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The College Board reported that \(60.3 \%\) of all students who took the 2016 AP Statistics exam earned scores of 3 or higher. One teacher wondered if the performance of her school was better. She believed that year's students to be typical of those who will take AP Stats at that school and was pleased when 34 of her 54 students achieved scores of 3 or better. a. How many standard deviations above the national rate did her students score? Does that seem like a lot? Explain. b. Can she claim that her school is better? Explain.

Short Answer

Expert verified
The students from the teachers class performed 0.405 standard deviations above the national average which is a moderate performance. Despite these results, the teacher cannot conclusively claim that her school is better based on this single data point as additional data and statistical significance analysis is required.

Step by step solution

01

Calculate z-score

Firstly, find out the proportion of students that scored 3 or better in her class. This can be done by dividing the number of students that scored 3 or over by the total number of students which is \( \frac{34}{54} = 0.629 \). The z-score is then calculated as follows: \[Z = \frac{P_s - P_n}{\sqrt{\frac{P_n(1-P_n)}{N}}}\]where \(P_s\) is the school's proportion, \(P_n\) is the national proportion, and \(N\) is the total number of students. This gives us: \(Z = \frac{0.629 - 0.603}{\sqrt{\frac{(0.603)(1-0.603)}{54}}} = 0.405\]
02

Evaluating the z-score

The calculated z-score says the teacher's students are 0.405 standard deviations above the national average. Without comparing this to a benchmark it's hard to say whether this is a lot or not. However, in general terms, a z-score between -1 and 1, as is this case, is considered to be within normal range meaning the score is not dramatically different from the national average.
03

Claiming the school is better

Based on the z score calculated, the teacher's class did perform slightly better than the national average. However, to make a statistically significant claim that the school performs better than the typical school, we require additional data and a larger sample size. Therefore, the teacher can't make a definitive claim her school is better based solely on this data, but it is an encouraging result.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Z-score
A Z-score, often referred to as the standard score, is a numerical measurement that expresses the value's relationship to the mean of a group of values. In simple terms, it tells us how many standard deviations a data point is from the mean. The formula for calculating a Z-score is:
  • Subtract the mean from the individual data point.
  • Divide the result by the standard deviation.
In the context of the exercise, a Z-score of 0.405 indicates that the performance of the students in the teacher's class is 0.405 standard deviations above the national average of 60.3%.
While it shows a slight improvement, it's important to note that a Z-score between -1 and 1 is typically considered as within the normal range, indicating that the class performance is not significantly different from the average.
This calculation is essential in statistics for understanding how extreme or typical a value is in a data set.
Clarifying Proportion
Proportion is a simple concept representing a part of a whole expressed as a fraction, percentage, or ratio. To find the proportion of students who achieved a score of 3 or higher, we calculate the ratio of students who scored 3 or above to the total number of students.
In this exercise, 34 out of 54 students scored 3 or better, giving us a proportion of 0.629.
This proportion serves as the basis for calculating the Z-score by comparing it to the national proportion or average of students who scored 3 or higher, which is 0.603, as given in the problem.
Understanding and calculating proportions are critical in making comparisons and evaluating differences in statistics.
Explaining Statistical Significance
Statistical significance is a determination of whether an observed effect is likely due to an actual difference rather than random chance. In simpler terms, it helps us understand if the difference in proportions or averages is meaningful.
  • To claim statistical significance, we often look at the p-value resulting from a statistical test, such as a Z-test or T-test.
  • A result is typically considered statistically significant if the p-value is below a certain threshold, often 0.05.
In the context of the teacher's classroom, even though the students performed slightly better than the national average, the Z-score of 0.405 is not large enough to make a statistically significant claim.
This means that the observed difference might be due to random variation rather than an indication of a genuine, superior performance.
Introduction to Standard Deviation
Standard deviation is a measure that quantifies the amount of variation or dispersion in a set of data values.
In practical terms, it measures how spread out the values in a data set are around the mean. A low standard deviation means that most of the numbers are close to the mean. Conversely, a high standard deviation means the numbers are more spread out.
  • In the provided exercise, standard deviation plays a crucial role in calculating the Z-score.
  • It functions as the denominator in the Z-score formula, helping to normalize different data points for comparison.
  • Understanding standard deviation is vital for interpreting statistical results, as it provides insights into the consistency of data.
Knowing standard deviation helps in making informed decisions in data analysis and interpreting how representative a data set is of actual trends.

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