Midterm 2 Solutions - Statistical Methods for Bioscience I | HORT 571, Assignments of Data Analysis & Statistical Methods

Download Midterm 2 Solutions - Statistical Methods for Bioscience I | HORT 571 and more Assignments Data Analysis & Statistical Methods in PDF only on Docsity! Stat/For/Hort 571 Midterm II, Fall 2004 Brief Solutions 1. (a) TRUE — From the tables, P (T ≤ 1.533) = 1 − P (T ≥ 1.533) = 0.90 (Check the directionality!!) Similarly, P (Z ≤ 1.533) = 1−P (Z ≥ 1.533) ≈ 0.938. Alter- natively, since the t distribution is known to have fatter tails than the Z, the prob- ability that t is above 1.533 (in the right hand tail) is smaller than the correspond- ing probability for Z. Considering the di- rectionality, the conclusion follows. (A pic- ture might be helpful!!) (b) FALSE — The given ’reasoning’ essen- tially is a ’dressed-up’ version of the argu- ment that scientific significance can out- weigh statistical significance. In practice, one needs both. A claim about what might have occurred means practically nothing. Inference rests on the data actually ob- tained. (Issues regarding df are not rel- evant here.) The most one can say for X is that if the potential scientific importance is great enough, perhaps a new experiment with larger n might be conducted. 2. This is a two-independent sample situation. (a) Let YA and YB be the concentrations on A and B respectively and let µA and µB be the respective population means. The needed summary statistics are ȳA = 6.1, s2A = 0.8733, ȳB = 4.9, and s 2 B = 0.74. Then, due to the balanced data, s2p = (s 2 A+ s2B)/2 = .8067. The CI for µA − µB is ȳA− ȳB ± t ∗ √ s2p(1/nA + 1/nB). Since s 2 p has 6 df, the appropriate t-value is 1.943. Thus, (−0.03 < µA − µB < 2.43). (b) The null hypothesis is written: H0 : µA − µB = 0.3. We notice that the value of 0.3 is contained within the CI above. Thus, we can conclude that the p− value > 0.10 for the given test. 3. The appropriate formula for the pooled variance is: s2p = ( ∑k i=1(ni−1)s2i )/(N−k) where k is the number of treatments and N is the total number of observations. You are given the standard de- viations so you must square them. s2p = 157.02. There are N − k = 46 degrees of freedom asso- ciated with this variance. This can be thought of as the addition of the number of df for each variety (9 + 9 + 6 + 5 + 8 + 9). 4. (a) The random variable, Y is the number of tigers with the bacteria present. A rea- sonable model (given the available infor- mation) is Y ∼ B(15, p). The hypothe- ses can be written: H0 : p = 0.05 vs HA : p > 0.05. Evidence against the null is obtained for ’large’ observed values of Y . Using the basic definition of p-values, p − val = P (Y ≥ 2). This can be written as 1− (P (Y = 0) + P (Y = 1)). Using the binomial formula results in p−val = 0.171. This means that there is no meaningful ev- idence against the claim that the rate of occurrence of this bacteria is 0.05 or less. (b) The random variable Y is the number of cats with bacteria present. A reasonable model is Y ∼ B(60, p). H0 : p = 0.15 vs HA : p ≥ 0.15. Since np(= 9) and n(1 − p)(= 51) are both larger than 5, the nor- mal approx may be used. Let p̂ = Y/60. Then, under H0, ˆpNA ∼ N(.15, (.0461)2). α = P (p̂ ≥ 0.25) = P (Z ≥ 2.17) = .015. 5. (a) The general form for a CI for the difference between two proportions is: (p̂A − p̂B) ± Zα/2 ∗ √ pA∗(1−pA) nA + pB∗(1−pB)nB . Because pA = 0.4 and pB = 0.6, both expressions of the form p(1 − p) are the same. Since Zα/2 affects all CIs equally (given the same 1 − α), we need to minimize √ .24 nA + .24nB . This is 0.163 for choice (1) and 0.107 for choice (2). Choice (2) has smaller width. (b) By looking at the expression above, the CI width is minimized when 1nA + 1 nB is as small as possible with nA + nB = 100. Based on part(a), the width is smaller for the values of nA and nB that are closer together. Thus the ’natural’ conjecture is that the CI width can be minimized if nA = nB = 50. This is the correct answer. It is primarily for this reason that most scientific studies comparing 2 (or more) groups have equal sample sizes. Grade Distribution 90-99:22 80-89:41 70-79:26 median = 79 60-69:17 50-59:13 below: 8