Applied Statistics and Probability for Engineers Sixth Edition Douglas C Montgomery George C Runger Chapter 9 Tests of Hypotheses for a Single Sample Copyright 2014 John Wiley Sons Inc All rights reserved 9 7 Testing for Goodness of Fit The test is based on the chi square distribution Assume there is a sample of size n from a population whose probability distribution is unknown Let Oi be the observed frequency in the ith class interval Let Ei be the expected frequency in the ith class interval The test statistic is X 02 Oi Ei 2 Ei i 1 k 9 16 Sec 9 7 Testing for Goodness of Fit Copyright 2014 John Wiley Sons Inc All rights reserved 2 9 7 Testing for Goodness of Fit EXAMPLE 9 12 Printed Circuit Board Defects Poisson Distribution The number of defects in printed circuit boards is hypothesized to follow a Poisson distribution A random sample of n 60 printed boards has been collected and the following number of defects observed Number of Defects 0 1 2 3 Observed Frequency 32 15 9 4 Sec 9 7 Testing for Goodness of Fit Copyright 2014 John Wiley Sons Inc All rights reserved 3 9 7 Testing for Goodness of Fit Example 9 12 The mean of the assumed Poisson distribution in this example is unknown and must be estimated from the sample data The estimate of the mean number of defects per board is the sample average that is 32 0 15 1 9 2 4 3 60 0 75 From the Poisson distribution with parameter 0 75 we may compute pi the theoretical hypothesized probability associated with the ith class interval Since each class interval corresponds to a particular number of defects we may find the pi as follows e 0 75 0 75 0 p1 P X 0 0 472 0 e 0 75 0 75 1 p 2 P X 1 0 354 1 e 0 75 0 75 2 p3 P X 2 0 133 2 p 4 P X 3 1 p1 p 2 p3 0 041 Sec 9 7 Testing for Goodness of Fit Copyright 2014 John Wiley Sons Inc All rights reserved 4 9 7 Testing for Goodness of Fit Example 9 12 The expected frequencies are computed by multiplying the sample size n 60 times the probabilities pi That is Ei npi The expected frequencies follow Number of Defects 0 1 2 3 or more Probability 0 472 0 354 0 133 0 041 Expected Frequency 28 32 21 24 7 98 2 46 Sec 9 7 Testing for Goodness of Fit Copyright 2014 John Wiley Sons Inc All rights reserved 5 9 7 Testing for Goodness of Fit Example 9 12 Since the expected frequency in the last cell is less than 3 we combine the last two cells Number of Defects 0 Observed Frequency 32 Expected Frequency 28 32 1 2 or more 15 13 21 24 10 44 The chi square test statistic in Equation 9 16 will have k p 1 3 1 1 1 degree of freedom because the mean of the Poisson distribution was estimated from the data Sec 9 7 Testing for Goodness of Fit Copyright 2014 John Wiley Sons Inc All rights reserved 6 9 7 Testing for Goodness of Fit Example 9 12 The seven step hypothesis testing procedure may now be applied using 0 05 as follows 1 Parameter of interest The variable of interest is the form of the distribution of defects in printed circuit boards 2 Null hypothesis H0 The form of the distribution of defects is Poisson 3 Alternative hypothesis H1 The form of the distribution of defects is not Poisson 4 Test statistic The test statistic is k i 1 oi Ei 2 Ei Sec 9 7 Testing for Goodness of Fit Copyright 2014 John Wiley Sons Inc All rights reserved 7 9 7 Testing for Goodness of Fit Example 9 12 5 Reject H0 if Reject H0 if the P value is less than 0 05 6 Computations 32 28 32 2 28 32 13 10 44 2 10 44 15 21 24 2 21 24 2 94 2 7 Conclusions We find from Appendix Table III that 2 71and 2 3 84 Because 2 94 lies between these values we conclude that the P value is between 0 05 and 0 10 Therefore since the P value exceeds 0 05 we are unable to reject the null hypothesis that the distribution of defects in printed circuit boards is Poisson The exact P value computed from Minitab is 0 0864 Sec 9 7 Testing for Goodness of Fit Copyright 2014 John Wiley Sons Inc All rights reserved 8 9 8 Contingency Table Tests Many times the n elements of a sample from a population may be classified according to two different criteria It is then of interest to know whether the two methods of classification are statistically independent Table 9 2 An r c Contingency Table Columns 1 2 c 1 O11 O12 O1c 2 O21 O22 O2c r Or1 Or2 Orc Rows Sec 9 8 Contingency Table Tests Copyright 2014 John Wiley Sons Inc All rights reserved 9 9 8 Contingency Table Tests We are interested in testing the hypothesis that the row and column methods of classification are independent If we reject this hypothesis we conclude there is some interaction between the two criteria of classification The exact test procedures are difficult to obtain but an approximate test statistic is valid for large n Let pij be the probability that a randomly selected element falls in the ijth cell given that the two classifications are independent Then pij uivj where ui is the probability that a randomly selected element falls in row class i and vj is the probability that a randomly selected element falls in column class j Now assuming independence the estimators of ui and vj are 1 u i n v j 1 n c Oij j 1 r Oij 9 17 i 1 Sec 9 8 Contingency Table Tests Copyright 2014 John Wiley Sons Inc All rights reserved 10 9 8 Contingency Table Tests Therefore the expected frequency of each cell is r 1 c Eij nu i v j Oij Oij n j 1 i 1 9 18 Then for large n the statistic r c Oij Eij 2 i 1 j 1 Eij 9 19 has an approximate chi square distribution with r 1 c 1 degrees of freedom if the null hypothesis is true We should reject the null hypothesis if the value of the test statistic is too large The P value would be calculated as the probability beyond on the r 1 c 1 distribution or P p r 1 c 1 For a fixed level test we would reject the hypothesis of independence if the observed value of the test statistic exceeded r 1 c 1 Sec 9 8 Contingency Table Tests Copyright 2014 John Wiley Sons Inc All rights reserved 11 9 8 Contingency Table Tests A company has to choose among three health insurance plans Management wishes to know whether the preference for plans is independent of …
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