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MASINDE MULIRO UNIVERSITY

OF SCIENCE & TECHNOLOGYCOMPUTER SCIENCE DEPARTMENT

PCT 911: ADVANCE RESEARCH METHODS

TASK

Chi-square goodness-of-fit test (x2) for Stolen

Vehicles

SUBMITTEDBY

NAME: NAHASON MATOKE

REGNO: SIT/H/004/10

SUBMITTEDTO:

DRG. WANYEMBI

KAKAMEGA

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Chi Sqr Assignment 2

Purpose:Test for distributional adequacythe chi-square test (Snedecor and Cochran, 1989) is used

to test if a sample of data came from a population with a specific distribution.

An attractive feature of the chi-square goodness-of-fit test is that it can be applied to any

univariate distribution for which you can calculate the cumulative distribution function.

The chi-square goodness-of-fit test is applied to binned data (i.e., data put into classes).

This is actually not a restriction since for non-binned data you can simply calculate a

histogram or frequency table before generating the chi-square test. However, the values of

the chi-square test statistic are dependent on how the data is binned. Another disadvantage

of the chi-square test is that it requires a sufficient sample size in order for the chi-square

approximation to be valid.

The chi-square test is an alternative to the Anderson-Darling and Kolmogorov-Smirnovgoodness-of-fit tests. The chi-square goodness-of-fit test can be applied to discrete

distributions such as the binomial and the Poisson. The Kolmogorov-Smirnov and

Anderson-Darling tests are restricted to continuous distributions.

DefinitionThe chi-square test is defined for the hypothesis:H0: The data follow a specified distribution.

Ha: The data do not follow the specified distribution.

Test Statistic: For the chi-square goodness-of-fit computation, the data are divided into kbins and the test statistic is defined as:

Where is the observed frequency for bin i and is the expected frequency for bin i.

The expected frequenciesTheKenya Evening Star, Nov. 7, 2009, reported the following information for a randomsample of 1000 stolen cars for the previous year:

170 were Fords, 300 Toyotas, 210 Nissans, 190 Hyundai's, and 130 Peugeots.

Using the X2= goodness-fit test and significance level of 0.01 to test the hypothesis that

proportions stolen are identical to population make proportions.Suppose it is established that 15% of all cars are Fords, 35% are Toyotas, 20% are Nissans,

15% are Hyundais, and 15% are Peugeots.

The Observed Stolen Vehicles.Ford Toyota Nissan Hyundai Peugeot Total

Stolen (Oij) 170 300 210 190 130 1000

Percentage of vehicles stolen for each make;

(Stolen make/Total stolen) * 100

Ford Toyota Nissan Hyundai Peugeot Total

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Stolen (Oij) % 17 30 21 19 13 100

Total vehicles

Total Vehicles=(stolen/percentage of stolen Vehicle)*100 = 5000

There fore

Expected Stolen Frequencies (Stolen Vehicle)

Given that15% of all cars are Fords, 35% are Toyotas, 20% are Nissans, 15% are Hyundais, and 15%

are Peugeots

15% Ford ofTotal Vehicles =

Ford Toyota Nissan Hyundai Peugeot

Eij % 15% 35% 20% 15% 15%Eij 150 350 200 150 150

Test the null hypothesis

Oij Eij Oij- Eij (Oij- Eij)2/ EijFord 170 150 20 2.666666667

Toyota 300 350 -50 7.142857143

Nissan 210 200 10 0.5

Hyundai 190 150 40 10.66666667

Peugeot 130 150 -20 2.666666667

23.64285714

That is, chi-square is the sum of the squared difference between observed (Oij) and theexpected (Eij) data (or the deviation, d), divided by the expected data in all possiblecategories

Assessing significance levels:

In the chi-square test for independence the degree of freedom is equal to the number ofcolumns in the table minus one multiplied by the number of rows in the table minus one.

Df: = (c-1) (r-1)

= (2-1) (5-1)

= 4

Thus the value calculated from the formula above is compared with values in the chi-

square distribution table (Bissonnette, 2006). We reject the null hypothesis if the chi-

squared value is greater than the critical value (what is called the upper critical value).

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ConclusionTherefore the chi square for these data is: 23.643 (4 degrees of freedom: (2-1) (5-1)). The

critical value at p =.01 is 13.277

Since 23.643 is larger than 13.277, what observed differs from these expectations is enough

to reject the null Hypothesis.

State the you can draw from the observations made

Test the null hypothesis

Set up the hypothesis for Chi-Square goodness of fit test:

H0. Null hypothesis: In Chi-Square goodness of fit test, the null hypothesis assumes thatthere is no significant difference between the observed and the expected value.

Ha. Alternative hypothesis: In Chi-Square goodness of fit test, the alternative hypothesisassumes that there is a significant difference between the observed and the expected value.

The calculated value of X2 (23.636) is much higher than the table value(13.277) which

means that the calculated value cannot be said to have been due to chance. It is significant

Hence, the hypothesis does not hold