20.008333 = 35.9162 0.5570833

βˆ0 = y − βˆ1 x = 39.83333 − 35.9162(0.9083333) = 7.2094 yˆ = 7.2094 + 35.9162 x

( ∑ y ) = 10, 255 − ( 239 ) = 734.8333 y − 2

b.

SS yy =

∑

2

2

n

6

SSE = SS yy − βˆ1SS xy = 734.8333 − (35.9162)(20.008333) = 16.210

s2 = c.

10.24

SSE 16.210 = = 4.0525 s = s 2 = 4.0525 = 2.013 6−2 n−2

We expect most of the drug release rates to fall within +2s = 4.026 of their least squares predicted values.

By Theorem 10.1,

χ2 =

(n − 2) s 2

σ2

⎡ (n − 2) s 2 ⎤ V χ2 = V ⎢ ⎥ = 2v = 2(n − 2) 2 ⎣ σ ⎦

( )

⇒

224

( n − 2) 2

σ4

V ( s 2 ) = 2(n − 2) = V ( s 2 ) =

2σ 4 (n − 2)

Chapter 10

10.26

a.

The 95% confidence interval for β1 is:

βˆ1 ± tα 2 s

SSxx

For confidence coefficient .95, α = 1 − .95 = .05 and α/2 = .05/2 = .025. From Table 7 in Appendix B, with df = n − 2 = 6 – 2 = 4, t.025 = 2.776. The 95% prediction interval is: 0.9665 ± 2.776(3.220)

79.85333

⇒ 0.9665 ± 1.0003 ⇒ (−0.0338, 1.9668) For every one millimeter increase in the mean pore diameter of a brick, we are 95% confident that the apparent porosity of the brick will do anything from decrease by .00338% to increase by 1.9668%. b.

To determine if the true slope of the line differs from 0, we test: H 0 : β1 = 0 H a : β1 ≠ 0 The test statistic is t =

βˆ1 s

SSxx

=

0.9665 = 2.682 3.220 79.8533

For a two tailed test, the rejection region requires α/2 = .05/2 = .025 in both tails of the t distribution with df = n − 2 = 6 – 2 = 4. From Table 7 in Appendix B, t.025 = 2.776.

The rejection region is t > 2.776 or t < −2.776.

Since the observed value of the test statistic does not fall in the rejection region (t = 2.682 >/ 2.776), H0 cannot be rejected. There is insufficient evidence to indicate that mean pore diameter and apparent porosity of a brick are linearly related at α = .10.

10.28

c.

Both the confidence interval and the test of hypothesis fail to conclude that the slope of the line differs from 0. Therefore, neither indicates that the straight-line model is useful in predicting the apparent porosity percentage of a brick.

a.

From Exercise 10.9, SSxx = 64.8173333 , βˆ1 = 1.022 , s = .8187 To determine if the rain gauge amount and artificial neural network rain amount are positively linearly related, we test: H0: β1 = 0 Ha: β1 > 0

Simple Linear Regression

225

The test statistic is t =

βˆ1 − 0 sβˆ

=

1

1.022 = 10.050 .8187 64.8173333

The rejection region requires α = .10 in the upper tail of the t distribution with df = n – 2 = 15 – 2 = 13. From Table 7, Appendix B, t.10 = 1.350. The rejection region is t > 1.350.

Since the observed value of the test statistic falls in the rejection region (t = 10.050 > 1.350), H0 is rejected. There is sufficient evidence to indicate the rain gauge amount and artificial neural network rain amount are positively linearly related at α = .10. b.

For confidence coefficient .90, α = 1 − .90 = .10 and α/2 = .10/2 = .05. From Table 7, Appendix B, with df = n – 2 = 15 – 2 = 13, t.05 = 1.771. The 90% confidence interval is:

βˆ1 ± t.05 sβˆ ⇒ βˆ1 ± t.05 1

s .8187 ⇒ 1.022 ± 1.771 SS xx 64.817333

⇒ 1.022 ± .180 ⇒ (.842, 1.202) We are 90% confident that the change in the mean value of rain gauge amount for each unit change in neural network rain is between .842 and 1.202.

