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Statistics Grade 9-12 Answer Key

Statistics: Chi-Square Tests

Goodness of fit, independence, and homogeneity

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Statistics: Chi-Square Tests

Goodness of fit, independence, and homogeneity

Statistics - Grade 9-12

Instructions: Read each problem carefully. Use a chi-square table or calculator when needed. Show your work in the space provided.
  1. 1

    A teacher rolls a six-sided die 120 times to see if each number is equally likely. The observed counts are: 1: 18, 2: 22, 3: 19, 4: 21, 5: 17, 6: 23. Identify the correct chi-square test to use.

    Look for whether the data involve one categorical variable or two categorical variables.

    The correct test is a chi-square goodness-of-fit test because one categorical variable is being compared to a claimed distribution.
  2. 2

    For the die-rolling situation with 120 rolls and 6 equally likely outcomes, find the expected count for each face of the die.

    The expected count for each face is 120 divided by 6, which equals 20. Each number is expected to occur 20 times.
  3. 3

    A bag of candy is advertised as 40% red, 30% blue, 20% green, and 10% yellow. In a sample of 100 candies, find the expected counts for each color.

    Convert each percent to a decimal and multiply by the sample size.

    The expected counts are 40 red, 30 blue, 20 green, and 10 yellow because each percentage is multiplied by 100.
  4. 4

    For a chi-square goodness-of-fit test with 5 categories, find the degrees of freedom.

    The degrees of freedom are 4 because a goodness-of-fit test has degrees of freedom equal to the number of categories minus 1.
  5. 5

    A chi-square goodness-of-fit test has observed counts 28, 22, and 30. The expected counts are 25, 25, and 30. Calculate the chi-square test statistic.

    Use the formula sum of (observed - expected)^2 divided by expected.

    The chi-square statistic is (28 - 25)^2/25 + (22 - 25)^2/25 + (30 - 30)^2/30 = 9/25 + 9/25 + 0 = 0.72.
  6. 6

    In a chi-square test, the p-value is 0.03 and the significance level is 0.05. State the decision and conclusion.

    Because 0.03 is less than 0.05, we reject the null hypothesis. There is statistically significant evidence against the null hypothesis.
  7. 7

    A school wants to test whether students prefer four lunch options equally: pizza, salad, tacos, and sandwiches. Write the null and alternative hypotheses for the chi-square test.

    The null hypothesis usually states that the claimed distribution is true.

    The null hypothesis is that students prefer the four lunch options equally. The alternative hypothesis is that the preferences are not equally distributed.
  8. 8

    A survey records grade level and favorite school subject to see whether the two variables are associated. Identify the correct chi-square test to use.

    The correct test is a chi-square test of independence because the survey examines whether two categorical variables are associated in one population.
  9. 9

    A two-way table compares pet ownership and whether a student plays a sport. The row total for students who own a pet is 60, the column total for students who play a sport is 50, and the grand total is 120. Find the expected count for the cell representing students who own a pet and play a sport.

    For a two-way table, expected count equals row total times column total divided by grand total.

    The expected count is (60 times 50) divided by 120, which equals 25. The expected count for that cell is 25 students.
  10. 10

    A two-way table has 3 rows and 4 columns. Find the degrees of freedom for a chi-square test of independence.

    The degrees of freedom are (3 - 1)(4 - 1) = 2 times 3 = 6.
  11. 11

    A chi-square test is planned for a table with expected counts 12, 9, 15, 6, 11, and 14. Explain whether the expected count condition is met.

    A common rule is that every expected count should be 5 or greater.

    The expected count condition is met because all expected counts are at least 5.
  12. 12

    A researcher compares music preference among random samples from three different grade levels. The categories are pop, rock, country, and classical. Identify whether this is a chi-square test of independence or a chi-square test of homogeneity, and explain why.

    Homogeneity compares distributions across different populations or groups.

    This is a chi-square test of homogeneity because the researcher is comparing the distribution of one categorical variable across separate samples from different grade levels.
LivePhysics™.com Statistics - Grade 9-12 - Answer Key