By Hans-Michael Kaltenbach
The textual content offers a concise advent into primary techniques in data. bankruptcy 1: brief exposition of chance conception, utilizing wide-spread examples. bankruptcy 2: Estimation in idea and perform, utilizing biologically influenced examples. Maximum-likelihood estimation in lined, together with Fisher details and tool computations. tools for calculating self belief durations and powerful possible choices to straightforward estimators are given. bankruptcy three: speculation trying out with emphasis on options, really type-I , type-II error, and studying attempt effects. a number of examples are supplied. T-tests are used all through, vital different exams and robust/nonparametric possible choices. a number of trying out is mentioned in additional intensity, and mixture of autonomous exams is defined. bankruptcy four: Linear regression, with computations completely in keeping with R. a number of workforce comparisons with ANOVA are lined including linear contrasts, back utilizing R for computations.
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Extra resources for A Concise Guide to Statistics
8 Visualizing Distributions 4 6 8 10 14 ● 4 Value ● 6 ● ● ● ● ● ● ● ● 8 ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● 10 ● ● ● ● ● ● ● ● ● 12 ● ● ● ● 14 ● 16 Value Fig. 9 Empirical (solid) and theoretical (dashed) density functions (left) and cumulative distribution functions (right) of 50 Norm(10, 6) sample points Hn (x) , Fˆn (x) = n where Hn (x) is the number of sample points smaller than x. This leads to a step function, which in the example quite closely follows the theoretical function.
Our goal is to infer one or more parameters θ of the distribution of the X i . For this, we construct an estimator θˆn by finding a function g, such that θˆn = g(X 1 , . . , X n ) is a “good guess” of the true value θ. Since θˆn depends on the data, it is a random variable. Finding its distribution allows us to compute confidence intervals that quantify how likely it is that the true value θ is close to the estimate θˆn . Example 10 Let us revisit the problem of sequence matching from Example 8 (p.
For k = 2 and θ = 2, the distribution has expectation μ = kθ = 4 and variance σ 2 = kθ 2 = 8; its density is shown in Fig. 8 (solid line). For comparison, a normal distribution with the same expectation and variance is plotted by a dashed line. As we can see, the density functions look very different, although both have the same mean and variance. For additionally capturing their different shapes, higher moments are needed (see Sect. 5). Example 8 Let us consider the following model of a random DNA sequence as introduced earlier: we assume independence among the nucleotides and in each position, the probabilities of having a particular nucleotide are p A , pC , pG , pT , respectively.