Statistics - Ch5 z Scores and standardized distributions

Raw score

original, unchanged datum that is the direct result of measurement

The distribution for scores on a history exam has µ = 70 and σ = 4. The distribution for a set of scores on a sociology exam has µ = 68 and σ = 4. Tenisha scored 76 on both exams. Which of the following statements best describes the grades Tenisha will receive on her exams? a. She will receive a higher grade on the sociology exam based on standardization. b. She will receive the same grades on both exams because her raw scores are the same. c. She will receive the same grades on both exams because σ is the same for both distributions. d. She will receive a higher grade on the history exam based on standardization.

a. She will receive a higher grade on the sociology exam based on standardization. Standardizing both distributions by converting both exam scores to z-scores, Tenisha's sociology score is z = +2.0 and her history score is z = +1.5. See 5.4: Using z-Scores to Standardize a Distribution.

What is the benefit of converting raw scores from a sample into z-scores? a. The new scores are easier to compare. b. The new scores are all relative to 100. c. The new scores are easier to average. d. The new scores are all higher than zero.

a. The new scores are easier to compare. Comparing z-scores relative to the mean is easier than comparing raw scores relative to the mean. Standardizing sample scores offers the same benefits as standardizing scores for a full population. See 5.4: Using z-Scores to Standardize a Distribution.

A distribution with µ = 61 and σ = 6 has been standardized to reflect a new mean of µ = 50 and a standard deviation of σ = 10. What is the new standardized score for a score of X = 52 in the original distribution? a. X = 35 b. X = 65 c. X = 40 d. X = 70

a. X = 35 For the original distribution, a score of 52 has a deviation score of −9 (X − µ = 52 - 61 = −9), and with σ = 6 produces a z-score of −1.50. In the new distribution, 1.5 × σ = 1.5 × 10 = 15. A score 15 points below the standardized mean of 50 is 35. See 5.5: Other Standardized Distributions Based on z-Scores.

Which of the following z-scores indicates an X value that falls farthest below the mean for the distribution? a. z = −1.50 b. z = −0.50 c. z = +2.00 d. z = +1.25

a. z = −1.50 The sign indicates whether a z-score is above or below the mean, and the number indicates the distance of the z-score from the mean. See 5.2: z-Scores and Locations in a Distribution.

A researcher has compiled a set of scores with a mean of µ = 39 and standard deviation of σ = 4. What is the z-score for the raw score X = 28? a. z = −2.75 b. z = +2.75 c. z = +0.60 d. z = −0.50

a. z = −2.75 X - µ produces a deviation score of −11 (28 - 39 = −11); the negative result indicates a negative z-score. Dividing the deviation score by σ = 4 provides a z-score of −2.75 (−11/4 = −2.75). See 5.2: z-Scores and Locations in a Distribution.

A test of life satisfaction has a mean of µ = 60. Jenna's test score of X = 72 has a z-score of +3.0. What is the standard deviation for this distribution of test scores? a. σ = 4 b. σ = 3 c. σ = 6 d. σ = 2

a. σ = 4 It is easy to see that Jenna's test score is 12 points above the mean. A z-score of +3.0 indicates that her score is three standard deviations away from the mean. Dividing 12 by 3 produces a standard deviation of 4. See 5.3: Other Relationships Between z, X, the Mean, and the Standard Deviation.

When Lorenzo finished his exam last week, he thought the test was over. But the instructor put z-scores on each student's paper and asked them to figure out their original score. The mean for the class is µ = 61 and standard deviation is σ = 8. Lorenzo's z-score is +1.75. What did he score on the exam? a. 47 b. 75 c. 69 d. 53

b. 75 The formula to determine a raw X score from the z-score is X = µ + zσ. Filling in the data, the equation is 61 + (1.75 × 8) = 75. See 5.2: z-Scores and Locations in a Distribution.

