Chi-Square Test of Independence

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Questions and Answers

In a Chi-Square test of independence, what does the null hypothesis ($H_0$) typically state?

  • The variables are dependent.
  • The variables are associated.
  • The variables are independent. (correct)
  • There is a significant relationship between the variables.

What does the alternative hypothesis ($H_A$) suggest in a Chi-Square test of independence?

  • There is no relationship between the variables.
  • The variables are dependent. (correct)
  • The variables are independent.
  • The variables have equal variances.

Which formula is used to calculate the degrees of freedom ($df$) in a Chi-Square test of independence for a two-way table?

  • $df = R \times C$
  • $df = (R - 1) \times (C - 1)$ (correct)
  • $df = (R - 1) + (C - 1)$
  • $df = (R + 1) \times (C + 1)$

The chi-squared test statistic formula is given by $X^2 = \sum \frac{(O - E)^2}{E}$. What do 'O' and 'E' represent in this formula?

<p>O = Observed frequency, E = Expected frequency (D)</p> Signup and view all the answers

In the context of a Chi-Square test, how is the expected count for a cell in a two-way table calculated?

<p>Expected Count = (Row total) * (Column total) / Table total (B)</p> Signup and view all the answers

What does a large p-value (e.g., greater than 0.05) indicate in a Chi-Square test of independence?

<p>Weak evidence against the null hypothesis (C)</p> Signup and view all the answers

What does the Chi-Square test of independence help to determine?

<p>Whether there is an association between two categorical variables (A)</p> Signup and view all the answers

Given a two-way table with 3 rows and 4 columns, what are the degrees of freedom ($df$) for the Chi-Square test?

<p>6 (A)</p> Signup and view all the answers

In a Chi-Square test, if the calculated test statistic is large and the p-value is small (e.g., less than 0.05), what decision should be made regarding the null hypothesis?

<p>Reject the null hypothesis (B)</p> Signup and view all the answers

Which of the following is a critical assumption for the Chi-Square test of independence to be valid?

<p>The expected counts in each cell should be sufficiently large (typically at least 5). (A)</p> Signup and view all the answers

If the p-value in a Chi-Square test is 0.65, what conclusion can be drawn?

<p>There is no evidence of dependence between the variables. (B)</p> Signup and view all the answers

In the example data provided regarding popular kids, what are the two categorical variables being analyzed for independence?

<p>Grade Level and Most Important Goal (C)</p> Signup and view all the answers

Using the 'Popular kids' dataset, if the calculated Chi-Square statistic is 1.3121 with 4 degrees of freedom, what is the correct interpretation based on the provided p-value range?

<p>The p-value is more than 0.3. (A)</p> Signup and view all the answers

Based on the Chi-Square test results from the 'Popular kids' dataset, what is the appropriate conclusion regarding the relationship between grade and goals?

<p>Goals do not vary by grade. (B)</p> Signup and view all the answers

Suppose a researcher aims to use a Chi-Square test of independence. Which type of data should they collect?

<p>Categorical data (D)</p> Signup and view all the answers

A study finds a significant association between smoking status and the occurrence of lung cancer using a Chi-Square test. What does this imply?

<p>Smoking and lung cancer are related, but causality cannot be determined from the test. (D)</p> Signup and view all the answers

In a Chi-Square test examining the relationship between political affiliation (Democrat, Republican, Independent) and opinion on a certain policy (Favor, Oppose, Neutral), how should the data be organized?

<p>As a two-way contingency table showing the counts for each combination of political affiliation and opinion (C)</p> Signup and view all the answers

If conducting a Chi-Square test of independence with a significance level of 0.05, and the calculated p-value is 0.03, what statistical decision should be made?

<p>Reject the null hypothesis. (A)</p> Signup and view all the answers

When is it most appropriate to use a Chi-Square test of independence rather than a t-test or ANOVA?