10.30 From Exercise 10.12, βˆ1 = 2.4264 and SSxx = 3.9532 . From Exercise 10.20 c, s = .454 The confidence interval for β1 is βˆ1 ± tα / 2 sβˆ where sβˆ = 1

1

s SSxx

For confidence coefficient .95, α = 1 − .95 = .05 and α/2 = .05/2 = .025. From Table 7 with df = n − 2 = 24 − 2 = 22, t.025 = 2.074. The 95% confidence interval is: 2.4264 ± 2.074

.454 ⇒ 2.4264 ± .4736 ⇒ (1.9528, 2.9000) 3.9532

We are 95% confident the mean heat transfer enhancement will increase from between 1.9528 and 2.9000 for each 1 unit increase in unflooded area ratio.

10.32

226

a.

Chapter 10

b.

Some preliminary calculations are:

∑ x = 2,053

∑ x = 602, 265.5 2

∑ y = 59.2

∑ y = 517.22

∑ xy = 17,326.7

2

( ∑ x ) = 602, 265.5 − (2,053) = 149.92857 x − 2

2

SS xx =

∑

SS xy =

∑ xy − ∑ n∑ = 17,326.7 −

x=

2

n

x

7

y

∑ x = 2,053 = 293.28571

βˆ1 =

n

SSxy SS xx

7

=

(2,053)(59.2) = −35.814286 7

y=

∑ y = 59.2 = 8.4571429 n

7

−35.814286 = −.2388757 ≈ −.239 149.92857

βˆ0 = y − βˆ1 x = 78.516 The least squares line is yˆ = 78.516 − .239 x c.

The least squares line is plotted on the graph above.

d.

To determine if temperature is a useful linear predictor of change in free energy, we test:

H 0 : β1 = 0 H a : β1 ≠ 0 The test statistic is t =

βˆ1 − 0 s / SSxx

=

−.239 − 0 = −2.31 1.265 / 149.9286

The rejection region requires α/2 = .01/2 = .005 in both tails of the t distribution with df = n − 2 = 7 − 2 = 5. From Table 7, Appendix B, t.005 = 4.032. The rejection region is t > 4.032 or t < −4.032.

Since the observed value of the test statistic does not fall in the rejection region (t = −2.31 </ −4.032), H 0 is not rejected. There is insufficient evidence to indicate that temperature is a useful linear predictor of change in free energy. e.

It looks like observation #5 is an outlier.

f.

Refitting the data yields a least squares line of yˆ = 139.759 − .4996 x .

Simple Linear Regression

227

10.34

If we make the four assumptions about ε in Section 10.2, then the sampling distribution of βˆ1 will be a normal distribution with mean β1 and standard deviation σ βˆ = σ / SSxx .

1

If we take βˆ1 and subtract its mean and divide by its standard deviation, we will form a normally distributed random variable with mean 0 and standard deviation 1.

10.36

P ( −tα / 2 < t < tα / 2 ) = 1 − α

⎡ ⎤ βˆ − β P ⎢ −tα / 2 < 1 1 < tα / 2 ⎥ = 1 − α s / SSxx ⎣⎢ ⎦⎥ ⎡ s s ⎤ < βˆ1 − β1 < tα / 2 P ⎢ −tα / 2 ⎥ = 1−α SSxx SSxx ⎥⎦ ⎢⎣

⎡ s s ⎤ < − β1 < − βˆ1 + tα / 2 P ⎢ − βˆ1 − tα / 2 ⎥ = 1−α SSxx SSxx ⎥⎦ ⎢⎣ ⎡ s s ⎤ < β1 < βˆ1 + tα / 2 P ⎢ βˆ1 − tα / 2 ⎥ = 1−α SSxx SSxx ⎥⎦ ⎣⎢ Therefore, a (1 − α)100% confidence interval for β1 is βˆ1 ± tα / 2 s / SSxx with df = n − 2.

10.38

βˆ0 = 175.4:

We estimate the mean time required to perform the task to equal 175.4 milliseconds when the Index of Difficulty is 0.

βˆ1 = 132.2:

For every one unit increase in the Index of Difficulty, we estimate the mean time to perform the task to increase 133.2 milliseconds.

b.

r = .951:

There is a strong positive linear relationship between the Index of Difficulty and the time required to perform the task.

c.

To determine if the Index of Difficulty is a useful linear predictor of time, we test:

a.

H 0: ρ = 0 Ha: ρ ≠ 0 The test statistic is t =

r n−2 1− r

2

=

.951 160 − 2 1 − .9512

= 38.662

The p-value for this is 2 P (t > 38.662) where t is based on df = n − 2 = 160 − 2 = 158.

