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A textbook of Computer Based Numerical and Statiscal Techniques part 53

A textbook of Computer Based Numerical and Statiscal Techniques part 53. By joining statistical analysis with computer-based numerical methods, this book bridges the gap between theory and practice with software-based examples, flow charts, and applications. Designed for engineering students as well as practicing engineers and scientists, the book has numerous examples with in-text solutions. | 506 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Sol. Given that n1 800 n2 1200 X1 800 4 P1 n1 1000 5 X2 800 2 P2 n2 1200 3 P p1n1 p2 n2 X1 X2 n1 n2 n1 n2 800 800 8 3 1000 1200 11 Q 11 Also Null hypothesis H0 p1 p2 i.e. there is no significant difference in the consumption of tea before and after increase of excise duty. H1 p1 p2 right tailed test The test statistic P1 - P2 0.8 - 0.6666 1 n1 n2 18 3 1 1 X 11 11 I. 1000 1200 6.842 Z Conclusion Since the calculated value of I ZI 1.645 also IZI 2.33 both the significant values of z at 5 and 1 level of significance. Hence H0 is rejected i.e. there is a significant decrease in the consumption of tea due to increase in excise duty. Example 12. In two large populations there are 30 and 25 respectively of fair haired people. Is this difference likely to be hidden in samples of 1200 and 900 respectivelyfrom the two populations. Sol. P1 proportion of fair haired people in the first population 30 0.3 P2 25 0.25 Q1 0.7 Q2 0.75. Here H0 Sample proportions are equal i.e. the difference in population proportions is likely to be hidden in sampling. H1 P1 A P2 P1- P2 0.3 - 0.25 Z 1 2 2.5376. P1Q1 P2Q2 0.3 x 0.7 0.25 x 0.75 Conclusion IZ I 1.96 the significant value of Z at 5 level of significance. H0 is rejected. However IZI 2.58 the significant value of Z at 1 level of signficance H0 is accepted. At 5 level these samples will reveal the difference in the population proportions. Example 13. 500 articles from a factory are examined and found to be 2 defective 800 similar articles from a second factory are found to have only 1.5 defective. Can it reasonably be concluded that the product of the first factory are inferior to those of second TESTING OF HYPOTHESIS 507 Sol. n1 500 n2 800 p1 proportion of defective from first factory 2 0.02 p2 proportion of defective from second factory 1.5 0.015 H0 There is no significant difference between the two products i.e. the products do not differ in quality. H1 P1 p2 one tailed test Under