Goodness-of-fit (GOF) Tests Testing for the distribution of the underlying population H 0 : The sample data came from a specified distribution. H 1 : The.

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Presentation transcript:

Goodness-of-fit (GOF) Tests Testing for the distribution of the underlying population H 0 : The sample data came from a specified distribution. H 1 : The sample data did not come from the specified distribution.

9-7 The Chi-Square GOF Test Assume there is a sample of size n from a population whose probability distribution is unknown. Let O i be the observed frequency in the ith class interval. Let E i be the expected frequency in the ith class interval. The test statistic is chi-square with df = k – p – 1 where p = number of estimated parameters

Chi-Square GOF Test where k = number of classes O i = observed number in the i th class E i = expected number in the i th class = n p i n = sample size p i = F(a i ) - F(a i-1 ) = probability of a failure occurring in the i th class if H 0 is true with df = k number of estimated parameters Hypothesized distribution i th class is defined by [a i-1, a i ) with a 0 = 0

Example 9-12

More of Example 9-12

Much More of Example 9-12

Still Example 9-12

Yes, Example 9-12

The last of Example 9-12

Is this Normal? Three years of daily high temperatures in Dayton during the month of August yields the following data: Enter the raw data Stats sample size mean variance std dev median st quartile rd quartile interquartile rg Mimimum Maximum Range21 Skewness Kurtosis

The observed count Sturges rule: k = 1+3.3*LOG(90,10) = ; 7 or 8 classes BinFrequency More0

The expected count H 0 : High temperatures during August in Dayton are normally distributed H 1 : High temperatures in August do not have a normal distribution mean variance std dev4.545

The Chi-Sq Statistic intervalUpper bndobservedexpected(O-E)^2/E infinity sum df = 9 – 2 – 1 = 6 Cannot reject the null hypothesis that the sample came from a normal distribution

combine cells intervalupperobservedexpected(O-E)^2/E infinity sum intervalUpper bndobservedexpected(O-E)^2/E infinity sum df = 6 – 2 – 1 = 3