From Table 7, Appendix B, p < .01.

Using α = .05 (α > p), we can reject H 0 . There is sufficient evidence to indicate the Fitt’s Law Model is statistically adequate for predicting performance time.

228

Chapter 10

d.

r 2 = (.951) 2 = .9044

90.44% of the sum of squares of deviation of the sample time values about their mean can be explained by using the Index of Difficulty as a linear predictor.

10.40

10.42

Radiata Pine:

84% of the variation in the sample stress values around their mean can be explained by the linear relationship with the natural logarithm of number of blade cycles.

Hoop Pine:

90% of the variation in the sample stress values around their mean can be explained by the linear relationship with the natural logarithm of number of blade cycles.

a.

There is a strong negative linear relationship between number of online courses and weekly quiz grade.

b.

To determine if there is a significant negative correlation, we test: H 0: ρ = 0 Ha: ρ < 0

The test statistic is t =

r n−2

1− r2

=

−.726 24 − 2 1 − (−.726) 2

= −4.952

For a one tailed test, the rejection region requires α = .05 in the lower tail of the t distribution with df = n − 2 = 24 – 2 = 22. From Table 7 in Appendix B, t.05 = 1.717. The rejection region is t < −1.717.

Since the observed value of the test statistic does fall in the rejection region (t = −4.952 < −1.717), H 0 can be rejected. There is sufficient evidence to indicate that a significant negative correlation exists between number of online courses and weekly quiz grade.

10.44

a.

βˆ1 = 450:

For each additional processing step required, we estimate the mean response time to increase 450 milliseconds.

b.

r 2 = .91:

91% of the sum of squares of deviations of the sample response times about their means can be explained by using the number of processing steps required as a linear predictor.

c.

H 0: ρ = 0 Ha: ρ ≠ 0

The test statistic is t =

Simple Linear Regression

r n−2

1− r

2

=

.954 8 − 2 = 7.79 1 − .91

229

The rejection region requires α/2 = .01/2 = .005 in both tails of the t distribution with df = n − 2 = 8 − 2 = 6. From Table 7, Appendix B, t.005 = 5.959. The rejection region is t > 5.959 or t < −5.959.

Since the observed value of the test statistic falls in the rejection region (t = 7.79 > 5.959), H 0 is rejected. There is sufficient evidence to indicate the number of processing steps required is a useful linear predictor of response time.

10.46

Show

βˆ1 s / SSxx

=

r n−2

1− r2

From Exercise 10.45, we know: SS yy

βˆ1 = r

⇒

=

SSxx

r

βˆ

1

s / SSxx

r SS yy s

and SSE = SS yy (1 − r 2 )

=

SS yy SSxx

s / SSxx

r SS yy

=

SS yy (1 − r 2 )

⎛ SSE ⎞ ⎜⎜ since s = ⎟ n − 2 ⎟⎠ ⎝

(n − 2) =

10.48

a.

r (1 − r 2 ) n−2

=

r n−2 1− r2

Statistix was used to find the desired confidence interval and the output is shown here: Predicted/Fitted Values of MASS Lower Predicted Bound Predicted value Upper Predicted Bound SE (Predicted Value)

2.7787 4.3085 5.8383 0.8890

Unusualness (Leverage) Percent Coverage Corresponding T

0.0755 90.0 1.72

Lower Fitted Bound Fitted Value Upper Fitted Bound SE (Fitted Value)

3.9032 4.3085 4.7138 0.2355

Predictor Values: TIME = 8.0000

We are 90% confident that the mean mass of all spills with an elapsed time of 8 minutes will fall between 3.9032 and 4.7138 pounds.

230

Chapter 10

b.

10.50

The confidence interval for E(y) is always narrowest at the predicted value x = x . So the interval would be narrower than the one found above.

Answers may vary. One possible answer is: The 90% confidence interval for x = 220.00 is (5.64898, 5.83848). We are 90% confident that the mean sweetness index of all orange juice samples will be between 5.64898 and 5.83848 parts per million when the pectin value is 220.00

10.52

a.

b.

