By North American Cambridge Classics Projec
The Cambridge Latin direction goals either to coach comprehension of the Latin language via interpreting perform, and to advance knowing of the social and political historical past of the Romans. This ebook, which accompanies Unit 1, includes standardized exams reflecting the course's ambitions. every one try starts off with a examining passage through comprehension and grammar questions within the kind of true/false statements, a number of selection questions or matching goods. the remainder questions are designed to guage the student's wisdom of grammar, derivatives and tradition. to supply kind, a number of tradition sections are interspersed thoroughout the exams. all the interpreting passages and questions are acceptable to the degrees of the corresponding levels, and pertain to fabric featured within the textbook. The checks might be graded both via Scantron machines, or via extra conventional tools.
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Additional info for Cambridge Latin Course, Unit 1: Machine Scored Stage Tests
The one-tailed hypothesis pairs are H0 : θ = θ0 vs. H1 : θ > θ0 H0 : θ = θ0 vs. H1 : θ < θ0 , and where the reference value θ0 is given. The pair H0 : θ = θ0 vs. H1 : θ > θ0 is tested at the level of significance α by calculating from the realized sample the lower 100(1 − α) % confidence limit θˆL for parameter θ in the manner presented earlier. The null hypothesis H0 is rejected if the reference value θ0 isn’t included in the obtained confidence interval, in other words if θ0 ≤ θˆL . Correspondingly, the pair H0 : θ = θ0 vs.
The hypothesis pair to be tested is thus H0 : σ12 = σ22 vs. H1 : σ12 = σ22 (and so k = 1). 10. 1789 (with 9 and 9 degrees of freedom) and the critical region consists of the values that aren’t included in that interval. 64, and it’s not in the critical region. No proof about the inequality of the variances was obtained, so H0 isn’t rejected. 14] Graphical Methods for Comparing Means A glimpse to a graphical display obtained from the population often gives quite a good image about the matter, at least when considering the expectations.
P(Tk ) = pk are supposed to be known, which is the null hypothesis H0 of the test. The alternative hypothesis H1 is that at least for one i P(Ti ) = pi . For the test, let’s take a sample with n elements, from which we determine the realized (absolute) frequencies f1 , . . , fk of the cases T1 , . . , Tk . These can be also considered to be random variables F1 , . . , Fk and E(Fi ) = npi . The test is based on the fact that the random variable k H= i=1 (Fi − npi )2 npi has nearly the χ2 -distribution with k − 1 degrees of freedom.