How To Test For Fit

Transcription

1 Tests for Goodness of Fit: General Notion: We often wish to know whether a particular distribution fits a general definition Example: To use t tests, we must suppose that the population is normally distributed If a sample is drawn from, say, a normal distribution, the sample values should be reflect the population distribution Allows us to state the number in the sample that should be in a particular range Example: 68% of a normal distribution is within +/- 1 standard deviation of the mean. About 68% of the values in a sample from a normal distribution should be within +/- 1 standard deviation of the mean Comparison of actual and expected numbers is the province of the distribution

2 Let O j be the number observed in the sample in range j Let E j be the number that would be expected if the population had a given distribution, as uniform, Poisson, normal, etc. Then ( O j E j ) E k j 1 where k is the number of categories degrees of freedom = k 1 m where m is the number of parameter estimates used in the calculation j

3 Example: Are the answers to Dr. Dinwiddie s multiple-choice tests random? If so, the answers should conform to a uniform distribution and P(A) = P(B) = P(C) = P(D) = ¼. (For the uniform distribution P(E) = 1/n, where n is the number of possible values.) On a recent exam there were sixty questions with correct answers: A-0, B-5, C-17, and D-18. H 0 : the distribution of answers is uniform H 1 : the distribution is not uniform

4 Correct Answer Observed Expected Squared Difference A 0 15 B 5 15 C D k ( O j E j ) E j 1 j Then = 9.07, and no parameters were estimated, so degrees of freedom = 4 1 = 3

5 Excel and the chi-square distribution CHIDIST(x value, df) returns the area in the right-hand tail of the chi-square distribution goodness of fit tests are all upper one-tail tests, so chidist gives the p-value of the test CHIINV(probability, df) gives the chi-square value for the upper tail of the probability entered use to find the critical value for a chi-square test For the Dinwiddie problem: CHIDIST(9.07, 3) gives the p-value of the test

6 EXAMPLE: Hamish suspects that the dice at Black Bart s are not fair, so he spirits one out of the casino one night. After rolling the stolen die 10 times, he has the following result: No. of Dots No. of Times What are the null and alternative hypotheses? Is Hamish right to be suspicious of Black Bart? k ( O j E j ) E j 1 j

7 Testing for normality suppose that nationally auto insurance has a mean price of $700 with standard deviation $135. We have a sample of 80 NC drivers, and we d like to know whether their insurance bills are normally distributed with the national parameters. how many would we expect in the range 700 to 835? HINT: how many standard deviations? What proportion are within that range of standard deviations?

8 answer: on a normal distribution, 0.34 are between the mean and +1 st dev, so we d expect to find 0.34 * 80 = 7. in that range Setting up a spreadsheet: use normsdist normsdist(-) gives the proportion more than two standard deviations below the mean normsdist(-1) normsdist(-) would give proportion between 1 and st devs below mean

9 Continuing in that fashion, we d have the following St Devs Range Prop. Expected freq < - < to to to to > >

10 To find the observed values in the sample, use the HISTOGRAM tool An elaborated solution appears under Study Aids on my web site. Click on the link to normaltest.xls Issue: how many degrees of freedom does the statistic have? df= k 1 m = = 5

11 Alternate technique: determine whether the sample was drawn from a normal population First, calculate sample mean and standard deviation and use those numbers in the calculation Issue: how many degrees of freedom does the statistic have? df = k 1 m = 6 1 = 3

12 A problem and an alternate solution Each cell should have expected frequency at least 5, otherwise chisquare value is not correct One solution: choose ranges with equal expected frequencies Divide data into, say, 10 ranges each expected to contain 8 observations So we define ranges that each contain 1/10 of total Remember NORMINV(probability, mean, standard deviation) displays the upper boundary of the given probability for the specified mean and standard deviation Example: NORMINV(.1, 300, 0) = % of this distribution is NORMINV(1/10, X, s) will find the boundary of the lowest 10% of the distribution NORMINV(4/10, X, s) finds the boundary of the lowest 40% and so on Look carefully at sheet of the workbook normaltest.xls as posted

13 The boundaries thus found are the bin range Each will have expected number equal to n/c where n is the amount of data and c the number of categories

14 Testing for conformity to an observed distribution: The national distribution of pets is as follows: Number of Pets Percentage of Households or more A marketing company wants to know whether Boone conforms to the national pattern. In a sample of 300 Boone households, they found the following:

15 No. of Pets or more 9 No. of Households Expected No. Squares k ( O j E j ) E j 1 j