Some preliminary calculations are:

∑ x = 655.1

∑ x = 12,711.47

∑ y = 644.7

∑ y = 11,824.45

∑ xy = 12,112.34

2

2

( ∑ x ) = 12,711.47 − 655.1 = 790.46972 x − 2

2

SS xx =

∑

SS xy =

∑ xy − ∑ n∑ = 12,112.34 −

2

n

36

x

y

655.1(644.7) = 380.59083 36

( ∑ y ) = 11,824.45 − 644.7 = 278.9475 y − 2

SS yy =

βˆ1 =

∑

SSxy SS xx

2

2

n

=

36

380.59083 = .481474268 ≈ .4815 790.46972

βˆ0 = y − βˆ1 x =

644.7 (655.1) − .481474268 = 9.146839081 ≈ 9.1468 36 36

The least squares line is yˆ = 9.1468 + .4815 x

Simple Linear Regression

231

c.

See the plot in part a.

d.

SSE = SS yy − βˆ1SS xy = 278.9475 − .481474268(380.59083) = 95.70280872 s2 =

SSE 95.70280872 = = 2.8147885 n−2 36 − 2

s = 2.8147885 = 1.6777

To determine if the breast height diameter is a useful predictor of tree height, we test:

H 0 : β1 = 0 H a : β1 ≠ 0 The test statistic is t =

βˆ1

s SS xx

−

.4815 = 8.069 1.6777 790.46972

The two-sided, small sample test requires α/2 = .05/2 = .025 in each tail of the t distribution with df = n − 2 = 36 − 2 = 34. From Table 7, Appendix B, t.025 ≈ 2.042.

The rejection region is t < − 2.042 or t > 2.042.

Since the observed value of the test statistic falls in the rejection region (t = 8.069 > 2.042), H 0 is rejected. There is sufficient evidence to indicate that breast height diameter is a useful predictor of tree height at α = .05. e.

For x = 20, ŷ = 9.1468 + .4815(20) = 18.7768.

The form of the confidence interval is yˆ ± tα / 2 s

1 ( x − x )2 + n SSxx

For confidence coefficient .90 = 1 − α ⇒ α = 1 − .90 = .10 and α/2 = .10/2 = .05. From Table 7, Appendix B, with df = n − 2 = 36 − 2 = 34, t.05 ≈ 1.697. The 90% confidence interval is: 18.7768 ± 1.697(1.6777)

1 (20 − 18.19722) 2 + ⇒ 18.7768 ± .5084 36 790.46972 ⇒ (18.2684, 19.2852)

We are 90% confident that the mean height of white spruce trees with breast diameter of 20 cm is between 18.2684 m and 19.2852 m.

232

Chapter 10

10.54

a.

The assumption that the E’s have a normal distribution causes the error of prediction to also have a normal distribution.

b.

E ( yp − yˆ ) = E ( yp ) = E ( yˆ ) = yp − yp = 0

V ( yp − yˆ ) = V ( yˆ ) = σ

⎡

2 ⎢1

⎢n ⎣

( x − x ) ⎤⎥ + 2

p

SSxx

⎥ ⎦

yp − yˆ

(

10.56

t=

xp − x 1 σ 1+ + n SSxx

)

2

(n − 2)s 2 / σ 2 ( n − 2) yp − yˆ

From Exercise 10.55, we know z =

1 n

σ 1+ + variable. From Theorem 6.11,

(n − 1) s 2

σ2

(x − x)

2

is a standard normal random

p

SSxx

is a chi-square distribution with v = (n − 2) degrees

of freedom. ⇒t =

10.58

z χ2 / v

which is a t distribution with (n − 2) df.

a.

Simple Linear Regression

233

b.

βˆ1 =

SSxy SSxx

=

−15.5528 = −.05334 ≈ −.0533 291.6

βˆ0 = y − βˆ1 x = 6.826 /10 − (−.053336076)(−2682 /10) = −13.622 The least squares line is yˆ = −13.622 − .0533x . c.

For each unit increase in temperature, we estimate the proportion of impurity passing through helium to decrease .0533.

d.

H 0 : β1 = 0 H a : β1 ≠ 0 The test statistic is t =

βˆ1 s / SSxx

s = SSE/(n − 2) SSE = SS yy − βˆ1SS xy = .9731464 − (−.05334)(−15.5528) = .14356 ⇒ s = .14356 /(10 − 2) = .13396 t=

−.05334 = −6.797 .13396 / 291.6

For a two-tailed test, the rejection region requires α/2 = .05/2 = .025 in each tail of the t distribution with df = n − 2 = 10 − 2 = 8 df. From Table 7, Appendix B, t.025 = 2.306 .

