14 Let’s return to the carbonation depth-strength data of 13 and calculate a

Example 12.14 Let’s return to the carbonation depth-strength data of Example 12.13 and calculate a

  95 PI for a strength value that would result from selecting a single core specimen whose depth is 45 mm. Relevant quantities from that example are

  yˆ 5 13.79 s Yˆ 5 .7582 s 5 2.8640

  For a prediction level of 95 based on n 2 2 5 16

  df, the t critical value is 2.120,

  exactly what we previously used for a 95 confidence level. The prediction interval is then

  Plausible values for a single observation on strength when depth is 45 mm are (at the

  95 prediction level) between 7.51 MPa and 20.07 MPa. The 95 confidence inter- val for mean strength when depth is 45 was (12.18, 15.40). The prediction interval

  is much wider than this because of the extra (2.8640) 2 under the square root. Figure

  12.18, the Minitab output in Example 12.13, shows this interval as well as the con- fidence interval.

  ■ The Bonferroni technique can be employed as in the case of confidence inter-

  vals. If a 100(1 2 a) PI is calculated for each of k different values of x, the simul- taneous or joint prediction level for all k intervals is at least 100(1 2 ka) .

  EXERCISES Section 12.4 (44–56)

  44. Fitting the simple linear regression model to the n 5 27

  a. Explain why s Yˆ is larger when x 5 60 than when x 5 40 .

  observations on x5 modulus of elasticity and y5 flexural

  b. Calculate a confidence interval with a confidence level of

  strength given in Exercise 15 of Section 12.2 resulted in yˆ 5

  95 for the true average strength of all beams whose

  7.592, s Yˆ

  5 .179 when and x 5 40 yˆ 5 9.741, s Yˆ

  modulus of elasticity is 40.

  for . x 5 60

  CHAPTER 12 Simple Linear Regression and Correlation

  c. Calculate a prediction interval with a prediction level of

  astringency that can be attributed to the model relation-

  95 for the strength of a single beam whose modulus of

  ship between astringency and tannin concentration.

  elasticity is 40.

  b. Calculate and interpret a confidence interval for the slope

  d. If a 95 CI is calculated for true average strength when

  of the true regression line.

  modulus of elasticity is 60, what will be the simultane-

  c. Estimate true average astringency when tannin concen-

  ous confidence level for both this interval and the inter-

  tration is .6, and do so in a way that conveys information

  val calculated in part (b)?

  about reliability and precision.

  45. Reconsider the filtration rate–moisture content data intro-

  d. Predict astringency for a single wine sample whose tan-

  duced in Example 12.6 (see also Example 12.7).

  nin concentration is .6, and do so in a way that conveys

  a. Compute a 90 CI for b 0 1 125b 1 , true average mois-

  information about reliability and precision.

  ture content when the filtration rate is 125.

  e. Does it appear that true average astringency for a tannin

  b. Predict the value of moisture content for a single experi-

  concentration of .7 is something other than 0? State and

  mental run in which the filtration rate is 125 using a 90

  test the appropriate hypotheses.

  prediction level. How does this interval compare to the

  47. The simple linear regression model provides a very good fit

  interval of part (a)? Why is this the case?

  to the data on rainfall and runoff volume given in Exercise

  c. How would the intervals of parts (a) and (b) compare to

  16 of Section 12.2. The equation of the least squares line is

  a CI and PI when filtration rate is 115? Answer without

  y 5 21.128 1 .82697x, r 2 5 .975 , and s 5 5.24 .

  actually calculating these new intervals.

  a. Use the fact that s Yˆ

  5 1.44 when rainfall volume is

  d. Interpret the hypotheses H 0 :b 0 1 125b 1 5 80 and

  40 m 3 to predict runoff in a way that conveys information

  H a :b 0 1 125b 1 , 80 , and then carry out a test at sig-

  about reliability and precision. Does the resulting inter-

  nificance level .01.

  val suggest that precise information about the value of

  46. Astringency is the quality in a wine that makes the wine

  runoff for this future observation is available? Explain

  drinker’s mouth feel slightly rough, dry, and puckery. The

  your reasoning.

  paper “Analysis of Tannins in Red Wine Using Multiple

  b. Calculate a PI for runoff when rainfall is 50 using the

  Methods: Correlation with Perceived Astringency” (Amer. J.

  same prediction level as in part (a). What can be said

  of Enol. and Vitic., 2006: 481–485) reported on an investi-

  about the simultaneous prediction level for the two in-

  gation to assess the relationship between perceived astrin-

  tervals you have calculated?

