By V. Rajagopalan
This publication may help them to interpret their information themselves in a greater demeanour. during this e-book, usually used statistical exams are offered in an easy and comprehensible means with actual existence examples and workouts.
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Extra resources for Selected Statistical Tests
Let µX is unknown population mean before the treatment and µY is the unknown population mean after the treatment. Assumptions (i) The observations for the two samples must be obtained in pair. (ii) The population from which, the sample drawn is normal. Null Hypothesis H0: The treatment applied, is ineffective. That is, there is no significant difference between before and after the treatment applied. , H0: µd = µX – µY = 0. Alternative Hypotheses H1(1) : µd ≠ 0 H1(2) : µd > 0 H1(3) : µd < 0 Level of Significance ( α ) and Critical Region: (As in Test 3) Parametric Tests 43 Test Statistic t= d − µd Sd / n ( Under H0 : µd = 0) n ∑d d = i i =1 n , d i = X i − Yi , S d2 ∑( 1 n d −d = n − 1 i =1 i ) 2 The statistic t follows t distribution with (n–1) degrees of freedom.
A sample of 20 6-yearolds has the following attention spans in minutes: 86 89 84 78 75 74 85 71 84 71 75 68 75 71 82 85 81 78 79 78. State explicit null and alternative hypotheses and test at 5% level. 2. 10 respectively. A sample of 10 office going women is selected whose daily expenditure is obtained as 35 33 40 30 25 28 35 28 35 40. Test whether the variance of the daily expenditure of office going women is 10 at 1% level of significance. TEST – 5 TEST FOR A POPULATION VARIANCE (Population Mean is Unknown) Aim To test the population variance σ 2 be regarded as σ 20 , based on a random sample.
Solution Aim: To test the proportion of male and female students are equal or not, in introducing CBCS system in their university. H0: The proportion of male (P 1) and female (P 2) students are equal, in favour of the proposal of introducing CBCS system in their university. , H0: P 1 = P 2. H1: The proportion of male and female students is not equal, in favour of the propasal of introducing CBCS system in their university. 53) + (400 × 0. 56 300 + 400 n1 + n2 ( p1 − p 2 ) − (P1 − P2 ) Z= (Under H0: P 1 = P 2) ∧ ∧ 1 1 P(1 − P ) + n1 n2 ∧ P= Test Statistic: Z= (0.
Selected Statistical Tests by V. Rajagopalan