Stat 503. Solutions to Homework #11 (115 points) (11) (19)


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1 Leucine (ng) Stat 503 Solutions to Homework #11 (115 points) Problem 1.3 (p. 536) In a study of protein synthesis in the oocyte (developing egg cell) of the frog Xenopus laevis, a biologist injected individual oocytes with radioactively labelled leucine. At various times after injection, he made radioactivity measurements and calculated how much of the leucine had been incorporated into protein. The results are given in the accompanying table; each leucine value is the content of labelled leucine in two oocytes. All oocytes were from the same female. (a) Use linear regression to estimate the rate of incorporation of the labelled leucine. The rate of incorporation is the slope of the regression line. b 1 = SP XY / SS X = 81.9/800 = (b) Plot the data and draw the regression line on your graph. b 0 = y b 1 x = 0.83 (0.093)(30) = Time (min) Leucine (ng) X Y mean SS SP XY 81.9 SS(resid) (11) (3) (6) (c) Calculate the residual standard deviation. Leucine Incorporation 1.5 y = 0.093x Time (min) s Y X = SS( resid ) n Problem 1.5 (p. 537) Twenty plots, each 10x4 metres, were randomly chosen in a large field of corn. For each plot, the plant density (number of plants in the plot) and the mean cob weight (gm of grain per cob) were observed. The results are given in the table. (19)
2 COB WEIGHT Y PLANT DENSITY COB WEIGHT PLANT DENSITY COB WEIGHT X Y X Y Preliminary calculations yield the following results: x = y = 4.1 SS X = 0,09.0 SS Y = 11,831.8 SP XY = 14,563.1 SS(resid) = (a) Calculate the linear regression of Y on X. b 1 = SP XY / SS X = /009.0 = b 0 = y  b 1 x = (0.706)(18.05) = The equation of the linear regression line is Y = X Regression Plot y = x PLANT DENSITY X (b) Plot the data and draw the regression line on your graph. (c) Interpret the value of the slope of the regression line, b 1, in the context of this setting. On average, a onecob increase in plant density will result in a reduction of 0.7 g in the average corn yield for each cob in that plot. (d) Calculate s Y and s Y X and specify the units of each. s Y = SS Y = 4.95 gm n 1 19 SS( resid ) s Y X = = 8.6 gm n 18 The units of both s Y and s Y X are the same as the units for Y, namely grams. (6)
3 Fungus Growth (mm) (e) Interpret the value of s Y X in the context of this setting. The standard deviation of the points from the line is 8.6. Approximately 68% of observations for Y will be within 8.6 of the value predicted by the line, and approximately 95% of observations for Y will be within 17.4 of the value predicted by the line. Problem 1.6 (p. 538) Laetisaric acid is a compound that holds promise for control of fungus diseases in crop plants. The accompanying data show the results of growing the fungus Pythium ultimum in various concentrations of laetisaric acid. Each growth value is the average of four radial measurements of a P. ultimum colony grown in a petri dish for 4 hours; there were two petri dishes at each concentration. Laetisaric Acid Fungus Growth Laetisaric Acid Fungus Growth Concentration Concentration X(g/ml) Y(mm) X(g/ml) Y(mm) mean SS X = 1,303 SS Y = SP XY = SS(resid) = (a) Calculate the linear regression of Y on X. b 1 = SP XY / SS X = / 1303 = 0.71 b 0 = y b 1 x = 3.64 ( 0.71)(11.5) = The linear regression is Y = X. (b) Plot the data and draw the regression line on your graph. (14) Problem y = 0.71x R = Laetisaric Acid Concentration (mcg/ml)
4 COB WEIGHT Y (c) Calculate s Y X. What are the units of s Y X? s Y X = SS( resid ) = 1.95 mm. n 1 (d) Draw a ruler line on your graph to show the magnitude of s Y X. (See Figure 1.7.) (see above graph) (3) Problem 1.11b) (p. 541) For the corn yield data of Exercise 1.5, prepare a plot like figure 1.8, showing the data, the fitted regression line, and two lines whose vertical distance above and below the regression line is s Y X. What percentage of the data points are within s Y X of the regression line? (7) [It's okay to put the lines on the same graph as that for 1.5] Regression Plot PLANT DENSITY X Of the 0 data points, 13 are within 8.6 of the straight line, and 7 are outside that range. Therefore 13/0 = 65% are within s Y X of the regression line.
