Free BPCC-108 Solved Assignment | July 2023-January 2024 | STATISTICAL METHODS FOR PSYCHOLOGICAL RESEARCH- II | IGNOU

BPCC-108 Solved Assignment

BPCC-108 Solved Assignment

1. Explain factorial designs with the help of suitable examples. Describe the types of interactional effect with the help of suitable diagrams.
2. Compute One Way ANOVA (Parametric statistics) for the following data:

3. Explain the procedure for testing hypothesis.
4. Explain Kruskal Wallis ANOVA test with a focus on its assumptions. Compare between Kruskal Wallis ANOVA test and one way ANOVA (parametric statistics)
5. Compute Chi-square for the following data:

6. Compute independent $t$$t$tt$t$ test for the following data :

7. Explain the concept of parametric statistics and differentiate it from nonparametric statistics.
8. Explain Statistical Package for Social Sciences (SPSS) with a focus on its data entry and coding.

Expert Answer:

Explain factorial designs with the help of suitable examples. Describe the types of interactional effect with the help of suitable diagrams.

Answer:

Factorial Designs

Example 1: Studying the Effect of Study Time and Type of Study Material on Exam Performance

• Factor 1: Study Time (2 levels: 1 hour vs. 3 hours)
• Factor 2: Type of Study Material (2 levels: Notes vs. Videos)

1. 1 hour of study with notes
2. 1 hour of study with videos
3. 3 hours of study with notes
4. 3 hours of study with videos

Main Effects

• Main Effect of Study Time: The average difference in exam performance between students who studied for 1 hour vs. 3 hours, across both types of study material.
• Main Effect of Study Material: The average difference in exam performance between students who used notes vs. videos, across both study times.

Interaction Effects

Types of Interaction Effects

1. No Interaction
   • If the effect of study time on performance is the same regardless of the study material (e.g., study time always improves performance by the same amount, whether students use notes or videos), then there is no interaction.
   • Diagram: Two parallel lines show the relationship between study time and performance for both notes and videos.

2. Synergistic Interaction (Positive Interaction)
   • A synergistic interaction occurs when the effect of two factors combined is greater than the sum of their individual effects. For example, studying for 3 hours using videos might lead to a significantly higher performance boost than expected from either study time or study material alone.
   • Diagram: The lines representing the two study materials are not parallel and diverge, indicating an enhanced effect when study time and study material interact.

3. Antagonistic Interaction (Negative Interaction)
   • An antagonistic interaction occurs when the effect of one factor is reduced by the presence of another factor. For example, studying for 3 hours with notes might actually result in lower performance than studying for just 1 hour with videos, suggesting that more study time with notes could lead to fatigue or confusion.
   • Diagram: The lines cross, showing that the effect of one factor (study time) is reversed depending on the other factor (study material).

Types of Factorial Designs

1. 2×2 Factorial Design: Two factors, each with two levels.
   • Example: Study Time (1 hour, 3 hours) and Study Material (Notes, Videos).
2. 3×3 Factorial Design: Two factors, each with three levels.
   • Example: Study Time (1 hour, 3 hours, 5 hours) and Study Material (Notes, Videos, Flashcards).
3. 2×3 Factorial Design: One factor with two levels and another with three levels.
   • Example: Study Time (1 hour, 3 hours) and Study Material (Notes, Videos, Flashcards).

Conclusion

Compute One Way ANOVA (Parametric statistics) for the following data:

Answer:

Step 1: Organize the Data

• Group 1: 2, 5, 6, 3, 6, 2, 1, 5, 6, 4
• Group 2: 5, 4, 6, 3, 4, 5, 6, 4, 3, 4
• Group 3: 5, 6, 3, 2, 6, 7, 3, 2, 3, 4

Step 2: Define Hypotheses

• Null Hypothesis (H₀): The means of all groups are equal.
• Alternative Hypothesis (H₁): At least one group mean is different from the others.

Step 3: Calculate Group Means

• Mean of Group 1 = $\frac{2+5+6+3+6+2+1+5+6+4}{10}=4.0$$\frac{2+5+6+3+6+2+1+5+6+4}{10}=4.0$(2+5+6+3+6+2+1+5+6+4)/(10)=4.0\frac{2 + 5 + 6 + 3 + 6 + 2 + 1 + 5 + 6 + 4}{10} = 4.0$\frac{2+5+6+3+6+2+1+5+6+4}{10}=4.0$
• Mean of Group 2 = $\frac{5+4+6+3+4+5+6+4+3+4}{10}=4.4$$\frac{5+4+6+3+4+5+6+4+3+4}{10}=4.4$(5+4+6+3+4+5+6+4+3+4)/(10)=4.4\frac{5 + 4 + 6 + 3 + 4 + 5 + 6 + 4 + 3 + 4}{10} = 4.4$\frac{5+4+6+3+4+5+6+4+3+4}{10}=4.4$
• Mean of Group 3 = $\frac{5+6+3+2+6+7+3+2+3+4}{10}=4.1$$\frac{5+6+3+2+6+7+3+2+3+4}{10}=4.1$(5+6+3+2+6+7+3+2+3+4)/(10)=4.1\frac{5 + 6 + 3 + 2 + 6 + 7 + 3 + 2 + 3 + 4}{10} = 4.1$\frac{5+6+3+2+6+7+3+2+3+4}{10}=4.1$

