Chi Square Analysis - The Open University
www open ac uk/socialsciences/spsstutorial/files/tutorials/chi-square pdf
the same as the expected frequencies (except for chance variation) observed frequency-distribution to a theoretical expected frequency-distribution
SPSS: Expected frequencies, chi-squared test In-depth example
www sfu ca/~jackd/Stat203_2011/Wk12_2_Full pdf
Most important things to know: - How to get the expected frequency from a particular cell - Chi-squared is a measure of how far the observed frequencies are
Chi-Square
www d umn edu/~rlloyd/MySite/Stats/Ch 2013 pdf
Step 1: Arrange data into a frequency/contingency table Step 2: Compute Expected Frequencies Based Upon Null Hypothesis
?2 Test for Frequencies
courses washington edu/psy315/tutorials/chi_2_test_frequencies_tutorial pdf
17 jan 2021 Like all statistical tests, the ?2test involves calculating a statistic that measures how far our observations are from those expected under the
2 X 2 Contingency Chi-square
web pdx edu/~newsomj/uvclass/ho_chisq pdf
examine the expected vs the observed frequencies The computation is quite similar, except that the estimate of the expected frequency is a little harder
Chi-Square Tests and the F-Distribution Goodness of Fit
www3 govst edu/kriordan/files/mvcc/math139/ pdf /lfstat3e_ppt_10 pdf
To calculate the test statistic for the chi-square goodness-of-fit test, the observed frequencies and the expected frequencies are used The observed frequency
1 4 Chi-squared goodness of fit test 1 Introduction 2 Example
www lboro ac uk/media/media/schoolanddepartments/mlsc/downloads/1_4_gofit pdf
estimated from the (sample) data used to generate the hypothesised distribution From these we can calculate the expected frequencies
Chi-Squared Tests
www thphys nuim ie/Notes/EE304/Notes/LEC14/ChiSlide pdf
If the 6-sided die is fair, then the expected frequency is on the null hypothesis and then compare the expected frequencies with the actual frequencies
Week 6: Frequency data and proportions - UBC Zoology
www zoology ubc ca/~whitlock/bio300/labs/LabManual/Week 2006 20-- 20FREQUENCY 20DATA pdf
categorical variable to the frequencies predicted by a null hypothesis than 25 of the expected frequencies are less than 5 and none is less than 1 )
Ex 8- Chi-squared Mapping Exercise pdf - webspace ship edu
webspace ship edu/pgmarr/Geo532/Ex 208- 20Chi-squared 20Mapping 20Exercise pdf
difference between the observed and expected frequencies ij is the expected frequency, R is the row, C is the column, and n total observations
100438_31_4_gofit.pdf
Statistics: 1.4 Chi-squared goodness of fit test
Rosie Shier. 2004.
1 Introduction
A chi-squared test can be used to test the hypothesis that observed data follow a particular distribution. The test procedure consists of arranging the n observations in the sample into a frequency table withkclasses. The chi-squared statistic is: χ
2=?(O-E)2E
The number of degrees of freedom isk-p-1 wherepis the number of parameters estimated from the (sample) data used to generate the hypothesised distribution.
2 Example: The Poisson Distribution
Let X be the number of defects in printed circuit boards. A random sample ofn= 60 printed circuit boards is taken and the number of defects recorded. The results were as follows:Number of Observed
Defects Frequency
0 32 1 15 2 9 3 4 Source: Applied Statistics and Probability for Engineers - Montgomery and Runger Does the assumption of a Poisson distribution seem appropriate as a model for these data?
The null hypothesis isHO:X≂Poisson
The alternative hypothesis isH1:Xdoes not follow a Poisson distribution. The mean of the (assumed) Poisson distribution is unknown so must be estimated from the data by the sample mean: ˆμ=?(32×0) + (15×1) + (9×2) + (4×3)?/60 = 0.75 Using the Poisson distribution withμ= 0.75 we can computepi, the hypothesised prob- abilities associated with each class. From these we can calculate the expected frequencies (under the null hypothesis): p
0=P(X= 0) =e-0.75(0.75)00!
= 0.472?E0= 0.472×60 = 28.32 1 p
1=P(X= 1) =e-0.75(0.75)11!
= 0.354?E1= 0.354×60 = 21.24 p
2=P(X= 2) =e-0.75(0.75)22!
= 0.133?E2= 0.133×60 = 7.98 p
3=P(X≥3) = 1-(p0+p1+p2) = 0.041?E3= 0.041×60 = 2.46
Note: The chi-squared goodness of fit test is not valid if the expected frequencies are too small. There is no general agreement on the minimum expected frequency allowed, but values of 3, 4, or 5 are often used. If an expected frequency is too small, two or more classes can be combined. In the above example the expected frequency in the last class is less than 3, so we should combine the last two classes to get:Number of Observed Expected
Defects Frequency Frequency
0 32 28.32
1 15 21.24
2 or more 9 10.44
The chi-squared statistic can now be calculated:
χ
2=?(O-E)2E
=(32-28.32)228.32+(15-21.24)221.24+(13-10.44)210.44= 2.94 The number of degrees of freedom isk-p-1. Here we havek= 3 classes and we have p= 1 because we had to estimate one parameter (the mean,μ) from the data. So, our chi-squared statistic has 3-1-1 = 1 df. If we look up 2.94 in tables of the chi-squared distribution with df=1, we obtain a p-value of 0.05< p <0.1. We conclude that there is no real evidence to suggest the the data DO NOT follow a Poisson distribution, although the result is borderline.
3 Goodness of fit test for other distributions
The chi-squared goodness of fit test can be used for any distribution. For a discrete distribution the procedure is as described above. For a continuous distribution it is necessary to group the data into classes. Common practice used to carry out the goodness of fit test is to choose class boundaries so that the expected frequencies are equal for each class. For example, suppose we decide to usek= 8 classes. For the standard normal distribution the intervals that divide the distribution into 8 equally likely segments are [0,0.32),[0.32,0.675),[0.675,1.15),[1.15,∞) and their four "mirror image" intervals below zero. For each intervalpi= 1/8 so the expected frequencies would beEi=n/8. 2