Purchase Solution

Chi-square test for goodness of fit: One-day Absences

Not what you're looking for?

Ask Custom Question

See attached file.

(13.17)

From the one-day work absences during the past year, the personnel director for a large firm has identified the day of the week for a random sample of 150 of the absences. Given the following observed frequencies, and for α = 0.01, can the director conclude that one-day absences during the various days of the week are not equally likely?

Monday Tuesday Wednesday Thursday Friday Total
Absences 42 18 24 27 39 150

Purchase this Solution

Solution Summary

The solution provides step by step method for the calculation of chi square test for goodness of fit. Formula for the calculation and Interpretations of the results are also included.

Solution Preview

Please see the attachments.

Answer
The Null hypothesis tested is
H0: One-day absences during the various days of the week are equally likely.
The Alternative Hypothesis is
H1: One-day absences during the various days of the week are not equally likely.
The ...

Purchase this Solution


Free BrainMass Quizzes
Know Your Statistical Concepts

Each question is a choice-summary multiple choice question that presents you with a statistical concept and then 4 numbered statements. You must decide which (if any) of the numbered statements is/are true as they relate to the statistical concept.

Measures of Central Tendency

This quiz evaluates the students understanding of the measures of central tendency seen in statistics. This quiz is specifically designed to incorporate the measures of central tendency as they relate to psychological research.

Terms and Definitions for Statistics

This quiz covers basic terms and definitions of statistics.

Measures of Central Tendency

Tests knowledge of the three main measures of central tendency, including some simple calculation questions.