Step 4: Calculate the Overall Mean

Step 5: Compute the Sum of Squares

a. Between-Groups Sum of Squares (SSB)

b. Within-Groups Sum of Squares (SSW)

Step 6: Compute the Degrees of Freedom

• Degrees of Freedom Between Groups (dfB): $k-1=3-1=2$$k-1=3-1=2$k-1=3-1=2k – 1 = 3 – 1 = 2$k-1=3-1=2$, where $k$$k$kk$k$ is the number of groups.
• Degrees of Freedom Within Groups (dfW): $N-k=30-3=27$$N-k=30-3=27$N-k=30-3=27N – k = 30 – 3 = 27$N-k=30-3=27$, where $N$$N$NN$N$ is the total number of observations.

Step 7: Compute the Mean Squares

• Mean Square Between (MSB):

• Mean Square Within (MSW):

Step 8: Compute the F-Statistic

Step 9: Find the p-Value

Step 10: Conclusion

• Since the p-value (0.849) is much larger than the significance level (usually 0.05), we fail to reject the null hypothesis.
• Conclusion: There is no statistically significant difference in the emotional intelligence scores between the three groups.

Explain the procedure for testing hypothesis.

Answer:

1. State the Hypotheses

• Null Hypothesis (H₀): This is the default assumption that there is no effect or no difference. It represents the status quo or the claim to be tested. For example, "There is no significant difference between the means of two groups."
• Alternative Hypothesis (H₁ or Hₐ): This is what you want to prove, suggesting that there is an effect, a difference, or a relationship. For example, "There is a significant difference between the means of two groups."

2. Set the Significance Level (α)

• This is the threshold for deciding whether to reject the null hypothesis, usually set at 0.05 or 5%. A significance level of 0.05 means there is a 5% risk of concluding that an effect exists when there is no actual effect (false positive).

3. Choose the Appropriate Test

• The test depends on the type of data and the hypothesis. Common tests include:
  • t-tests: Compare means between two groups (independent or paired).
  • ANOVA: Compare means among more than two groups.
  • Chi-square test: For categorical data to test relationships.
  • Regression analysis: For testing relationships between variables.
• Other tests include z-tests, non-parametric tests, etc., depending on the data distribution and sample size.

4. Calculate the Test Statistic

• Perform the test based on your data. Each test will generate a test statistic (e.g., t-value, z-value, F-value) that measures how far the sample data deviates from the null hypothesis.

5. Find the p-value

• The p-value indicates the probability of obtaining the observed results if the null hypothesis were true. A small p-value (usually less than 0.05) suggests that the observed data is unlikely under the null hypothesis, leading to its rejection.

6. Make a Decision

• Compare the p-value to the significance level (α):
  • If p-value ≤ α, reject the null hypothesis. This means there is enough evidence to support the alternative hypothesis.
  • If p-value > α, fail to reject the null hypothesis. This means there isn’t enough evidence to support the alternative hypothesis.

7. Draw Conclusions

• Based on the decision, you either accept the evidence supporting the alternative hypothesis or conclude that the data does not provide sufficient evidence to reject the null hypothesis.

8. Report the Results

• Clearly report the hypothesis, test results, p-value, confidence intervals, and your conclusion. Also, discuss the implications of these results in the context of the research.

Explain Kruskal Wallis ANOVA test with a focus on its assumptions. Compare between Kruskal Wallis ANOVA test and one way ANOVA (parametric statistics).

Answer:

Key Points of Kruskal-Wallis ANOVA:

1. Purpose: It tests whether there are statistically significant differences between the medians of three or more independent groups.
2. Test Statistic: The test is based on the rank sums of the data rather than the actual values.
3. Hypotheses:
   • Null hypothesis ($H_0$$H_0$H_(0)H_0$H_0$): The population medians of all groups are equal.
   • Alternative hypothesis ($H_1$$H_1$H_(1)H_1$H_1$): At least one group has a different median.

