Down to earth Measurements.


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of a Monkey Shine Restaurant is comprised of 30% Western agents, ... at the Kowloon Monkey Shine discovered 150 Western. agents, 190 Chinese representatives, 100 ...
Transcripts
Slide 1

Down to earth Statistics Chi-Square Statistics

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There are six insights that will answer 90% of all inquiries! Clear Chi-square Z-tests Comparison of Means Correlation Regression

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Chi-square: Chi-square is a basic test for numbers… ..

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Chi-square: Chi-square is a straightforward test for checks… .. Which implies: ostensible information and… in the event that a few cases… Ordinal information

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Chi-square: There are three sorts: Test for populace change Test of "decency of-fit" Contingency table examination

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Chi-square: There are three sorts: Test for populace difference

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Chi-square: There are three sorts: Test for populace fluctuation Test of "integrity of-fit" Where o = recurrence of real perception, and e = recurrence you anticipated that would discover

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According to showcasing research, the customer base of a Monkey Shine Restaurant is comprised of 30% Western specialists , 30% ladies who stop in while shopping, 30% Chinese representatives , and 10% visitors . An irregular specimen of 600 clients at the Kowloon Monkey Shine discovered 150 Western agents , 190 Chinese specialists , 100 sightseers , and 65 ladies who were shopping. Is the customers at this foundation unique in relation to the standard of the this organization?

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= 5.00 + 0.56 + 2.22 + 26.67 = 34.45 With (4-1) degrees of flexibility

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The chi-square circulation is very skewed and ward upon what number of degrees of opportunity ( df ) an issues has.

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The chi-square for the eatery issue was: Chi-square = 34.45, df = 3 By looking in a table , the basic estimation of Chi-square with df = 3 is 7.82. The likelihood that the explored recurrence breaks even with the recurrence found in the MR task was p < .001 . http://www.fourmilab.ch/rpkp/tests/examination/chiCalc.html

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By taking a gander at the investigation, clearly the biggest commitment to chi-square originated from the sightseers. = 5.00 + 0.56 + 2.22 + 26.67 = 34.45 df = 3 Hence, the Kowloon property is pulling in more traveler than what might be normal at the Monkey Shine.

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Chi-square: There are three sorts: Test for populace difference Test of "decency of-fit" Contingency table investigation Where o = recurrence of genuine perception, and e = recurrence you anticipated that would discover

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A possibility table is a table with numbers gathered by recurrence.

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A possibility table is a table with numbers assembled by recurrence. There are three gatherings: brand faithful clients, standard purchasers, and infrequent purchasers. Each is inquired as to whether they like the essence of new item over the old. They reply with a "yes" or a "no."

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A possibility table would appear as though this:

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A possibility table is a table with numbers gathered by recurrence. Every one of the numbers in the table are " watched " frequencies ( o ). Along these lines, what are the normal qualities ?

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The normal qualities ( e ) would be an irregular circulation of frequencies.

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The normal qualities ( e ) would be an arbitrary circulation of frequencies. These can be computed by increasing the line recurrence by the segment recurrence and separating by the aggregate number of perceptions.

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For instance, the normal qualities ( e ) of "faithful" And "yes" would be (150 X 90)/270 = 50

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For instance, the normal qualities ( e ) of "customary" And "no" eventual (120 X 100)/270 = 44.4

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The normal qualities ( e ) for the whole table would be:

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The chi-square esteem is figured for each cell, and afterward summed over every one of the cells.

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The chi-square esteem is computed for each cell: For Cell A : (50-50)^2/50 = 0 For Cell D : (40-44.4)^/44.4 = 0.44

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The chi-square esteem is ascertained for each cell:

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The chi-square esteem is figured for each cell: Chi-square = 0 + 0 + .35 + .44 + .44 + .54 = 1.77 The df = (r-1)(c-1) = 1 X 2 = 2

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A chi-square with a df = 2 has a basic estimation of 5.99, this chi-square = 1.77, so the outcomes are nonsignificant . http://www.fourmilab.ch/rpkp/tests/examination/chiCalc.html The likelihood = 0.4127 . This implies the dissemination is arbitrary , and there is no relationship between client sort And taste inclination.

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A chi-square with a df = 2 has a basic estimation of 5.99, this chi-square = 1.77, so the outcomes are nonsignificant. This implies the dispersion is arbitrary, and there is no relationship between client sort And taste inclination. Note : This kind of chi-square is a trial of affiliation utilizing only include (recurrence); VERY valuable business research.

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