The rejection region is t < −2.306 or t > 2.306.

Since the observed value of the test statistic does fall in the rejection region (t = −6.797 < −2.306, H 0 is rejected. There is sufficient evidence to indicate x and y are linearly related at α = .05. e.

The form of the confidence interval is:

βˆ1 ± tα / 2 s / SSxx For confidence coefficient .90, α = .10 and α/2 = .10/2 = .05. From Table 7, Appendix B, with df = n − 2 = 10 − 2 = 8, t.05 = 1.860. The 90% confidence interval is: −.053 ± 1.860

(.13396) ⇒ −.053 ± .015 ⇒ (−.068, − .038) 291.6

For every one degree increase in temperature, we are 90% confident that the change in mean proportion of impurity passing through helium will be between −.068 and −.038.

234

Chapter 10

f.

r=

SSxy SSxx SS yy

=

−15.5528 = −.923 291.6(.9731464)

For r = −.923, there is a strong negative linear relationship between the proportion of impurity and temperature. g.

r 2 = (−.923) 2 = .825 82.5% of the total sum of squares of deviations of the proportion of impurities about their mean can be attributed to the linear relationship between the proportion of impurities and temperature.

h.

The 99% prediction interval is: 1 (x − x) yˆ ± tα / 2 s 1 + + SSxx n

2

.769 ± 3.355(.13396) 1 +

1 ( −270 − (−268.2) 2 + 10 291.6

⇒ .769 ± .474 ⇒ (.295, 1.243) i.

When x = −270, yˆ = −13.622 − .0533(−270) = .769. For confidence coefficient .99, α = 1 − .99 = .01 and α/2 = .01/2 = .005. From Table 7, Appendix B, t.005 = 3.355 with df = n − 2 = 10 − 2 = 8. The 99% confidence interval for the mean proportion is: 1 (x − x) + , where s = SSE /(n − 2) SSxx n 2

yˆ ± tα / 2 s

SSE = SS yy − βˆ1SS xy = .9731464 − (−.05334)(−15.5528) = .14356 ⇒ s = .14356 /(10 − 2) = .13396 The confidence interval is:

.769 ± 3.355(.13396)

Simple Linear Regression

1 (−270 − (−268.2) 2 + ⇒ .769 ± .150 ⇒ (.619, .919) 10 291.6

235

10.60

Statistix was used to conduct a simple linear regression analysis between the two ratios. We used y = IR/Green ratio and x = UV/Green ratio. The results are shown below: Unweighted Least Squares Linear Regression of IR Predictor Variables Constant UV

Coefficient 2.48553 -1.55300

Std Error 0.08173 0.10273

R-Squared Adjusted R-Squared

0.9122 0.9082

Source Regression Residual Total

SS 0.66212 0.06374 0.72586

DF 1 22 23

Cases Included 24

T 30.41 -15.12

P 0.0000 0.0000

Resid. Mean Square (MSE) Standard Deviation MS 0.66212 0.00290

F 228.53

0.00290 0.05383

P 0.0000

Missing Cases 0

To determine if a linear relationship exists between the two ratios, we test: H 0 : β1 = 0 H a : β1 ≠ 0

The test statistic is t = −15.12.

The p-value is p = 0.0000.

At α = .05, there is sufficient evidence to reject H0. There is sufficient evidence to indicate that a linear relationship exists between the two ratios.

10.62

Some preliminary calculations are:

∑ x = 285 ∑ y = −1964

∑x ∑ xy = −110,030 2

∑ y = 771,994 2

SSxy = ∑ xy −

∑ x∑ y = −110,030 − 285(−1964) = 1918

SSxy = ∑ x 2 −

( ∑ x ) = 25,025 − 285 = 8780

SSyy = ∑ y 2 −

( ∑ y ) = 771,994 − (−1964) = 534.8

n

5

2

5

n

2

x=

n

∑ x = 285 = 57 n SSxy

2

2

5 ∑ y = −1964 = −392.8 y= n 5

5 1918 βˆ1 = = = .218450125 =.21845 SSxx 8780 βˆ = y = βˆ x = −392.8 − .218451025(57) = −405.2517084 ≈ −405.2517 0