Assumptions of Kruskal-Wallis ANOVA:

1. Independent Observations: The groups are independent of each other, and the observations within each group are also independent.
2. Ordinal or Continuous Data: The test can handle data that is ordinal or continuous, but it doesn’t require the data to follow a normal distribution.
3. Similar Distribution Shapes: The distributions of the groups should have the same shape. The Kruskal-Wallis test only compares medians, so if the shape of the distributions differs, the test might lead to incorrect conclusions.

Kruskal-Wallis ANOVA Test Process:

1. Rank all data points together, regardless of their group.
2. Sum the ranks for each group.
3. Use these ranks to calculate the Kruskal-Wallis test statistic ($H$$H$HH$H$).
4. Compare the test statistic to a chi-squared distribution to determine significance.

Kruskal-Wallis ANOVA vs. One-Way ANOVA:

When to Use Each:

• Kruskal-Wallis ANOVA is preferred when:
  • The data is not normally distributed.
  • You have ordinal data.
  • The variances between groups are unequal.
• One-Way ANOVA is preferred when:
  • The data is normally distributed.
  • The groups have similar variances.
  • The data is continuous and measured on an interval or ratio scale.

Compute Chi-square for the following data:

Answer:

Chi-Square Test of Independence – Detailed Steps

Step 1: Data Observation

Step 2: Hypotheses

• Null Hypothesis (H₀): There is no association between the type of manager and their response (i.e., the responses are independent of the manager level).
• Alternative Hypothesis (H₁): There is an association between the type of manager and their response (i.e., the responses are not independent).

Step 3: Calculate Row and Column Totals

Step 4: Calculate Expected Values

• For Junior Managers (Agree):
  $$E=\frac{(15)\times (9)}{30}=4.5$$$$E=\frac{(15)\times (9)}{30}=4.5$$E=((15)xx(9))/(30)=4.5E = \frac{(15) \times (9)}{30} = 4.5$$E=\frac{(15)\times (9)}{30}=4.5$$
• For Junior Managers (Undecided):
  $$E=\frac{(15)\times (10)}{30}=5.0$$$$E=\frac{(15)\times (10)}{30}=5.0$$E=((15)xx(10))/(30)=5.0E = \frac{(15) \times (10)}{30} = 5.0$$E=\frac{(15)\times (10)}{30}=5.0$$
• For Junior Managers (Disagree):
  $$E=\frac{(15)\times (11)}{30}=5.5$$$$E=\frac{(15)\times (11)}{30}=5.5$$E=((15)xx(11))/(30)=5.5E = \frac{(15) \times (11)}{30} = 5.5$$E=\frac{(15)\times (11)}{30}=5.5$$
• For Senior Managers (Agree):
  $$E=\frac{(15)\times (9)}{30}=4.5$$$$E=\frac{(15)\times (9)}{30}=4.5$$E=((15)xx(9))/(30)=4.5E = \frac{(15) \times (9)}{30} = 4.5$$E=\frac{(15)\times (9)}{30}=4.5$$
• For Senior Managers (Undecided):
  $$E=\frac{(15)\times (10)}{30}=5.0$$$$E=\frac{(15)\times (10)}{30}=5.0$$E=((15)xx(10))/(30)=5.0E = \frac{(15) \times (10)}{30} = 5.0$$E=\frac{(15)\times (10)}{30}=5.0$$
• For Senior Managers (Disagree):
  $$E=\frac{(15)\times (11)}{30}=5.5$$$$E=\frac{(15)\times (11)}{30}=5.5$$E=((15)xx(11))/(30)=5.5E = \frac{(15) \times (11)}{30} = 5.5$$E=\frac{(15)\times (11)}{30}=5.5$$

Step 5: Calculate the Chi-Square Statistic

• $O$$O$OO$O$ is the observed frequency.
• $E$$E$EE$E$ is the expected frequency.

