Chapter 3: Problem 50
A student took two national aptitude tests. The mean and standard deviation were 475 and 100 , respectively, fol the first test, and 30 and 8 , respectively, for the second test The student scored 625 on the first test and 45 on the second test. Use \(z\) -scores to determine on which exam the student performed better relative to the other test takers. (Hint: See Example \(3.18 .)\)
Short Answer
Expert verified
The student performed better on the second test relative to the other test takers with a z-score of 1.875 compared to a z-score of 1.5 for the first test.
Step by step solution
01
Identify the given information
For Test 1:
Mean (μ_1) = 475
Standard Deviation (σ_1) = 100
Student's Score (X_1) = 625
For Test 2:
Mean (μ_2) = 30
Standard Deviation (σ_2) = 8
Student's Score (X_2) = 45
02
Calculate the z-score for Test 1
Using the formula mentioned above, we can calculate the z-score for Test 1:
\( z_1 = \frac{(X_1 - \mu_1)}{\sigma_1} \)
Plug in the values:
\( z_1 = \frac{(625 - 475)}{100} \)
\( z_1 = \frac{150}{100} \)
\( z_1 = 1.5 \)
The z-score for Test 1 is 1.5.
03
Calculate the z-score for Test 2
Now let's calculate the z-score for Test 2:
\( z_2 = \frac{(X_2 - \mu_2)}{\sigma_2} \)
Plug in the values:
\( z_2 = \frac{(45 - 30)}{8} \)
\( z_2 = \frac{15}{8} \)
\( z_2 = 1.875 \)
The z-score for Test 2 is 1.875.
04
Compare the z-scores
Now that we have the z-scores for both tests, we can compare them to determine on which exam the student performed better relative to the other test takers:
- Test 1: z-score = 1.5
- Test 2: z-score = 1.875
The student's z-score is higher for Test 2, meaning they performed better on Test 2 relative to the other test takers compared to their performance on Test 1.
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Standard Deviation
Understanding the standard deviation is critical when analyzing data sets like test scores. It represents the amount of variation or dispersion of a set of values. A low standard deviation indicates that the values tend to be close to the mean of the set, while a high standard deviation indicates that the values are spread out over a wider range.
When we say a test has a standard deviation of 100, like in the first test from our example, it signifies wider variability in test scores. Conversely, the second test's standard deviation of 8 denotes much tighter grouping around the mean. Evaluating student performance without considering standard deviation can lead to misleading conclusions, as the context of the scores' distribution is lost.
When we say a test has a standard deviation of 100, like in the first test from our example, it signifies wider variability in test scores. Conversely, the second test's standard deviation of 8 denotes much tighter grouping around the mean. Evaluating student performance without considering standard deviation can lead to misleading conclusions, as the context of the scores' distribution is lost.
Test Performance Evaluation
The z-score is a powerful tool used in test performance evaluation. It indicates how many standard deviations an element is from the mean. A positive z-score suggests a score above the mean, while a negative z-score indicates a score below the mean. The magnitude of the z-score reflects the degree of performance relative to the group.
In our example, to evaluate whether the student performed better on the first or second test, we must look at the z-scores. Higher z-scores represent better performance. This method of comparison is particularly useful when the tests have different scales and means, as it provides a normalized measure to accurately compare results. Therefore, interpreting z-scores enables us to assess student performance across different test metrics effectively.
In our example, to evaluate whether the student performed better on the first or second test, we must look at the z-scores. Higher z-scores represent better performance. This method of comparison is particularly useful when the tests have different scales and means, as it provides a normalized measure to accurately compare results. Therefore, interpreting z-scores enables us to assess student performance across different test metrics effectively.
Mean Scores
Mean scores are the average of all the scores in a distribution. They serve as a measure of central tendency, giving us a reference point to gauge individual scores against the collective performance of the group. For instance, in the scenario given, the first test has a mean of 475, and the second test has a much lower mean of 30.
However, comparing mean scores directly between different tests can be deceptive because they may not account for the scale and variability of the tests. It is the reason why z-scores are a more effective way to compare individual results—they adjust for these differences by incorporating the mean and standard deviation into their calculation, creating a standardized score that can be compared across different datasets.
However, comparing mean scores directly between different tests can be deceptive because they may not account for the scale and variability of the tests. It is the reason why z-scores are a more effective way to compare individual results—they adjust for these differences by incorporating the mean and standard deviation into their calculation, creating a standardized score that can be compared across different datasets.