236

1

Chapter 10

SSE = SSyy = βˆ1SSxy = 534.8 − .218451025(1918) = 115.8109339

s2 =

SSE 115.8109339 = = 38.60364463 n−2 5−2

s=

38.60364463 = 6.21318

To determine if annealing time is a useful linear predictor of passivation potential, we test:

H0: β1 = 0 Ha: β1 ≠ 0 The test statistic is t =

βˆ1 s / SSxx

=

.21845 = 3.294 6.21318 / 8780

The rejection region requires α/2 = .05/2 = .025 in both tails of the t-distribution with df = n − 2 = 5 − 2 = 3.

From Table 7 in Appendix B, t.025 = 3.182. The rejection region is t > 3.182 or t < −3.182.

Since the observed value of the test statistic falls in the rejection region (t = 3.29 > 3.182), H0 is rejected. There is sufficient evidence to indicate that annealing time is a useful linear predictor of passivation potential.

The 99% confidence interval for the mean passivation potential of crystallized alloy when annealing time is 30 minutes is 1 ( xp − x ) + yˆ ± tα / 2 s SSxx n ŷ = βˆ + βˆ x = −405.2517 + .21845(30) = −398.6982 2

0

1

For conficence coefficient .90, α = 1 − .90 = .10 and α/2 = .10/2 = .05. From Table 7, Appendix B with df = n − 2 = 5 − 2 = 3, t.05 = 2.353. The 90% confidence interval is −398.6982 ± 2.353(6.21318)

1 (30 − 57) 2 + ⇒ −398.6982 ± 7.7777 5 8780 ⇒ (−406.4759, −390.9205)

We are 90% confident that when the annealing time is 30 minutes, the mean passivation potential of crystallized alloy is between −406.4759 and −390.9205.

10.64

a.

There appears to be a somewhat positive linear relationship.

b.

If there was very little snowfall in an area, then the erosion will not be typical. Thus, it seems reasonable to remove these data points.

c.

For confidence level .90, α = .10 and α/2 = .10/2 = .05. From Table 7, Appendix B, with df = n − 2 = 47 − 2 = 45, t.05 ≈ 1.684. The confidence interval is:

βˆ1 ± t.05 sβˆ ⇒ 1.39 ± 1.684(.06) ⇒1.39 ± .101 ⇒ (1.289, 1.491) 1

Simple Linear Regression

237

d.

10.66

We are 90% confident that the change in the mean McCool winter-adjusted rainfall erosivity index for each one unit change in the once-in-5-year snowmelt runoff amount is between 1.289 and 1.491.

Statistix was used to create the following simple linear regression printouts: Unweighted Least Squares Linear Regression of load Predictor Variables Constant age_days

Coefficient 14192.1 -148.978

Std Error 878.684 34.1510

R-Squared Adjusted R-Squared

0.7311 0.6927

Source Regression Residual Total

SS 4039390 1485858 5525248

DF 1 7 8

Cases Included 9

T 16.15 -4.362

P 0.0000 0.0033

Resid. Mean Square (MSE) Standard Deviation MS 4039390 212265

F 19.03

212265 460.720

P 0.0033

Missing Cases 0

Predicted/Fitted Values of load Lower Predicted Bound Predicted value Upper Predicted Bound SE (Predicted Value)

8852.3 10021 11189 494.11

Unusualness (Leverage) Percent Coverage Corresponding T

0.1502 95.0 2.36

Lower Fitted Bound Fitted Value Upper Fitted Bound SE (Fitted Value)

9598.5 10021 10443 178.55

Predictor Values: age_days = 28.000

βˆ1 = −148.98 :

For every one-unit increase in age, we estimate mean load to decrease by −148.908 lb/ft.

βˆ0 = 14192 :

For age = 0, we estimate mean load to be 14192 (not very practical).

s = root MSE = 460.72: We expect most of the observed values to fall within ±2s = ±2(460.72) = 921.44 of their predicted values.

r 2 = .7311: b.

There is a 73.11% reduction in the sum of squared deviations of the sample y values about their mean when we use age as a linear predictor of load.

To determine if x and y are linearly related, we test:

H 0 : There is no linear correlation between x and y H a : The variables x and y are linearly related The test statistic is t = −4.362.

The p-value is .0033.

