Answered

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A company manager wishes to test a union leader's claim that absences occur on the different week days with the same frequencies. Test this claim at the 0.05 level of significance if the following sample data have been compiled.

Day Monday Tuesday Wednesday Thursday Friday
Absences 37 15 12 23 43

Does the data support her claim?

Sagot :

fichoh

Answer:

We reject H0 and conclude that difference ls occur in the number of Absences

Step-by-step explanation:

Given the data :

Day Monday Tuesday Wednesday Thursday Friday

Absences 37 15 12 23 43

H0: The proportions are all equal

H1: The proportions are not equal

We use the Chisquare test:

The test statistic :

χ² = Σ(observed - Expected)² / Expected

Expected value is the same for all the days, since they have the same frequency :

Expected value = percentage * sample size

Sample size = 37 + 15 + 12 + 23 + 43 = 130

Expected value = (100/5)% * 130

Expected value = 26

χ² = Σ(observed - Expected)² / Expected

χ² = (37-26)²/26 + (15-26)²/26 + (12-26)²/26 + (23-26)²/26 + (43-26)²/26

χ² = 28.31

The Pvalue using Pvalue from Chisquare test score calculator, df = 4

Pvalue = 0.000011

α = 0.05

Pvalue < α ; We reject H0 and conclude that difference ls occur in the number of Absences

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