# Statistics Chi Square

Topics: Variance, Chi-square distribution, Statistical hypothesis testing Pages: 8 (1340 words) Published: August 14, 2013
Objectives: Learn the uses of Chi-Square test in making inferences for given population(s)

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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%.

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A portion of 2 distribution

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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
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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?

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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 whether a frequency  

 

distribution fits a specific pattern of preference Ho should be stated as there is no difference or no change. Assumptions: The data are obtained from a random sample and the expected frequency for each category must be 5 or more. If an expected frequency of a class < 5, the class can be combined with another class. The test is always one-tailed right (O  E ) 2 Formula: 2   E with d.f.: c – 1 where: c = number of categories O = observed (actual) frequency E = expected frequency = N/c BUSSTAT prepared by CSANDIEGO

Example 2:
A bank manager wishes to see whether there is any difference in the time customers visit the bank. Six hours are selected and the numbers of customers visiting the bank during each hour are shown below. Do the customers show a preference for a specific time? Use  =0.05 Time No. of Customers 9:00 26 10:00 33 11:00 42 12:00 36 1:00 24 2:00 19

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Solution:
Statement of the Hypotheses Ho: The customers show no preference for a specific time. H1: The customer show a preference for a specific time. (claim) II. Level of Significance and Critical Value  = 0.05 d.f.: c – 1 = 6 – 1=5 c.v.: 11.071 III. Decision Rule Reject Ho if F value > 11.071, otherwise, do not reject Ho. I.

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(O  E ) 2 2 IV. Test Statistics    E

E=180/6 = 30 (O – E)2 16 9 144 (O – E)2 E 0.53 0.30 4.80

O 26 33 42 36

E 30 30 30

O–E -4 3 12

24 19

30 30 30

6 -6 -11

36 36 121

1.20 1.20 4.03 2  = 12.07

V. Decision

Reject Ho, since 12.07>11.071. This means that the customers show a prepared by CSANDIEGO a specific time. preference for BUSSTAT

Chi-square Independence Test
 Used to test the independence of two variables, when data

are tabulated in table form in terms of frequencies.  The table is called contingency table which is made of R rows and C columns.  The expected frequencies for each block called cell is computed as: 1. Find...

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