As part of a clinical research project, a sample population has been using an experimental drug intended to improve memory. Study participants completed a comprehensive memory test following 6 weeks of treatment. The results were standardized and compared to mean results for the full population. Which of the following participants indicates promising results for this drug treatment? a. Javier, a 48-year-old man with a memory test z-score of −0.25 b. Ellie, a 52-year-old woman with a memory test z-score of +2.75 c. Liming, a 34-year-old woman with a memory test z-score of +1.25 d. Ahmad, a 61-year-old man with a memory test z-score of +0.50

b. Ellie, a 52-year-old woman with a memory test z-score of +2.75 With a z-score almost 3 standard deviations above the mean for the general population, Ellie's test results are significant. The lower scores for male participants may indicate the drug is more effective on females. See 5.5: Other Standardized Distributions Based on z-Scores.

Professor Chao's engineering final exam has µ = 79 and σ = 7. After standardizing the exam scores, µ = 80 and σ = 5. How did Professor Chao arrive at these new standardized scores for the exam? a. factoring in how many students took the exam b. deciding what values would be simple to calculate c. selecting the nearest values that are multiples d. comparing z-scores across all exams in the college

b. deciding what values would be simple to calculate The new mean and standard deviation were predetermined by the professor to simplify working with the data by changing scores to whole round numbers after standardization. These values are arbitrary and simply selected for ease of use. See 5.5: Other Standardized Distributions Based on z-Scores.

What is necessary to determine the z-score for a raw score in a particular set of scores? a. the highest, lowest, and mean scores for the set b. the mean and standard deviation for the set c. the median X score and the mean for the set d. the number of scores and standard deviation for the set

b. the mean and standard deviation for the set A z-score uses the mean and standard deviation to determine the exact location of a raw score within the full set. See 5.1: Introduction.

A distribution of exam scores has a mean of µ = 52 and standard deviation of σ = 6. What is the z-score for the exam score X = 61? a. z = −1.5 b. z = +1.5 c. z = +2.0 d. z = +0.50

b. z = +1.5 X - µ produces a deviation score of 9 (61 - 52 = 9); the positive result indicates a positive z-score. Dividing the deviation score by σ = 6 provides a z-score of 1.5 (9/6 = 1.5). See 5.2: z-Scores and Locations in a Distribution.

A distribution has µ = 50 and σ = 6. In a graphical representation of the distribution, where would the score X = 20 appear? a. just left of the peak in the curve b. just right of the peak in the curve c. in the left tail of the curve d. in the right tail of the curve

c. in the left tail of the curve The deviation score for the raw score of 20 is −30 (X − µ = 20 - 50 = −30). Dividing the deviation score by the standard deviation σ = 6 provides a z-score of −5.0. At five standard deviations from the mean, a raw score of 20 will appear in the left tail of the graph. See 5.3: Other Relationships Between z, X, the Mean, and the Standard Deviation.

A calculus exam has a mean of µ = 73 and a standard deviation of σ = 4. Trina's score on the exam was 79, giving her a z-score of +1.50. The teacher standardized the exam distribution to a new mean of µ = 70 and standard deviation of σ = 5. What is Trina's z-score for the standardized distribution of the calculus exam? a. z = +2.25 b. z = +0.50 c. z = +1.50 d. z = +0.75

c. z = +1.50 After scores in a distribution are standardized, the revised z-scores are always the same as the original z-scores. See 5.5: Other Standardized Distributions Based on z-Scores.

Within a population having standard deviation of σ = 20, the raw score X = 75 has a z-score of −1.25. What is the mean for this population? a. µ = 90 b. µ = 50 c. µ = 100 d. µ = 125

c. µ = 100 The negative z-score indicates that raw score X = 75 is lower than the mean. Multiplying the z-score by the standard deviation (z × σ) produces a deviation score of 25 (1.25 × 20 = 25). If a score of 75 is 25 points below the mean, then µ = 100. See 5.3: Other Relationships Between z, X, the Mean, and the Standard Deviation.

