final2009

# final2009 - AMS572.01 Final Exam Fall 2009...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: AMS572.01 Final Exam Fall, 2009 Name ___________________________ID ________________Signature____________________ AMS Major? ______ Instruction: This is a close book exam. Anyone who cheats in the exam shall receive a grade of F. Please enter “Yes” or “No” for “AMS Major”. Please provide complete solutions for full credit. The exam goes from 2:15 - 4:45pm. Good luck! 1. ( for all students ) How to become an art sleuth? Like all creative artists, composers of music develop certain personal characteristics in their works. One such characteristic is the number of melody notes in each bar of music. Now suppose you buy an old unsigned manuscript of a waltz which you suspect is an unknown work by Johann Strauss, and if so, very valuable. You count the number of melody notes per bar of several genuine Strauss waltzes and compare frequency distribution with a similar count of the unknown work. Would the following results support your high hopes? Use α = 0.05. No. of melody notes per bar 1 2 3 4 5 ≥ 6 Total Strauss waltzes 5 32 133 114 67 22 15 388 Unknown waltz 6 60 62 96 33 7 18 282 SOLUTION: This is inference on several population proportions following a multinomial distribution. If the unknown work was from Johann Strauss, then we will expect the following frequency distribution of melody notes per bar: No. of melody notes per bar 1 2 3 4 5 ≥ 6 Expected relative frequency ( i p ) 5/388 32/388 133/388 114/388 67/388 22/388 15/388 Expected frequency (count) ( i E ) 282*5/388 ≈ 3.63 282*32/388 ≈ 23.26 282*133/388 ≈ 96.66 282*114/388 ≈ 82.86 282*67/388 ≈ 48.70 282*22/388 ≈ 15.99 282*15/388 ≈ 10.90 Observed frequency ( i O ) 6 60 62 96 33 7 18 The large sample chi-square test can be applied to test: : , 1, ,7 i i H p p i = = L versus : a H H is not true. The chi-square test statistic is: ( 29 ( 29 ( 29 ( 29 2 2 2 2 7 2 1 6 3.63 60 23.26 18 10.90 88.83 3.63 23.26 10.90 i i i i O E E χ =---- = = + + + ≈ ∑ L Since 2 2 6, 0.05, 88.83 12.59 upper α χ χ = ≈ = , we reject the null hypothesis at the significance level of α = 0.05 and conclude that it is not likely that the unknown waltz was written by Strauss. 2A. ( for AMS majors ) Suppose we have two independent random samples from two normal populations i.e., ( 29 1 2 1 2 1 1 , , , ~ , n X X X N μ σ K , and ( 29 2 2 1 2 2 2 , , , ~ , n Y Y Y N μ σ K ....
View Full Document

## This note was uploaded on 01/31/2011 for the course AMS 572 taught by Professor Weizhu during the Fall '10 term at SUNY Stony Brook.

### Page1 / 5

final2009 - AMS572.01 Final Exam Fall 2009...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online