• Junior Managers (Agree):
  $$\frac{(4-4.5{)}^2}{4.5}=\frac{0.25}{4.5}\approx 0.0556$$$$\frac{(4-4.5{)}^2}{4.5}=\frac{0.25}{4.5}\approx 0.0556$$((4-4.5)^(2))/(4.5)=(0.25)/(4.5)~~0.0556\frac{(4 – 4.5)^2}{4.5} = \frac{0.25}{4.5} \approx 0.0556$$\frac{(4-4.5{)}^2}{4.5}=\frac{0.25}{4.5}\approx 0.0556$$
• Junior Managers (Undecided):
  $$\frac{(5-5.0{)}^2}{5.0}=0$$$$\frac{(5-5.0{)}^2}{5.0}=0$$((5-5.0)^(2))/(5.0)=0\frac{(5 – 5.0)^2}{5.0} = 0$$\frac{(5-5.0{)}^2}{5.0}=0$$
• Junior Managers (Disagree):
  $$\frac{(6-5.5{)}^2}{5.5}=\frac{0.25}{5.5}\approx 0.0455$$$$\frac{(6-5.5{)}^2}{5.5}=\frac{0.25}{5.5}\approx 0.0455$$((6-5.5)^(2))/(5.5)=(0.25)/(5.5)~~0.0455\frac{(6 – 5.5)^2}{5.5} = \frac{0.25}{5.5} \approx 0.0455$$\frac{(6-5.5{)}^2}{5.5}=\frac{0.25}{5.5}\approx 0.0455$$
• Senior Managers (Agree):
  $$\frac{(5-4.5{)}^2}{4.5}=\frac{0.25}{4.5}\approx 0.0556$$$$\frac{(5-4.5{)}^2}{4.5}=\frac{0.25}{4.5}\approx 0.0556$$((5-4.5)^(2))/(4.5)=(0.25)/(4.5)~~0.0556\frac{(5 – 4.5)^2}{4.5} = \frac{0.25}{4.5} \approx 0.0556$$\frac{(5-4.5{)}^2}{4.5}=\frac{0.25}{4.5}\approx 0.0556$$
• Senior Managers (Undecided):
  $$\frac{(5-5.0{)}^2}{5.0}=0$$$$\frac{(5-5.0{)}^2}{5.0}=0$$((5-5.0)^(2))/(5.0)=0\frac{(5 – 5.0)^2}{5.0} = 0$$\frac{(5-5.0{)}^2}{5.0}=0$$
• Senior Managers (Disagree):
  $$\frac{(5-5.5{)}^2}{5.5}=\frac{0.25}{5.5}\approx 0.0455$$$$\frac{(5-5.5{)}^2}{5.5}=\frac{0.25}{5.5}\approx 0.0455$$((5-5.5)^(2))/(5.5)=(0.25)/(5.5)~~0.0455\frac{(5 – 5.5)^2}{5.5} = \frac{0.25}{5.5} \approx 0.0455$$\frac{(5-5.5{)}^2}{5.5}=\frac{0.25}{5.5}\approx 0.0455$$

Step 6: Degrees of Freedom

Step 7: P-value and Conclusion

• If the p-value is less than the significance level (typically 0.05), we reject the null hypothesis.
• Since the p-value (0.904) is much larger than 0.05, we fail to reject the null hypothesis.

Conclusion:

Compute independent $t$$t$tt$t$ test for the following data:

Answer:

1. State the Hypotheses:

• Null Hypothesis $H_0$$H_0$H_(0)H_0$H_0$: There is no significant difference between the means of Group 1 and Group 2.
• Alternative Hypothesis $H_1$$H_1$H_(1)H_1$H_1$: There is a significant difference between the means of Group 1 and Group 2.

2. Data Overview:

• Group 1: $[3,3,2,4,6,7,5,4,2,3]$$[3,3,2,4,6,7,5,4,2,3]$[3,3,2,4,6,7,5,4,2,3][3, 3, 2, 4, 6, 7, 5, 4, 2, 3]$[3,3,2,4,6,7,5,4,2,3]$
• Group 2: $[3,4,5,1,2,4,5,3,4,4]$$[3,4,5,1,2,4,5,3,4,4]$[3,4,5,1,2,4,5,3,4,4][3, 4, 5, 1, 2, 4, 5, 3, 4, 4]$[3,4,5,1,2,4,5,3,4,4]$

3. Assumptions:

• Independence of samples: The two groups are independent.
• Normality: The data in each group is approximately normally distributed.
• Equal Variances: The variances of the two groups are equal (which we assume here for the independent $t$$t$tt$t$-test).

4. Formula for the Independent $t$$t$tt$t$-Test:

• ${\overline{X}}_1$${\overline{X}}_1$bar(X)_(1)\bar{X}_1${\overline{X}}_1$ and ${\overline{X}}_2$${\overline{X}}_2$bar(X)_(2)\bar{X}_2${\overline{X}}_2$ are the means of Group 1 and Group 2.
• $s_1^2$$s_1^2$s_(1)^(2)s_1^2$s_1^2$ and $s_2^2$$s_2^2$s_(2)^(2)s_2^2$s_2^2$ are the variances of Group 1 and Group 2.
• $n_1$$n_1$n_(1)n_1$n_1$ and $n_2$$n_2$n_(2)n_2$n_2$ are the sample sizes for Group 1 and Group 2.

5. Step-by-Step Calculation:

a) Calculate the Means:

b) Calculate the Variances:

c) Calculate the Standard Error:

d) Calculate the $t$$t$tt$t$-statistic:

6. Degrees of Freedom (df):