  gency and tannin concentration using various analytic

  48. The catch basin in a storm-sewer system is the interface

  methods. Here is data provided by the authors on x tan-

  between surface runoff and the sewer. The catch-basin insert

  nin concentration by protein precipitation and y perceived

  is a device for retrofitting catch basins to improve pollutant-

  astringency as determined by a panel of tasters.

  removal properties. The article “An Evaluation of the Urban Stormwater Pollutant Removal Efficiency of Catch Basin

  Inserts” (Water Envir. Res., 2005: 500–510) reported on

  tests of various inserts under controlled conditions for which inflow is close to what can be expected in the field.

  Consider the following data, read from a graph in the arti-

  .336 cle, for one particular type of insert on x .221 .898 .836 amount filtered

  (1000s of liters) and y total suspended solids removed.

  .090 1.132 .538 1.098 .581 .862 .551 Summary quantities are

  5 1251, g x i 5 199,365, g y i 5 250.6,

  Relevant summary quantities are as follows:

  g y 2 i 5 9249.36, g x i y i 5 21,904.4

  gx i 5 19.404, g y

  i

  5 2.549, g x i 5 13.248032,

  g y 2

  a. Does a scatter plot support the choice of the simple lin-

  i

  5 11.835795, g x i y i 5 3.497811

  ear regression model? Explain.

  b. Obtain the equation of the least squares line.

  S

  5 13.248032 2 (19.404) xx 2 32 5 1.48193150,

  c. What proportion of observed variation in removed can

  S yy 5 11.82637622

  be attributed to the model relationship? d. Does the simple linear regression model specify a useful

  S xy 5 3.497811 2 (19.404)(2.549)32

  relationship? Carry out an appropriate test of hypotheses

  using a significance level of .05.

  a. Fit the simple linear regression model to this data. Then

  e. Is there strong evidence for concluding that there is at least

  determine the proportion of observed variation in

  a 2 decrease in true average suspended solid removal

  12.4 Inferences Concerning m Y • x and the Prediction of Future Y Values

  associated with a 10,000 liter increase in the amount fil-

  b. Use the Minitab output from the example to calculate

  tered? Test appropriate hypotheses using a 5 .05 .

  a 95 CI for expected rolling speed when test

  f. Calculate and interpret a 95 CI for true average

  speed

  removed when amount filtered is 100,000 liters. How

  c. Use the Minitab output to calculate a 95 PI for a single

  does this interval compare in width to a CI when amount

  value of rolling speed when test speed = 47.

  filtered is 200,000 liters?

  g. Calculate and interpret a 95 PI for removed when

  52. Plasma etching is essential to the fine-line pattern transfer in

  amount filtered is 100,000 liters. How does this interval

  current semiconductor processes. The article “Ion Beam-

  compare in width to the CI calculated in (f) and to a PI

  Assisted Etching of Aluminum with Chlorine” (J. of the

  when amount filtered is 200,000 liters?

  Electrochem. Soc., 1985: 2010– 2012) gives the accompa- nying data (read from a graph) on chlorine flow (x, in

  49. You are told that a 95 CI for expected lead content when

  SCCM) through a nozzle used in the etching mechanism

  traffic flow is 15, based on a sample of n 5 10 observa-

  and etch rate (y, in 100 Amin).

  tions, is (462.1, 597.7). Calculate a CI with confidence level

  99 for expected lead content when traffic flow is 15.

  x

  50. Silicon-germanium alloys have been used in certain types of

  y

  solar cells. The paper “Silicon-Germanium Films Deposited by Low-Frequency Plasma-Enhanced Chemical Vapor Deposition” (J. of Material Res., 2006: 88–104) reported on a

  The summary statistics are gx i 5 24.0, g y i 5 312.5, 2 study of various structural and electrical properties. Consider 2 gx i 5 70.50, g x i y i 5 902.25, g y i 5 11,626.75, bˆ 0 5

  the accompanying data on x Ge concentration in solid phase

  6.448718, bˆ 1 5 10.602564 .

  (ranging from 0 to 1) and y Fermi level position (eV):

  a. Does the simple linear regression model specify a useful

  x

  0 .42 .23 .33 .62 .60 .45 .87 .90 .79 1 relationship between chlorine flow and etch rate? b. Estimate the true average change in etch rate associated

  y .62 .53 .61 .59 .50 .55 .59 .31 .43 .46 .23 .22 .19

  with a 1-SCCM increase in flow rate using a 95 confi- dence interval, and interpret the interval.

