Answered

Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Experience the ease of finding accurate answers to your questions from a knowledgeable community of professionals. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

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

Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.