Purchase Solution

Chi-square Test

Not what you're looking for?

Ask Custom Question

The Springfield Emergency Medical Service keeps records of its calls. A study of 150 five-minute time intervals resulted in the distribution of number of calls. Use a .01 level of significance and do a chi-square goodness of fit analysis for the data. Analyze the results and interpret what they mean. Use the 5-step hypothesis testing procedure and determine whether this distribution is Poisson

Number of Calls
(per 5-minute interval) Frequency
-------------------------------------------
0 18
1 28
2 47
3 21
4 16
5 11
6 or more 9
-------------------------------------------

Purchase this Solution
Solution Summary

The solution tests the goodness of fit of a Poisson Distribution.

Purchase this Solution
Free BrainMass Quizzes
Measures of Central Tendency

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

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.

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.

View More Free Quizzes
Related BrainMass Solutions

  • Chi-square statistics and non-parametric tests

    Non-parametric test associated with chi square: The non-parametric test associated with chi square is Kolmogorov Smirnov goodness of fit test. This solution discusses chi-square statistics.

  • Why would you use the chi-square statistic?

    374801 Explain a chi-square test. Why would you use the chi-square statistic? Type of data? Provide an explanation of a chi-square test. Why would you use the chi-square statistic? What type of data is used with chi square analysis?

  • "Goodness-of-fit test" vs "Chi-square test for independence"

    In fact, chi-squared test can be used as a goodness-of-fit test, as well as a test for independence. In other words, chi-square tests are used for: Chi-square test for independence" and "Chi-square goodness-of-fit test".

  • Multiple choice questions from Chi-Square test

    198114 Chi square, independence and other concepts 1. In a chi-square test, the sample data are called observed frequencies. True False 2.

  • Statistics - Hypothesis Testing - Chi-square

    of Expected Frequencies Figure 15.2 Critical value of chi-square from Appendix E Figure 15.3 Right-tailed chi-square test for ν = 6 Figure 15.4 MegaStat's chi-square test for Web page data TABLE 15.4 Should Noninstrument Rated

  • Statistics: Chi Square Goodness of Fit Test

    418579 Statistics: Chi Square Goodness of Fit Test In a chi-square goodness-of-fit test when there is close agreement between the observed frequency and the expected frequency, the chi-square test value will be small.

  • Statistics: Hypothesis Tests and Chi Square Tests

    What is the degree of freedom for this hypothesis test? What is the numerical chi square critical value? What is the numerical value of the Chi Square test statistic? What is the numerical Chi Square critical value?

  • Chi Square Tests

    The chi-square test should not be used if the expected frequency in any cell exceeds 30. 3) Multiple Choice: The chi-square test compares the __________ frequency distribution of the data to the frequency distribution that would be __________ if

  • Four assumptions of Chi square tests, goodness of fit, criti

    Why is there just one critical value for a chi-square test even when the hypothesis is a two-tailed test? 1.

  • P value for Chi square test

    211249 P value for Chi square test A chi-square test for independence with 8 degrees of freedom results in a test statistic of 18.21.

View More Related Solutions