  A scatter plot shows a substantial linear relationship. Here

  c. Calculate a 95 CI for m Y 3.0 , the true average etch rate

  is Minitab output from a least squares fit. [Note: There are

  when flow

  5 3.0. Has this average been precisely

  several inconsistencies between the data given in the paper,

  estimated?

  the plot that appears there, and the summary information

  d. Calculate a 95 PI for a single future observation on

  about a regression analysis.]

  etch rate to be made when flow 5 3.0 . Is the prediction

  The regression equation is

  likely to be accurate?

  Fermi pos 5 0.7217 2 0.4327 Ge conc

  e. Would the 95 CI and PI when flow 5 2.5 be wider or S 5 0.0737573 R – Sq 5 80.2 R – Sq(adj) 5 78.4 narrower than the corresponding intervals of parts

  (c) and (d)? Answer without actually computing the intervals.

  Analysis of Variance

  f. Would you recommend calculating a 95 PI for a flow

  of 6.0? Explain.

  53. Consider the following four intervals based on the data of

  a. Obtain an interval estimate of the expected change in

  Exercise 12.17 (Section 12.2):

  Fermi-level position associated with an increase of .1 in

  a. A 95 CI for mean porosity when unit weight is 110

  Ge concentration, and interpret your estimate.

  b. A 95 PI for porosity when unit weight is 110

  b. Obtain an interval estimate for mean Fermi-level position

  c. A 95 CI for mean porosity when unit weight is 115

  when concentration is .50, and interpret your estimate.

  d. A 95 PI for porosity when unit weight is 115

  c. Obtain an interval of plausible values for position result-

  Without computing any of these intervals, what can be said

  ing from a single observation to be made when concen-

  about their widths relative to one another?

  tration is .50, interpret your interval, and compare to the

  54. The decline of water supplies in certain areas of

  interval of (b).

  the United States has created the need for increased

  d. Obtain simultaneous CIs for expected position when

  understanding of relationships between economic factors

  concentration is .3, .5, and .7; the joint confidence level

  such as crop yield and hydrologic and soil factors. The

  should be at least 97.

  article “Variability of Soil Water Properties and Crop

  51. Refer to Example 12.12 in which x test track speed and y

  Yield in a Sloped Watershed” (Water Resources Bull.,

  rolling test speed.

  1988: 281–288) gives data on grain sorghum yield (y, in

  a. Minitab gave s bˆ 0 1bˆ 1 (45) 5 .120 and s bˆ 0 1bˆ 1 (47) 5 .186 . Why

  gm-row) and distance upslope (x, in m) on a sloping

  is the former estimated standard deviation smaller than

  watershed. Selected observations are given in the accom-

  the latter one?

  panying table.

  CHAPTER 12 Simple Linear Regression and Correlation

  a. Is it plausible that yield load is normally distributed?

  b. Estimate true average yield load by calculating a confi- dence interval with a confidence level of 95, and inter-

  pret the interval.

  c. Here is output from Minitab for the regression of yield

  a. Construct a scatter plot. Does the simple linear regres-

  load on torque. Does the simple linear regression model

  sion model appear to be plausible?

  specify a useful relationship between the variables?

  b. Carry out a test of model utility. c. Estimate true average yield when distance upslope is 75

  Predictor

  Coef

  SE Coef

  T

  P

  by giving an interval of plausible values.

  55. Verify that V(bˆ 0 1 bˆ 1 x) is indeed given by the expression in

  the text. [Hint: V(g d i Y i )5gd 2 i V(Y i ) .]

  S 5 73.2141 R – Sq 5 53.6 R – Sq(adj) 5 50.0

  56. The article “Bone Density and Insertion Torque as

  F Predictors of Anterior Cruciate Ligament Graft Fixation P

  Strength” (The Amer. J. of Sports Med., 2004: 1421–1429)

  Residual Error 13 69684

  gave the accompanying data on maximum insertion torque

  Total

  (N m) and yield load (N), the latter being one measure of

  d. The authors of the cited paper state, “Consequently, we

  graft strength, for 15 different specimens.

  cannot but conclude that simple regression analysis- based methods are not clinically sufficient to predict Torque 1.8 2.2 1.9 1.3 2.1 2.2 1.6 2.1 individual fixation strength.” Do you agree? [Hint:

  Load

  Consider predicting yield load when torque is 2.0.]