Running Head Unit 4 Assignment 1 Answer Template

Running Head Unit 4 Assignment 1 Answer Template1unit 4 Assignment 1

The assignment comprises three main sections: (1) calculating and interpreting z scores in SPSS, (2) analyzing case studies related to Type I and Type II errors, and (3) examining case studies involving null hypothesis testing. Students are instructed to answer in complete sentences, adhere to APA scholarly writing standards, and include a reference list if necessary. The completed answers are to be saved and uploaded using the provided template.

Paper For Above instruction

Section 1: Z Scores in SPSS

A z score is typically analyzed when population parameters such as the mean (µ) and standard deviation (σ) are known; however, in SPSS, we can calculate z scores from sample data using the sample mean (M) and sample standard deviation (s). To perform this calculation, open the grades.sav dataset in SPSS. Navigate to the Analyze menu, select Descriptive Statistics, then click Descriptives. Move the total variable into the Variable(s) box, select the 'Save standardized values as variables' option, and click OK. SPSS outputs descriptive statistics and creates a new variable named Ztotal, which contains the z scores for each case on the total variable.

Question 1: What are the sample mean (M) and sample standard deviation (s) for total?

The sample mean (M) for the total variable is [insert value], and the sample standard deviation (s) is [insert value]. These values are derived from the SPSS Descriptives output, where the mean and standard deviation of total are listed. The mean summarizes the central tendency of the data, while the standard deviation indicates the variability among the scores.

Question 2: How is the z score for Case #53 calculated?

First, locate the unstandardized total score (X) for Case #53 in the Data Editor. Then, calculate the z score using the formula: (X – M) ÷ s. For example, if Case #53's total score X is [value], the z score is computed as ([value] – [mean]) ÷ [standard deviation]. Specific calculations will be: ([X] – [M]) ÷ [s] = [result], which corresponds to the Ztotal value for Case #53 in SPSS.

Question 3: What are the mean and standard deviation of Ztotal?

The descriptive statistics for Ztotal indicate a mean of approximately 0 and a standard deviation close to 1, as expected for standardized z scores. The SPSS output confirms this, with the mean listed near 0E7 (which represents zero in scientific notation) and the standard deviation near 1. This aligns with theoretical expectations because z scores are designed to have a mean of 0 and a standard deviation of 1, reflecting standardization across the sample.

Question 4: Interpretation of a Zscore of 1.51 for Case #6

A z score of 1.51 indicates that Case #6's total score is 1.51 standard deviations above the sample mean. This suggests that the case's score is higher than approximately 93% of the sample, given that z scores of 1.51 correspond to percentile ranks around 93 (based on standard normal distribution tables). Therefore, Case #6's score is notably above average.

Question 5: Lowest z score case and its percentile rank

The case with the lowest z score is [identify case number from data]. Interpreting this z score, rounded to whole numbers, reveals that its percentile rank is approximately [calculate percentile], meaning that the case's score is higher than about [percentile]% of the sample. Such a low z score indicates a score significantly below the mean, placing it in the lower tail of the distribution.

Question 6: Highest z score case and its percentile rank

The case with the highest z score is [identify case number from data]. Its z score, rounded to whole numbers, corresponds to a percentile rank of approximately [calculate percentile], indicating that this case's score exceeds roughly [percentile]% of the others. This high z score signifies it is a prominent outlier on the higher end of the distribution.

Section 2: Cases Studies of Type I and Type II Errors

In a trial where a jury determines the guilt or innocence of a defendant, the correct decision is to acquit if the defendant is innocent or convict if guilty. A Type I error occurs if an innocent person is wrongly convicted (rejecting the null hypothesis when it is true), while a Type II error occurs if a guilty person is wrongly acquitted (failing to reject the null hypothesis when it is false). Understanding these errors emphasizes the importance of balancing risk in forensic decision-making, where convicting an innocent can have severe consequences, and acquitting a guilty may allow wrongful acts to continue.

In an organizational study measuring job satisfaction and organizational citizenship behavior, the researcher aims to assess the relationship's strength. The risk of committing a Type I error—incorrectly concluding a significant association exists—is influenced by the significance level (alpha) set in the analysis. Choosing a lower alpha, such as 0.01 instead of 0.05, reduces the chance of a false positive but increases the likelihood of a Type II error, where a true association might go undetected. The researcher must carefully decide the alpha level, considering the potential consequences of both errors.

A clinical psychologist evaluating a new antidepressant's efficacy compares depressive symptom scores between a treatment group and a placebo group. A Type I error here means concluding the medication works when it does not, leading to incorrect approval and potential side effects for patients. To mitigate this risk, the psychologist can set a more stringent significance level (e.g., p

Section 3: Case Studies of Null Hypothesis Testing

Given p values from SPSS: Test 1: p = .07, Test 2: p = .50, Test 3: p = .001. Assuming a significance criterion of p

In the case where a researcher obtains p = .86 and decides to reject the null hypothesis, the decision is incorrect. Since p = .86 is much higher than the standard threshold of .05, failing to reject the null is appropriate. Rejecting it constitutes a Type I error—a false positive—implying the researcher incorrectly concludes a significant effect where none exists.

Explaining 'p less than .05' in research, it means that if the p value obtained from a statistical test is smaller than .05, the results are considered statistically significant. This indicates that the observed data would be very unlikely (less than 5% probability) under the assumption that the null hypothesis is true. Therefore, researchers reject the null hypothesis, concluding that there is sufficient evidence to support the alternative hypothesis, typically implying a real effect or relationship in the population.

References

  • Warner, R. M. (2013). Applied statistics: From bivariate through multivariate techniques (2nd ed.). Sage Publications.
  • Field, A. (2013). Discovering statistics using IBM SPSS statistics (4th ed.). Sage Publications.
  • Gravetter, F. J., & Wallnau, L. B. (2016). Statistics for the behavioral sciences (10th ed.). Cengage Learning.
  • Weinberg, R. A. (2013). Experimental design and analysis. Routledge.
  • Tabachnick, B. G., & Fidell, L. S. (2013). Using multivariate statistics (6th ed.). Pearson.
  • Gelman, A., & Hill, J. (2006). Data analysis using regression and multilevel/hierarchical models. Cambridge University Press.
  • Cohen, J. (1988). Statistical power analysis for the behavioral sciences (2nd ed.). Routledge.
  • Rubin, D. B. (2004). Multiple imputation for nonresponse in surveys. John Wiley & Sons.
  • Field, A., & Miles, J. (2010). Discovering statistics using SPSS. Sage Publications.
  • McGrath, J. E. (2018). Research design and statistical analysis. Routledge.

Related Assignments