STAT 572 Assignment 1 Answers: Hypothesis Testing and Regression Analysis, Assignments of Data Analysis & Statistical Methods

Download STAT 572 Assignment 1 Answers: Hypothesis Testing and Regression Analysis and more Assignments Data Analysis & Statistical Methods in PDF only on Docsity! STAT 572 Assignment 1 - Answers Due: February 2, 2007 1. An experiment is planned to evaluate the mean number of song birds in a certain tropical forest. The researchers will sample several 1km2 plots. In each plot, they will count the number of bird species present through song recording. They will test the null hypothesis that the mean number of bird species per km2 equals some fixed value. With their previous experience, they already have a good idea of the variability in species number. Therefore, they made the power curve below. (a) µ0 = 26 (b) α = 0.10 (c) Approximately 28.7− 26 = 2.7 birds (d) The dotted curve. 2. It is now well known that mothers can manipulate sex ratios at birth in many species, and researchers wanted to know if fathers can also influence offspring sex ratio. A study was conducted on red deers (Cervus elaphus). Sperm from 15 wild males was collected. 360 hinds (females) were kept under similar environmental conditions and provided with unlimited food supply. Each hind was artificially inseminated once. Sperm quality was measured as the percentage of morphologically normal spermatozoa. The number of sons and daughters each male had was counted in order to obtain the percentage of male offspring for each male. The data is presented here. Calculations yield x = 79.4, y = 49.27, ∑ (xi−x)2 = 1105.6, ∑ (yi − y)2 = 2916.93, and ∑ (xi − x)(yi − y) = 1192.4. Also, SSErr = 1630.92. Male 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Sperm Quality (x ) 65 72 68 73 71 77 93 77 87 85 80 83 87 78 95 Prop. of male offsprings (y) 25 33 38 42 43 41 44 42 49 57 56 58 70 74 69 (a) Slope: β̂1 = ∑ (xi−x)(yi−y)∑ (xi−x)2 = 1192.4 1105.6 = 1.0785 Intercept: β̂0 = y − β̂1 × x = 49.27− 1.0785× 79.4 = -36.3636 Least Squares Regression Line: (Proportion of male offsprings) = −36.3636 + 1.0785×(Sperm Quality) (b) We are testing the hypothesis Ho : β1 = 0 vs. Ha : β1 6= 0 Method 1: t = β̂1−β1[ √ SSErr n−2√∑ (xi−x)2 ] = 1.0785−0[√ 1630.92 15−2√ 1105.6 ] = 3.20 1 STAT 572 Assignment 1 - Answers Due: February 2, 2007 According to the t-table with n− 2 = 13 degrees of freedom, 0.002 < p < 0.01. There- fore we reject the null hypothesis with high confidence. We conclude that there is evidence to support the alternative hypothesis that the slope is significantly different than 0. Method 2: Recall that the values of an ANOVA table can be calculated as: Source df SS MS F Treatment k − 1 ∑ ni(xi − x)2 SSTr ÷ (k − 1) MSTr ÷MSErr Error n− k ∑ (ni − 1)si2 SSErr ÷ (n− k) Total n− 1 ∑ ∑ (xij − x)2 where k is the number of treatments and n is the total sample size. Note that SSTot = SSTr + SSErr. Therefore: Source df SS MS F p-value Treatment 1 1286 1286 10.25 0.005 < p < 0.01 Error 13 1630.92 125.45 Total 14 2916.93 Again, our p-value is significantly small. Therefore we can reject the null hypothesis. (c) (β̂0 + β̂1x ∗)± tα 2 ,df=n−2S √ 1 + 1 n + (x ∗−x)2 Sxx = (−36.3636+1.0785×65)±2.160×11.201×√ 1 + 1 15 + (65−79.4) 2 1105.6 = (6.643, 60.834) 3. Individuals from the fish species Telmatherina sarasinorum are of 3 different colors: females are grey, whereas males are either blue or yellow. It is hypothesized that fish with different colors may prefer to stay in different environments. Two types of environments were sur- veyed: shallow beach sites (yellow environment), and deeper sites with overhanging roots (blue environment). Many sites from each type were surveyed the same day, and at the same hour of the day, and fish from each color were counted. Below are the data. Fish Color Environment blue yellow grey Total blue 14 22 36 72 yellow 25 12 31 68 Total 39 34 67 140 (a) The degrees of freedom are calculated by taking (number of rows – 1)×(number of columns – 1) = (2 − 1) × (3 − 1) = 2. Therefore given a χ2 value of 6.3077, 0.025 < 2 STAT 572 Assignment 1 - Answers Due: February 2, 2007 7. True/False Questions. (a) False (b) False (c) False (d) True (e) True (f) False (g) False (h) True 8. In a study of the tufted titmouse (Parus bicolor), an ecologist captured seven male birds, measured their wing lengths, and then marked and released them. During the ensuing winter, he repeatedly observed the birds as they foraged for insects and seeds on tree branches, and noted the diameters of the branches. The average branch diameter for each bird is tabulated below. Bird 1 2 3 4 5 6 7 Wing Length (mm) 79.0 80.0 81.5 84.0 79.5 82.5 83.5 Branch Diameter (cm) 1.02 1.04 1.20 1.51 1.21 1.56 1.29 (a) There does appear to be a relatively strong, positive, linear relationship between wing length and branch diameter. See part (b) for graph. (b) The equation of the regression line is (Branch Diameter in cm) = −5.660 + 0.085×(Wing length in mm). The scatterplot and corresponding regression line for the data can be seen below. 5 STAT 572 Assignment 1 - Answers Due: February 2, 2007 (c) Using commands in R, the mean and standard deviations were found to be: Mean Standard Deviation Wing Length (mm) 81.42857 1.988060 Branch Diameter (cm) 1.261429 0.2103512 The correlation coefficient is further calculated to be: r = 1 n−1 ∑ (xi−x sx )(yi−y sy ) = (1 6 )4.820084 = 0.8034. This agrees with our initial observation in part (a). To verify the coefficients found in part (b): Slope: β̂1 = r sy sx = 0.80340.2103512 1.988060 = 0.08500557 Intercept: β̂0 = y − β̂1 × x = 1.261429− 0.08500557× 81.42857 = -5.660454 6 STAT 572 Assignment 1 - Answers Due: February 2, 2007 R Code: wing = c(79.0,80.0,81.5,84.0,79.5,82.5,83.5) branch = c(1.02,1.04,1.20,1.51,1.21,1.56,1.29) mw = mean(wing) sw = sd(wing) mb = mean(branch) sb = sd(branch) r = cor(wing,branch) c(mw,sw,mb,sb,r) bird.lm = lm(branch ∼ wing) coef(bird.lm) b1 = r*sb/sw b1 b0 = mb - b1*mw b0 7