Math Problem Statement

Betty recorded the number of visitors at her art gallery during the week:

Day

Visitor

Tuesday

28

Wednesday

20

Thursday

18

Friday

14

She expected to have 20 visitors each day. To determine whether the number of visitors follows a uniform distribution, a chi-square test for goodness of fit should be performed (alpha = 0.10).

Using the data above, what is the chi-square test statistic (answer choices are rounded to the hundredths place)?

A.)

5.2

B.)

0.83

C.)

0.20

D.)

12.5

Solution

To calculate the chi-square test statistic for goodness of fit, we need to follow these steps:

  1. Determine the expected frequency: Betty expected 20 visitors each day, so the expected frequency for each day is 20.

  2. Write the observed frequencies: From the dataTo calculate the chi-square test statistic for goodness of fit, we need to follow these steps:

  3. Determine the expected frequency: Betty expected 20 visitors each day, so the expected frequency for each day is 20.

  4. Write the observed frequencies: From the

    • Tuesday: 28
    • Wednesday: 20
    • Thursday: 18
    • Friday: 14
  5. Apply the chi-square formula: The chi-square statistic is calculated using the formula: χ2=(OiEi)2Ei\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} where:

    • OiO_i is the observed frequency for each category (each day).
    • EiE_i is the expected frequency for each category (20 visitors per day in this case).

    We compute this for each day:

    • Tuesday: (2820)220=6420=3.2\frac{(28 - 20)^2}{20} = \frac{64}{20} = 3.2
    • Wednesday: (2020)220=020=0\frac{(20 - 20)^2}{20} = \frac{0}{20} = 0
    • Thursday: (1820)220=420=0.2\frac{(18 - 20)^2}{20} = \frac{4}{20} = 0.2
    • Friday: (1420)220=3620=1.8\frac{(14 - 20)^2}{20} = \frac{36}{20} = 1.8
  6. Sum the values: χ2=3.2+0+0.2+1.8=5.2\chi^2 = 3.2 + 0 + 0.2 + 1.8 = 5.2

So, the chi-square test statistic is 5.2.

Answer: A.) 5.2


Would you like more details or have any questions? Here are some related questions to expand on this:

  1. What is the critical value for this chi-square test at α=0.10\alpha = 0.10?
  2. How would the test result differ with an alpha of 0.05?
  3. What are the degrees of freedom in this problem?
  4. How do we interpret the result of a chi-square goodness of fit test?
  5. Can this chi-square test be applied to larger datasets, and how does it scale?

Tip: For a chi-square test, the degrees of freedom are usually calculated as the number of categories minus one.

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Math Problem Analysis

Mathematical Concepts

Statistics
Chi-Square Test
Goodness of Fit

Formulas

Chi-square test statistic formula: χ² = Σ ((O_i - E_i)² / E_i)

Theorems

Chi-square distribution
Goodness of fit test

Suitable Grade Level

Grades 11-12