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This deck focuses on Chi Square Goodness Of Fit Test, giving you a quick way to review the definitions, rules, and examples that matter most for AP Statistics.
Study Chi Square Goodness Of Fit Test in AP Statistics with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
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What is the consequence if expected frequencies are less than 5?
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Chi-Square test may not be valid. Violates assumptions needed for valid statistical inference.
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This deck focuses on Chi Square Goodness Of Fit Test, giving you a quick way to review the definitions, rules, and examples that matter most for AP Statistics.
Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.
Answer: Chi-Square test may not be valid. Violates assumptions needed for valid statistical inference.
Answer: 0.05. Most commonly used threshold for statistical significance.
Answer: Data must be categorical. Chi-square tests require frequency counts of categories.
Answer: To determine whether to reject the null hypothesis. Compare with test statistic to make statistical decision.
Answer: To compare against observed frequencies. Baseline for measuring deviation from null hypothesis.
Answer: Data must be categorical. Chi-square tests require frequency counts of categories.
Answer: It becomes more symmetric. Approaches normal distribution as degrees of freedom increase.
Answer: Increases the power of the test. Larger samples better detect true differences from null.
Answer: Chi-Square Goodness of Fit test. Tests if sample matches theoretical distribution pattern.
Answer: Continuous data. Chi-square requires categorical frequency data only.
Answer: Degrees of freedom = 6. Formula: df=k−1 where k is categories.
Answer: Chi-Square Goodness of Fit test. Tests if sample matches theoretical distribution pattern.
Answer: Chi-Square statistic > critical value. Decision rule for hypothesis testing with chi-square.
Answer: Increases the power of the test. Larger samples better detect true differences from null.
Answer: Degrees of freedom = 4. Formula: df=k−1 where k is number of categories.
Answer: Higher Chi-Square statistic typically results in a lower p-value. Larger deviations produce higher statistics and lower p-values.
Answer: Expected frequency should be at least 5. Ensures valid approximation to chi-square distribution.
Answer: Fail to reject the null hypothesis. Insufficient evidence to reject null hypothesis.
Answer: Sample size should be large enough to ensure expected frequencies are at least 5. Prevents invalid approximation to chi-square distribution.
Answer: Higher Chi-Square statistic typically results in a lower p-value. Larger deviations produce higher statistics and lower p-values.
Answer: To determine if observed frequencies match expected frequencies. Tests if data follows a specific distribution pattern.
Answer: Goodness of Fit and Test of Independence. Both test distribution assumptions but in different ways.
Answer: Null hypothesis. Reject when evidence strongly contradicts null assumption.
Answer: Categorical data. Chi-square applies to frequency counts of categories.
Answer: To determine whether to reject the null hypothesis. Compare with test statistic to make statistical decision.
Answer: Greater evidence against the null hypothesis. Higher values indicate greater deviation from expected.
Answer: Sample size should be large enough to ensure expected frequencies are at least 5. Prevents invalid approximation to chi-square distribution.
Answer: Degrees of freedom = 4. Formula: df=k−1 where k is number of categories.
Answer: Continuous data. Chi-square requires categorical frequency data only.
Answer: df=k−1. Where k is the number of categories being tested.
Answer: To compare against observed frequencies. Baseline for measuring deviation from null hypothesis.
Answer: The observed frequencies match the expected frequencies. Assumes the theoretical distribution is correct.
Answer: Ei=1Total Observed×Proportion. Multiply total sample size by theoretical probability.
Answer: Degrees of freedom = 9. Formula: df=k−1 where k is categories.
Answer: Chi-Square test may not be valid. Violates assumptions needed for valid statistical inference.
Answer: df=k−1. Where k is the number of categories being tested.
Answer: Chi-Square distribution. Theoretical distribution for chi-square test statistic.
Answer: Ei=1Total Observed×Proportion. Multiply total sample size by theoretical probability.
Answer: The observed frequencies match the expected frequencies. Assumes the theoretical distribution is correct.
Answer: 7.815. From chi-square table with α=0.05 and df=3.
Answer: Fail to reject the null hypothesis. Insufficient evidence to reject null hypothesis.
Answer: Right-skewed. Chi-square values are always non-negative with long right tail.
Answer: Strong evidence against the null hypothesis. Small p-values suggest data unlikely under null hypothesis.
Answer: Degrees of freedom = 6. Formula: df=k−1 where k is categories.
Answer: Ei represents the expected frequency. The subscript i denotes each category under null hypothesis.
Answer: To determine if observed frequencies match expected frequencies. Tests if data follows a specific distribution pattern.
Answer: χ2=∑Ei(Oi−Ei)2. Compares observed and expected frequencies for each category.
Answer: Chi-Square distribution. Theoretical distribution for chi-square test statistic.
Answer: 0.05. Standard alpha level for hypothesis testing.
Answer: Chi-Square distribution table. Provides critical values based on degrees of freedom.
Answer: Ei represents the expected frequency. The subscript i denotes each category under null hypothesis.
Answer: Observed frequencies differ from expected frequencies. Provides evidence against null hypothesis of equal distribution.
Answer: It becomes more symmetric. Approaches normal distribution as degrees of freedom increase.
Answer: 7.815. From chi-square table with α=0.05 and df=3.
Answer: Strong evidence against the null hypothesis. Small p-values suggest data unlikely under null hypothesis.
Answer: Degrees of freedom = 9. Formula: df=k−1 where k is categories.
Answer: P-value indicates the probability of observing a test statistic as extreme. Area under chi-square curve beyond observed statistic.
Answer: Categorical data. Chi-square applies to frequency counts of categories.
Answer: Goodness of Fit and Test of Independence. Both test distribution assumptions but in different ways.
Answer: Expected frequency should be at least 5. Ensures valid approximation to chi-square distribution.
Answer: P-value indicates the probability of observing a test statistic as extreme. Area under chi-square curve beyond observed statistic.
Answer: The observed frequencies do not match the expected frequencies. At least one category differs from expected distribution.
Answer: Oi represents the observed frequency. The subscript i denotes each category in the data set.
Answer: The observed frequencies do not match the expected frequencies. At least one category differs from expected distribution.
Answer: 0.05. Most commonly used threshold for statistical significance.
Answer: Null hypothesis. Reject when evidence strongly contradicts null assumption.
Answer: Chi-Square distribution table. Provides critical values based on degrees of freedom.
Answer: Oi represents the observed frequency. The subscript i denotes each category in the data set.
Answer: Chi-Square statistic > critical value. Decision rule for hypothesis testing with chi-square.
Answer: 0.05. Standard alpha level for hypothesis testing.
Answer: Observed frequencies differ from expected frequencies. Provides evidence against null hypothesis of equal distribution.
Answer: Greater evidence against the null hypothesis. Higher values indicate greater deviation from expected.