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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 124 PM
Frnz 1178 Prof. K. D. Stolzenbach
5732J Boelter Hall, 2067624
stolzenb@ucla.edu Problem Set 7 Solutions
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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 34
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 Pvalue is between 0.05 and 0.1 using Table IV. From Minitab the Pvalue =
0.0571. 2. 1. The variable of interest is calls by surgicalmedical patients.
2. H0: Calls by surgicalmedical patients are independent of Medicare status.
3. H1: Calls by surgicalmedical 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 surgicalmedical patients and Medicare status are dependent. Pvalue = 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. ...
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This note was uploaded on 10/22/2011 for the course ENG 101 taught by Professor Yukov during the Spring '11 term at UCLA.
 Spring '11
 Yukov
 Environmental Engineering

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