CEE 110 11 PS7 Solutions

CEE 110 11 PS7 Solutions - UNIVERSITY OF CALIFORNIA, LOS...

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: UNIVERSITY OF CALIFORNIA, LOS ANGELES Civil and Environmental Engineering Department CEE 110 Introduction to Probability and Statistics for Engineers Spring Quarter 2011 MW 12-4 PM Frnz 1178 Prof. K. D. Stolzenbach 5732J Boelter Hall, 206-7624 stolzenb@ucla.edu Problem Set 7 Solutions _____________________________________________________________________________ 1. Expected Frequency is found by using the Poisson distribution Value Observed Frequency Expected Frequency 0 24 1 30 2 31 3 11 30.12 36.14 21.69 4 4 8.67 2.60 Since value 4 has an expected frequency less than 3, combine this category with the previous category: Value Observed Frequency Expected Frequency 0 24 1 30 2 31 30.12 36.14 21.69 3-4 15 11.28 The degrees of freedom are k − p − 1 = 4 − 0 − 1 = 3 a) 1) The variable of interest is the form of the distribution for X. 2) H0: The form of the distribution is Poisson 3) H1: The form of the distribution is not Poisson 4) The test statistic is for α = 0.05 5) Reject H0 if ( 24 − 30.12 ) + ( 30 − 36.14 )2 + ( 31 − 21.69 )2 + (15 − 11.28 )2 = 7.52 = 2 6) χ 2 0 30.12 36.14 21.69 11.67 7) Because 7.52 < 7.81 fail to reject H0. We are unable to reject the null hypothesis that the distribution of X is Poisson. b) The P-value is between 0.05 and 0.1 using Table IV. From Minitab the P-value = 0.0571. 2. 1. The variable of interest is calls by surgical-medical patients. 2. H0: Calls by surgical-medical patients are independent of Medicare status. 3. H1: Calls by surgical-medical patients are not independent of Medicare status. 4. The test statistic is: for α = 0.01 5. The critical value is 6. The calculated test statistic is 7. Because fail to reject H0. The evidence is not sufficient to claim that surgical-medical patients and Medicare status are dependent. P-value = 0.85 3. 1) The parameters of interest are the variances of concentration, 2) H0 : 3) H1 : 4) The test statistic is 5) Reject the null hypothesis if f0 < f0.975 ,15 , 9 where f0.975 ,15 , 9 = 0.320 or f0 > = 3.12 for α = 0.05 6) 10 16 4.7 5.8 where 7) Conclusion: Because 0.320 < 0.657 < 3.12 fail to reject the null hypothesis. There is insufficient evidence to conclude that the two population variances differ at the 0.05 level of significance. 4. Analysis of Variance for CIRCUIT TYPE Source DF SS MS CIRCUITT 2 260.9 130.5 Error 12 390.8 32.6 Total 14 651.7 Reject H0 F 4.01 P 0.046 5. The following table is from an Excel analysis not the Matlab analysis. The critical F value is for a 95% confidence interval (alpha = 5%). The P value is very low, so we reject the hypothesis that the means are equal. ...
View Full Document

This note was uploaded on 10/22/2011 for the course ENG 101 taught by Professor Yukov during the Spring '11 term at UCLA.

Ask a homework question - tutors are online