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Exploring Two-Paired Samples T-Tests: Analysis and Application in Real-World Scenarios

August 31, 2023
Vern Branson
Vern Branson
🇺🇸 United States
SPSS
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Key Topics
  • Problem Description:
    • Part I: General Concepts
    • Part II: SPSS Application
    • Part III: Cumulative
  • Conclusion
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This comprehensive assignment provides students with an in-depth understanding of the fundamental concepts and practical applications of two-paired samples t-tests. It covers essential topics such as counterbalancing, statistical power, and effect sizes. Through real-world scenarios, students learn to analyze data related to exercise's impact on anxiety, assess coping abilities, and understand the differences between in-person and virtual meetings. Practical exercises in SPSS, graph creation, and APA-style reporting further enhance students' proficiency in using these statistical tools to derive meaningful insights from data.

Problem Description:

This assignment focuses on Two-Paired Samples T-Tests, exploring their principles and practical applications. It covers concepts like counterbalancing and statistical power, delves into real-world scenarios analyzing exercise's impact on anxiety, individuals' coping abilities, and the effect of in-person versus virtual meetings on information retention. Students perform calculations, use SPSS for data analysis, create graphs, and report results in APA-style. The assignment enhances understanding of statistical tools and their real-world applications.

Part I: General Concepts

Overview: This section provides a deep understanding of the foundational concepts of two-paired samples t-tests.

  1. Counterbalancing: Counterbalancing is a meticulous technique that involves randomizing the order in which participants receive interventions. This minimizes the potential bias stemming from the order of interventions, enhancing data quality and mitigating the influence of extraneous factors such as practice or fatigue.
  • Question 1:Explain what counterbalancing and how to achieve it.

Answer: Counterbalancing involves randomizing the order in which participants will receive a given intervention, so that the order the interventions are received will be identical amongst all participants. It minimizes the influence of extraneous factors, such as practice or fatigue, on experimental results.

  • Statistical Power: In comparison to independent-groups t-tests, correlated-groups t-tests exhibit superior statistical power. The latter leverages related groups, which can significantly reduce variations between samples, thereby making it a more powerful statistical tool.
    • Question 2:Discuss how correlated-groups t-tests have more statistical power in comparison to an independent-groups t test.

    Answer: Correlated-groups t test has more statistical power than independent-groups t test because while Correlated-groups t test use the same participants or related groups and this reduces, if not eliminates, completely the variation between the samples as compared to independent-groups t test where groups are independent.

  • Strengths and Weaknesses of Within-Subjects Design: Within-subjects designs present distinct advantages and limitations. One key strength is their efficiency in terms of sample size as they require fewer participants. However, one notable weakness is the time factor, as the collection of data takes longer due to the administration of multiple treatment conditions.
    • • Question 3: Discuss one strength and one weakness of a within-subjects design using your own words.

    Answer:One strength of a within-subjects design is sample size. The design does not require a large sample size since its subjects or participants provide repeated measures for each condition. One weakness of a within-subjects design is time. It takes a longer time to gather data since more than one treatment condition is given to one participant, as opposed to between-subjects designs.

    Scenario: A research study investigates whether exercise has an impact on anxiety in women. Thirty women are subjected to an anxiety assessment before and after six months of exercise. The mean difference in anxiety scores is 3.4, with a standard deviation of 1.8.

  • Correlated-Groups t-test:Utilizing the provided data, we calculate a t-value of 10.346, indicating a statistically significant difference in anxiety levels.
    • Question 4: Calculate the correlated-groups t test to determine the t_obs.

    Work: t_obs = x̅_d / (S_D⁄√n) = (3.4) / ((1.8)⁄√30) = 10.34587

    Answer: t obtained: 10.34587

  • Effect Size (Cohen's d):The effect size is 1.89, signifying a very large effect based on conventions.
    • Question 5:Calculate the effect size using Cohen’s d.

    Work: d = x̅_d / S_D = (3.4) / (1.8) = 1.89

    Answer: 1.89

  • Directionality and Critical Value: This is a non-directional study, and the t critical value is determined as 2.0452.
    • Question 6: Is this a directional or non-directional study, and what is the t_cv?

    Work: It is a non-directional study. df = n-1 = 30-1 = 29 t_cv = t_(1-α/2, df) = t_(1-(0.05)/2, 29) = t_(0.975, 29) = 2.0452

    Answer: 2.0452

  • 95% Confidence Intervals: Employing the two-tailed t_cv, we establish a 95% confidence interval of (2.727879, 4.072121).
    • Question 7:Calculate the 95% confidence intervals (make sure you use the two-tailed t_cv).

    Work: 95% C.I. = n x̅_d ± t_CV * S_D⁄√n = 3.4 ± 2.0452 * 0.3286335 = (2.727879, 4.072121)

    Answer: (2.727879, 4.072121)

  • APA-Style Results: The results indicate a significant impact of exercise on anxiety in women (t(28) = 10.346, p = 0.0000, d = 1.88, 95% CI [2.7279, 4.072121]).
    • Question 8:Write an APA-style Results section based on your analyses.