A distribution of 1,000 test scores has a mean of µ = 287. Following standardization, what is the z-score for the standard deviation for this distribution? a. σ = 25 b. σ =14 c. σ = 1 d. Not enough data to calculate

c. σ = 1 Regardless of the original distribution, after standardizing to z-scores, the standard deviation is always 1. See 5.4: Using z-Scores to Standardize a Distribution.

standardized distribution

composition of data used to make dissimilar distributions comparable

Marisol received a score of X = 70 on her chemistry exam and a score of X = 59 on her geography exam. The mean for the chemistry exam is µ = 66, and the mean for the geography exam is µ = 50. On which exam will Marisol receive a higher grade? a. Geography; this score is higher above the mean than her chemistry score. b. Chemistry; her exam score of 70 is higher than the geography score of 59. c. Unknown; the total number of raw scores for each exam is required. d. Unknown; the standard deviation for each set of exam scores is required.

d. Unknown; the standard deviation for each set of exam scores is required. To correctly assess how a score compares to all others, the mean and the standard deviation are required. See 5.1: Introduction.

For a population with µ = 69 and σ = 4, which of the following X scores will convert to a positive z-score of magnitude greater than 1.0? a. X = 65 b. X = 72 c. X = 62 d. X = 74

d. X = 74 With a mean of µ = 69, any score above 69 will have a positive z-score. The raw score of 74 is 5 points higher than the mean. With a standard deviation of σ = 4, this score produces a z-score of magnitude greater than 1. See 5.2: z-Scores and Locations in a Distribution.

A sample has a mean of µ = 64 and standard deviation of σ = 3. What is the z-score for a sample score of X = 70? a. z = +1.00 b. z = −0.75 c. z = +1.25 d. z = +2.00

d. z = +2.00 Determining the z-score from a sample follows the same process as for a population. Calculate the deviation score by subtracting the mean from the raw score (X - µ = 70 - 64 = 6). Divide the deviation score by the sample standard deviation (6/3 = 2). The positive deviation value produces a positive z-score of +2.00. See 5.2: z-Scores and Locations in a Distribution.

A distribution of scores for a test of life stressors has a mean of µ = 125 and standard deviation of σ = 15. The researcher calculates z-scores to standardize the distribution. What are the mean and the standard deviation for the z-scores in this distribution? a. µ = 100, σ = 25 b. µ = 50, σ = 1 c. µ = 0, σ = 10 d. µ = 0, σ = 1

d. µ = 0, σ = 1 Regardless of the values for the original distribution, after standardization the mean for z-scores is always zero and the standard deviation is always 1. See 5.4: Using z-Scores to Standardize a Distribution.

In a distribution, a score of X = 60 has a z-score of −3.0. A score of X = 85 has a z-score of +2.0. What are the mean and the standard deviation for this population? a. µ = 50, σ = 10 b. µ = 70, σ = 2 c. µ = 68, σ = 4 d. µ = 75, σ = 5

d. µ = 75, σ = 5 The total distance between the two raw scores is 25 points (85 - 60 = 25). The total number of standard deviations in this distance is 5 (3 + 2 = 5). Dividing 25 by 5 produces one standard deviation of σ = 5. If raw score X = 60 is 3 standard deviations below the mean, then counting up 3 × 5 (15 points) from 60 produces a mean of µ = 75. See 5.3: Other Relationships Between z, X, the Mean, and the Standard Deviation.

X = µ + zσ

equation used to transform z-scores back into X values, for a population

X = M + zs

equation used to transform z-scores back into X values, for a sample

z = (X-µ)/σ

formula used to transform X values into z-scores, for a population

z = (X-M)/s

formula used to transform X values into z-scores, for a sample

z score transformation

relabeling of X values in a population into precise X-value locations within a distribution

standardized score

result from relabeling data into new table with positive, whole-number predetermined mean and standard deviation

z score

specification of the precise location of each X value within a distribution