Important Questions for IGNOU MAPC MPC006 Exam with Main Points for Answer - Block 4 Unit 3 Kruskal Wallis Analysis of Variance

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Block 4 Unit 3 Kruskal Wallis Analysis of Variance

For Numericals, check the questions and examples in your study material.

1) Under what circumstances does the chi-square distribution provide an appropriate characterisation of the sampling distribution of the Kruskal–Wallis H statistic?

The chi-square distribution provides an appropriate characterisation of the sampling distribution of the Kruskal-Wallis H statistic when the sample sizes are large (generally n > 5 for each group). As the values of k (number of groups) and N (total sample size) increase, the chi-square distribution becomes a more accurate approximation of the exact Kruskal-Wallis distribution.

2) Data were collected from three populations A, B, and C by means of a completely randomized design. The following describes the sample data:

nA = nB = nC = 15
RA = 235 RB = 439 RC = 361

a) Specify the null and alternative hypotheses that should be used in conducting a test of hypothesis to determine whether the probability distributions of populations A, B, and C differ in location.

The null and alternative hypotheses for testing if the probability distributions of populations A, B, and C differ in location are:

H₀: The population medians are equal (Median_A = Median_B = Median_C).
H₁: At least one population median is different.

b) Conduct the test of part

To conduct the Kruskal-Wallis test:

Calculate the Kruskal-Wallis test statistic (H) = \[12 / *N*(*N*+1)\] \[Σ((*ΣR*)²/ *n*)\] – 3(*N* + 1)

Where: N = Total number of cases (45), n = Number of cases in each group (15), ΣR = Sum of ranks for each group

H = \[12 / 45(45+1)\] \[(235²/15) + (439²/15) + (361²/15)\] – 3(45 + 1)

H ≈ 14.35

Determine the degrees of freedom (df) = k - 1 = 3 - 1 = 2

Compare the calculated H value with the critical chi-square value: Using a chi-square table with df = 2 and a significance level of α = 0.05, the critical value is 5.991.

Conclusion: Since the calculated H value (14.35) is greater than the critical chi-square value (5.991), we reject the null hypothesis. Therefore, we conclude that there is a statistically significant difference in the location (medians) of at least one of the populations.

3) What are the assumptions of ANOVA?

The assumptions of ANOVA are:

Data is obtained through random sampling.
The populations are normally distributed.
The variances within each group are equal (homogeneity of variances).

4) Why are multiple t-tests not preferred when we have to compare more than 2 means?

Multiple t-tests are not preferred when comparing more than two means because it increases the risk of Type I error (false positive). Each t-test has a probability of making a Type I error (usually set at α = 0.05). When conducting multiple t-tests, the overall probability of at least one Type I error increases, leading to an inflated chance of falsely rejecting a true null hypothesis. ANOVA controls for this inflated error rate by conducting a single overall test to compare all group means simultaneously.

5) State the null and alternative hypothesis for Kruskal Wallis ANOVA.

The null and alternative hypotheses for the Kruskal-Wallis ANOVA are:

H0: The population medians are equal.
H1: At least one population median is different.

6) Enumerate the assumptions of Kruskal Wallis ANOVA

The assumptions of the Kruskal-Wallis ANOVA are:

Random sampling: Each sample has been randomly selected from the population it represents.
Independent samples: The k samples are independent of one another.
Continuous random variable: The dependent variable (which is subsequently ranked) is a continuous random variable.
Identical distribution shapes: The underlying distributions from which the samples are derived are identical in shape.

7) What are the Steps of the Kruskal-Wallis Test?

Rank all the numbers in the entire data set from smallest to largest (using all samples combined); in the case of ties, use the average of the ranks that the values would have normally been given.
Total the ranks for each of the samples; call those totals T1, T2, . . ., Tk, where k is the number of populations.
Calculate H.
Find the p-value.
Make your conclusion about whether you can reject H0 by examining the p-value.

Important Points

i) We would use the t-test if we are testing a hypothesis of μ1 = μ2, and the ANOVA test when μ1 = μ2 = μ3 = μ4, if the populations under consideration are normally distributed.

ii) ANOVA was developed by British Statistician Sir Ronald Fisher.

iii) ANOVA is used when k (the number of groups being compared) is greater than 2.

iv) ANOVA compares multiple means but the logic behind ANOVA is similar to the t-test that compares two independent means.

v) ANOVA is a parametric statistic. Its equivalent non-parametric statistic is the Kruskal-Wallis test.

vi) The Kruskal Wallis ANOVA was developed by William Kruskal and W. Allen Wallis in 1952.

vii) ANOVA compares the means of more than two groups, whereas the medians of more than two groups are compared by Kruskal Wallis ANOVA.

viii) One of the assumptions in Kruskal Wallis ANOVA is that the dependent variable (which is subsequently ranked) is a continuous random variable.

ix) Kruskal Wallis ANOVA can be viewed as ANOVA based on rank transformed data.

x) As the values of k and N increase, the chi-square distribution provides a more accurate estimate of the exact Kruskal-Wallis distribution.

xi) Use of the chi-square distribution for small sample sizes will generally result in a slight decrease in the power of the test.

xii) When the critical value of H is more than the actual obtained value of H, we fail to reject the null hypothesis.

xiii) When the critical value of H is less than the actual obtained value of H, we reject the null hypothesis.

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