238

Chapter 10

Since α = .05 > p-value = .0033, H 0 is rejected. There is sufficient evidence to indicate x and y are linearly related at α = .05. c.

10.68

From the printout, we see the 95% confidence interval for the mean load required before the first crack is observed for 28-day-old pipes is (9598.5, 10443).

Some preliminary calculations are:

∑ x = 21.8

∑ y = 243.81

∑ x = 51.635

∑ y = 6144.1421

2

∑

( ∑ x ) = 51.635 − (21.8) = 4.111 x −

SS yy =

∑

( ∑ y ) = 6144.1421 − (243.81) = 199.81049 y −

SSxy =

∑ xy − ∑ n∑ = 559.92 −

2

SSxx =

2

2

n

10

2

a.

∑ xy = 559.92

2

βˆ1 =

SSxy SSxx

2

2

n

x

=

10

y

(21.8)(243.81) = 28.4142 10

28.4142 = 6.91175 4.111

βˆ0 = y − βˆ1 x =

243.81 ⎛ 21.8 ⎞ = −(6.91175) ⎜ ⎟ = 9.3134 10 ⎝ 10 ⎠

The least squares line is yˆ = 9.3134 + 6.91175 x

Simple Linear Regression

239

b.

c.

d.

βˆ0 = 9.3134

When total oxidation products is 0%, we estimate the amount of oxidative compounds to be 9.3134%.

βˆ1 = 6.91175

For every 1% increase in total oxidation products, we estimate the amount of oxidation compounds to increase 6.91175%.

SSE = SS yy − βˆ1SS xy = 199.81 − 6.91175(28.4142) = 3.4186

s2 = e.

10.70

SSE 3.4186 = = .4273 n − 2 10 − 2

s = s 2 = .4273 = .6537

We expect most of the observed amount of oxidation compound values to fall within ±2s = 1.3074 percentage of their least squares predicted value.

Some preliminary calculations are:

∑ x = 280

∑ x = 14550

∑ y = 103

∑ y = 1959

SSxy =

2

∑ xy = 5330

2

∑ xy − ∑ n∑ = 5330 − x

y

(280)(103) = 523.33 6

∑

( ∑ x ) = 14550 − (280) = 1483.33 x −

∑

( ∑ y ) = 1959 − (103) = 190.83 y −

2

SSxx =

2

2

n

6

2

SS yy =

240

2

2

n

6

Chapter 10

a.

SSxy SSxx

=

523.33 = .353 1483.33

βˆ0 = y − βˆ1 x =

103 280 − (.353) i = .702 6 6

The least squares line is yˆ = .702 + .353 x b.

c.

r=

SSxy SSxx SS yy

=

523.33 = .9836 1483.33(190.83)

For r = .9836, there is a strong positive linear relationship between the amount of insulation and temperature rise above ambient.

r 2 = (.9836) 2 = .9675 96.75% of the total sum of squared deviation of the temperature rise above ambient about their mean can be attributed to the linear relationship with insulation. d.

To determine if the linear model is useful, we test:

H 0 : β1 = 0 H a : β1 ≠ 0 The test statistic is t =

βˆ1 s/ SSxx

s = SSE /(n − 2)

Simple Linear Regression

241

SSE = SS yy − βˆ1SS xy = 190.83 − (.353)(523.33) = 6.0945

s = 6.0945 /(6 − 2) = 1.234 ⇒t =

.353 = 10.92 1.234 / 1483.33

The rejection region for the two-tailed test requires α/2 = .01/2 = .005 in both tails of the t distribution with df = n − 2 = 6 − 2 = 4. From Table 7, Appendix B, t.005 = 4.604.

The rejection region is t > 4.604 or t < −4.604.

Since the observed value of the test statistic does fall in the rejection region (t = 10.92 > 4.604), H 0 is rejected. There is sufficient evidence to indicate the model is useful in predicting temperature rise above ambient at α = .01. e.

The form of the confidence interval is: 1 (x − x) yˆ ± tα / 2 s + SSxx n

2

where x = 35, yˆ = .702 + .353(35) = 13.05 .

The 99% confidence interval is: 13.05 ± 4.604 ⋅ 1.234

242

1 (35 − 46.7) 2 + ⇒ 13.05 ± 2.91 ⇒ (10.14, 15.96) 6 1483.33

Chapter 10