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 standardi zed 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].

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