<p>When analyzing the relationship between two categorical variables (C)</p> Signup and view all the answers

A researcher wants to determine if there is an association between gender (male, female) and preferred mode of transportation (car, bus, train). Which statistical test is most appropriate?

<p>Chi-Square test of independence (B)</p> Signup and view all the answers

What is the purpose of calculating expected counts in a Chi-Square test of independence?

<p>To compare what the counts would be if the variables were independent to the observed counts (D)</p> Signup and view all the answers

In interpreting a Chi-Square test result, practical significance differs from statistical significance. What does practical significance refer to?

<p>Whether the observed effect is large enough to be useful in the real world (C)</p> Signup and view all the answers

Considering the 'Popular kids' dataset, which values are used to calculate the Chi-Square test statistic?

<p>Both observed and expected counts (B)</p> Signup and view all the answers

In a Chi-Square test of independence, if all observed and expected counts are exactly the same, what would be the value of the Chi-Square statistic ($X^2$)?

<p>It would be equal to 0. (A)</p> Signup and view all the answers

Flashcards

Chi-Square Test of Independence

A statistical test to determine if there is a significant association between two categorical variables.

Null Hypothesis (Hâ‚€) in Chi-Square Test

The hypothesis that the variables are independent; goals do not vary by grade.

Alternative Hypothesis (Há´€) in Chi-Square Test

The hypothesis that the variables are dependent; goals vary by grade.

Chi-Square Test Statistic

A measure of the difference between observed and expected values in a contingency table.

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Degrees of Freedom (df) Formula

where k = number of cells, R = number of rows, C = number of columns in a contingency table.

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P-value

The probability of obtaining results as extreme as, or more extreme than, the observed results, assuming the null hypothesis is true.

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Purpose of the Chi-Square Test in the Example

Used to determine if there is evidence to suggest goals vary by grade.

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Expected Count Formula

f Expected Count = (row total) times (column total) / table total

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Expected Count for a Particular Cell

It can be determined by 176 x 141 / 478 = 52

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Conclusion based on a Large P-value

Since the p-value is large, we fail to reject Hâ‚€. There is no convincing evidence that grade and goals are dependent. It doesn't appear that goals vary by grade.

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Study Notes

Chi-Square Test of Independence

  • This test is used to determine if there's a statistically significant association between two categorical variables.
  • Students in grades 4-6 were surveyed about whether good grades, athletic ability, or popularity was most important to them.
  • A two-way table separates students by grade and by their choice of most important factor.

Hypotheses

  • Null hypothesis (H0): Grade and goals are independent and goals do not vary by grade.
  • Alternative hypothesis (HA): Grade and goals are dependent and goals vary by grade.

Test Statistic Calculation

  • The test statistic is calculated using the formula: X²df = Σ [(O - E)² / E]
    • k represents the number of cells.
    • R is the number of rows.
    • C is the number of columns.
  • Degrees of freedom (df) for the test statistic calculation = (R - 1) × (C - 1).
  • Note: The calculation of df differs for one-way and two-way tables.

P-Value

  • The p-value represents the area under the X²df curve, above the calculated test statistic.

Expected Counts in Two-Way Tables

  • Expected Count is calculated as: (row total) × (column total) / table total

Example Calculation

  • The expected count for the highlighted cell is: (176 x 141) / 478 = 52.
  • This implies more than the expected number of 5th graders consider being popular as their goal.

Calculating the Test Statistic with Observed and Expected Counts

  • With degrees of freedom df = (R – 1) × (C – 1) = (3 – 1) × (3 – 1) = 2 × 2 = 4.
  • Χ² = Σ [(63 − 61)² / 61] + [(31 − 35)² / 35] +…+ [(32 – 34)² / 34] = 1.3121

Determining the P-Value

  • For a test statistic X²df = 1.3121 with df = 4, the p-value is more than 0.3.

Conclusion

  • Since the p-value is large, the null hypothesis (H0) is not rejected.
  • There is no convincing evidence that goals vary by grade.

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