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Statistics Chi Square

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Statistics Chi Square
Objectives: Learn the uses of Chi-Square test in making inferences for given population(s)

BUSSTAT prepared by CSANDIEGO

Chi-Square

2, 

used  to test hypotheses concerning variances  for test concerning frequency distributions  to test the independence of two variables  The chi-square variable cannot be negative and the distributions are positively skewed. At about 100 d.f., the distribution becomes symmetrical. The area under each chi-square distribution is equal to 1 or 100%.

BUSSTAT prepared by CSANDIEGO

A portion of 2 distribution

BUSSTAT prepared by CSANDIEGO

Test for a Single Variance
 Used to test a claim about a single variance, should be

perform to verify consistency  Assumptions: The sample must be selected from a normally distributed population and observations must be independent of each other  The test can be one-tailed or two-tailed. 2  Formula: (n  1) s 2

with d.f. = n – 1 where n = sample size

 

2

s 2  sample var iance

 2  population var iance
BUSSTAT prepared by CSANDIEGO

Example 1:
A company claims that the variance of the sugar content of its yogurt is less than or equal to 25 (mg/oz)2. A sample of 20 servings is selected and the sugar content is measured. The variance of the sample was found to be 36. At =0.05, is there enough evidence to reject the claim?

BUSSTAT prepared by CSANDIEGO

I. Statement of the Hypotheses H0: 2  25 (claim) Ha: 2 > 25 (one-tailed right) II. Level of Significance and Critical Value  = 0.05 d.f. = 20 -1 = 19 c.v.: 30.144 III. Decision Rule: Reject Ho if 2 value is > 30.144, otherwise, do not reject Ho. IV. Test Statistics (n  1) s 2 (20  1)(36) 2    27.36 2 V. Decision  25 Do not reject Ho, since 27.36 < 30.144. There is enough evidence to support the claim that the sugar content of the yogurt is less than or equal to 25 mg/oz.
BUSSTAT prepared by CSANDIEGO

Chi-Square Goodness-of-fit
 Test for Goodness of Fit is used to see

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