    Answer: There was a significant effect of exercise on anxiety in women t(28) = 10.346, p = 0.0000 d=1.88, 95% CI [2.7279, 4.072121].

    Part II: SPSS Application

    Overview: This segment involves applying SPSS for analyzing data related to people's self-rating and peer rating of their coping abilities in challenging situations.

    Scenario: Participants rated their mood in response to a negative event and imagined how the same event would affect a peer.

  • Correlated-Groups t-test:SPSS output confirms a significant difference in self-rating and peer rating (t(11) = 4.180, p = 0.002).
    • Question 9: Conduct a correlated groups t-test. Paste the appropriate SPSS output below.

    Answer:

    Paired Samples Statistics
    MeanNStd. DeviationStd. Error Mean
    Pair 1Self_rating5.58121.929
    Peer_rating4.83121.697
    Paired Samples Test
    MeanStd. DeviationStd. Error Mean95% Confidence Interval of the Difference
    Lower
    Pair 1Self_rating - Peer_rating.750.622.179
    • Table 1:Paired Samples statistics

  • Effect Size (r^2): The effect size is calculated as 0.61366, indicating a medium effect, following conventions for effect size interpretation.
    • Question 10:Use the output to calculate r^2 (show your work in the space provided and remember if a number is negative, when squaring it will lose its sign and be positive).

    Work: r^2 = t^2 / (t^2 + df) = 〖4.180〗^2 / (〖4.180〗^2 + 11) = 0.61366

    Answer: 0.61366 - This implies a medium effect size.

  • SPSS Graph:An appropriate graph is created and included in the document.
    • Question 11:Paste an appropriate SPSS graph (make sure you use the procedures outlined in this module’s SPSS video– and don’t forget to label your y axis “Coping Ability”.

    Part III: Cumulative

    Overview:In this part, you'll analyze data concerning information retention in meetings, comparing in-person and virtual attendance.

    Scenario: Participants were given the choice to attend quarterly meetings either in person or via Google Meet. Afterward, they underwent an assessment of information retention.

  • Independent and Dependent Variables:In this scenario, the independent variable is the mode of attendance (in person or virtual), and the dependent variable is the amount of information retained.
    • Question 12:Identify the independent and dependent variables.

    Answer: The independent variable in the study is the mode of attendance retained (either in person or virtually), and the dependent variable is the amount of information.

  • Graph:An appropriate graph, complete with axes labels and error bars, is created.
    • Question 13:Create an appropriate graph based on the data and paste it below (make sure to have axes labels and error bars).

  • APA-Style Results:There is a statistically significant difference in information retention between in-person (M=74, SD=7.90218) and virtual (M=82.8, SD=7.88529) meetings (t(18) = -2.493, p = 0.023, d = 0.26, 95% CI [-16.21665, -1.38335]). As the p-value is below the significance level (0.05), the null hypothesis is rejected.
    • Question 14:Write an APA-style Results section based on your analyses.

    Answer: There was a statistically significant difference in the amount of information retained for In-person (M=74, SD=7.90218) and Virtual (M=82.8, SD=7.88529). t(18)= -2.493, p=0.023, d=0.26, 95% CI [-16.21665, -1.38335]. We reject the null hypothesis since p-value is less than the significance level (0.05).

  • Internal Validity Threat:In this research scenario, one potential threat to internal validity is selection bias. As participants chose their mode of attendance, the study lacked random assignment, potentially introducing bias into the results.
    • Question 15: Discuss one threat to internal validity as it specifically relates to this research scenario.

    Answer: In relation to this research scenario, Selection is a threat to internal validity, since participants got to choose whether they attend a quarterly meeting in person or via Google Meet, there was already a selection bias as random assignment was not done.

    Use this information to answer the following questions: Age at onset of dementia was determined in the general population to be µ = 70 and σ = 7.0.

  • Based on the data above:
    • Question 16: What is the z score for someone being diagnosed with dementia at age 65 (can round to two decimal places)?
  • Work: Z = (x - µ) / σ = (65 - 70) / 7 = -0.7142857

    P(Z < -0.7142857) = 0.2375

    23.75% of people might start to show signs of dementia before 65.

    Answer:

    • Z score = -0.7142857
    • Percentage of people with dementia = 23.75%

    Conclusion

    In conclusion, this assignment on Two-Paired Samples T-Tests equips students with the knowledge and practical skills necessary to harness the power of statistical analysis in real-world scenarios. From understanding the intricacies of counterbalancing to conducting SPSS analyses, the journey through these concepts is both enlightening and essential. By applying these techniques to scenarios like exercise's influence on anxiety or information retention in meetings, students gain a deeper appreciation for the practical applications of statistics. This assignment serves as a stepping stone towards becoming proficient data analysts and informed consumers of research.

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