A z score is typically analyzed when population mean (µ) and population standard deviation (σ) are known. However, in SPSS, we can still calculate z scores with the grades.sav data using the sample mean (M) and sample standard deviation (s). To do this, open grades.sav in SPSS. On the Analyze menu, point to Descriptive Statistics, and then click Descriptives…
You will be calculating and interpreting z scores for the total variable. In the Descriptives dialog box, move the total variable into the Variable(s) box. Select the Save standardized values as variables option and click OK.
SPSS provides descriptive statistics for total in the Output window. SPSS also creates a new variable in the far right column, labeled Ztotal, in the Data Editor area. Ztotal provides a z score for each case on the total variable. You are now prepared to answer the following Section 1 questions.
Question 1
[What is the sample mean (M) and sample standard deviation (s) for total? You will use these values in Question 2 below.
Mean = 100.06
Std. Deviation = 14.19]
Question 2
A z-score for this sample is calculated as [(X– M) ÷ s]. Locate Case #53’s unstandardized total score (X)in the Data Editor. In the formula below, replace X, M, s, to show how the z score in Ztotal is derived for Case #53.
[ (75-100.06) + 14.190] = -10.87
(X – M) ÷ s = -10.87
Question 3
Run Descriptives on Ztotal. What are the mean and standard deviation of Ztotal? (Hint: “0E7” in SPSS is scientific notation for 0). Are the mean and standard deviation what you would expect? Justify your answer.
[Its standard deviation is equal to 1.0 and its mean is equal to 0.0.
Yes, that is what I expected because it is standardized normal variate; therefore, it should have a mean equal to zero and a standard deviation equal to 1.]
Question 4
Case number 6 has a Ztotal score of 1.19. What does a z value of 1.19 represent?
[It means that when we standardize the normal variate then the probability is 0.4192.The observed z value is 1.19. A z value greater than zero shows the total deviations that some element is from the mean. In the given case it is 0.004192%.]
Question 5
Identify the case with the lowest z score. Refer to Appendix A in the Warner (2013) text. Interpret the percentile rank of this z score rounded to whole numbers.
[The lowest z score is of case # 66 which is -3.24007. So according to attached appendix it shows a probability of 0.5006].
Question 6
Identify the case with the highest z score. Refer to Appendix A in the Warner (2013) text. Interpret the percentile rank of this z score rounded to whole numbers.
[The highest z score is of case # 10 which is -1.53133. So according to attached appendix it shows a probability of 0.4370].