5 PLANT COB is resid >8.6? DENSITY WEIGHT X Y ŷ residual 0 = no, 1 = yes Problem 1.14 (p. 547) Refer to the cob weight data of Exercise 1.5. Assume that the linear model holds. (a) Estimate the mean cob weight to be expected in a plot containing (i) 100 plants; (ii) 10 plants. Use the linear regression equation to find the predicted Y for those values of X. (i) With x = 100, ŷ = = 44 g (ii) With x = 10, ŷ = = 30 g (b) Assume that each plant produces one cob. How much grain would we expect to get from a plot containing (i) 100 plants? (ii) 10 plants? The total amount of corn from a plot is the mean weight per cob times the number of cobs. (i) 44 g 100 = 4,400 g = 4.4 kg of corn (ii) 30 g 10 = 7,600 g = 7.6 kg of corn (8) Problem 1.1 (p. 55) Refer to the cob weight data of Exercise 1.5. Construct a 95% confidence interval for 1. b 1 = syx 8.6 SE b1 = SS 009 = X df = 0  = 18 ; t.05 = % CI for 1 is (.101)(0.0606) = = [0.848, ] We are 95% confident that < 1 <
6 Problem 1. (p. 55) Refer to the fungus growth data of Exercise 1.6. (a) Calculate the standard error of the slope, b 1. SS X = 1303 and s Y X = 1.9 from Problem 1.5 above syx 1.95 SE b1 = SS 1303 = X (b) Consider the null hypothesis that laetisaric acid has no effect on growth of the fungus. Assuming that the linear model is applicable, formulate this as a hypothesis about the true regression line, and test the hypothesis against the alternative that laetisaric acid inhibits growth of the fungus. Let = Does laetisaric acid inhibit fungus growth? Let 1 be the slope of the true regression line. H 0 : 1 = 0 ; laetisaric acid has no effect on growth of the fungus H A : 1 < 0 ; laetisaric acid inhibits growth of the fungus Use a directional ttest. t s = b 1 / SE b1 has a tdistribution with 10 df under H 0. With = 0.05, critical value is Will reject H 0 if t s < b 1 = 0.71 (from 1.6 earlier) t s = 0.71 / = < 1.81 so reject H 0. This study provides evidence at the 0.05 significance level that laetisaric acid inhibits growth of the fungus. (16) (14) Problem 1.8 (p. 564) Proceed as in Exercise 1.7, but use the cob weight data of Exercise 1.5 (a) Calculate the correlation coefficient. SPXY r = = 0.94 SS SS X Y (b) Calculate s Y and s Y X ; specify the units for each. Verify the approximate relationship between s Y, s Y X, and r. s Y = 4.95 g (from last HW) s Y X = 8.6 g (from last HW) s Y X / s Y = 4.95/8.6 = ( ) r = The approximate relationship is verified since (c) Calculate the regression line of Y on X. The regression line is Y = X from last HW. (13) (d) Construct a scatterplot of the data and draw the regression line on your graph. Place ruler lines on the scatterplot to show the magnitudes of s Y and s Y X. (See Figure 1.16.)
7 Regression Plot y = x C 60 O 50 B 40 W 30 EI 0 G 10 HT 00 Y s Y s Y X PLANT DENSITY X Problem 1.30 (p. 564) In a study of,669 adult men, the correlation between age and systolic blood pressure was found to be r = The SD of systolic blood pressures among all,669 men was 19.5 mm Hg. Assuming that the linear model is applicable, estimate the SD of systolic blood pressures among men 50 years old. Find s Y X. Since s Y X / s Y 1 r, therefore s Y X = s Y 1 r = = The SD of systolic blood pressures among 50yearold men is approximately 17.6 mm Hg. Problem 1.3 (p. 564) A veterinary anatomist measured the density of nerve cells at specified sites in the intestine of nine horses. Each density value is the average of counts of nerve cells in five equal sections of tissue. The results are given in the accompanying table for site I (midregion of jejunum) and site II (mesenteric region of jejunum). (These data were also given in Exercise 9.8). (a) Calculate the correlation coefficient between the densities at the two sites. SPXY r = SS SS = X Y (b) Construct a scatterplot of the data. Animal Site 1 Site mean SS SP XY (8) (3)
8 Site Nerve Cell Density Site 1 (c) Four potential sources of variation in these data are: (i) errors in counting the nerve cells, (ii) sampling error due to choosing certain slices for counting, (iii) variation from one horse to another, and (iv) variation from site to site within a horse. Which of these sources of variation would tend to produce positive correlation between the sites? [Hint: Ask yourself how each source contributes to the appearance of the scatterplot.] Source (iii) would tend to produce positive correlation between the sites. Variation in nerve cell density from one horse to another would tend to separate points for different horses but would presumably affect each horse in the same way at both points, that is, move X and Y together, creating positive correlation. Problem 1.33 Refer to Exercise 1.3. Test the hypothesis that the true (population) correlation coefficient is zero against the alternative that it is positive. Let =0.05. Are nerve cell densities between the two sites correlated? Let be the true population correlation coefficient. H 0 : = 0 ; nerve cell densities at the two sites are uncorrelated H A : > 0 ; nerve cell densities at the two sites are positively correlated n Use a directional ttest. t s = r has a tdistribution with 7 df under H 0. 1 r With = 0.05, the critical value is H 0 will be rejected if t s > r = 0.81 (from previous question) 9 t s = 0.81 = Since > 1.895, H 0 is rejected. This study provides evidence at the 0.05 significance level that nerve cell densities are correlated between the two sites. (3) (10)
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