Given The Following Students' Test Scores: 95, 92, 90, 90, 8
Given The Following Students Test Scores 95 92 90 90 83 83 83
Given the following students' test scores (95, 92, 90, 90, 83, 83, 83, 74, 60, and 50), identify the mean, median, mode, range, variance, and standard deviation for the sample. Write a 1000-word summary and analysis discussing the results of your calculations. State your results for the sample: the mean, median, mode, range, variance, and standard deviation. Explain which method is best for this data set and why. Conduct a one-sample t-test and interpret the results (using a population mean of 70). In what situations would this information be useful? Prepare this assignment according to the guidelines found in the APA Style Guide, located in the Student Success Center. An abstract is not required. Must pass Turnitin by being less than 20%.
Paper For Above instruction
Introduction
Understanding descriptive and inferential statistics is essential in analyzing educational data, such as student test scores. This paper aims to evaluate a specific set of test scores through various statistical measures—mean, median, mode, range, variance, and standard deviation—and determine the most appropriate measure for this data. Additionally, a one-sample t-test will be performed to compare the sample mean against a hypothetical population mean of 70, providing insights into the data's significance. Such statistical analysis is crucial in educational assessments, curriculum development, and policy-making, as it helps educators interpret student performance and make informed decisions.
Descriptive Statistics Calculation
The provided scores are: 95, 92, 90, 90, 83, 83, 83, 74, 60, and 50. Calculating the mean involves summing all scores and dividing by the total number of observations. The median is the middle value when scores are ordered. The mode is the score(s) that appear most frequently. The range is the difference between the highest and lowest scores. Variance measures the average squared deviations from the mean, and the standard deviation is the square root of variance.
The mean score is calculated as:
\[
\text{Mean} = \frac{95 + 92 + 90 + 90 + 83 + 83 + 83 + 74 + 60 + 50}{10} = \frac{850}{10} = 85
\]
To find the median, the scores are ordered:
50, 60, 74, 83, 83, 83, 90, 90, 92, 95
Since there are ten scores, the median is the average of the 5th and 6th scores:
\[
\text{Median} = \frac{83 + 83}{2} = 83
\]
The mode is 83, as it appears three times, more than any other score. The range is:
\[
95 - 50 = 45
\]
Calculating variance involves finding the squared deviations from the mean for each score, summing them, and dividing by \(n-1\) for sample variance:
\[
\text{Variance} = \frac{\sum (x - \bar{x})^2}{n-1}
\]
The squared deviations are:
- (95 - 85)^2 = 100
- (92 - 85)^2 = 49
- (90 - 85)^2 = 25
- (90 - 85)^2 = 25
- (83 - 85)^2 = 4
- (83 - 85)^2 = 4
- (83 - 85)^2 = 4
- (74 - 85)^2 = 121
- (60 - 85)^2 = 625
- (50 - 85)^2 = 1225
Sum: 100 + 49 + 25 + 25 + 4 + 4 + 4 + 121 + 625 + 1225 = 2206
Variance:
\[
\frac{2206}{10 - 1} = \frac{2206}{9} \approx 245.11
\]
Standard deviation:
\[
\sqrt{245.11} \approx 15.65
\]
Analysis of Measures
The mean score of 85 suggests a high overall student performance; however, the median of 83 indicates that half the students scored below and half above 83. The mode being 83 highlights that this score is the most common, reflecting a potential concentration around this point. The range of 45 points reveals considerable variability, especially with the lowest score at 50 and the highest at 95.
When considering measures of central tendency, the mean is sensitive to outliers; in this case, the lowest score of 50 significantly affects the average. The median provides a better center point for the data, especially with outliers present. The mode emphasizes the most frequently achieved score, which could indicate a common achievement level among students.
The variance and standard deviation quantify the spread of scores; a standard deviation of approximately 15.65 points indicates moderate dispersion around the mean. Given the data's distribution, the median and mode are more representative of typical scores than the mean due to the presence of an outlier at 50.
Best Method for this Data Set
Considering the distribution, the median is the most appropriate measure of central tendency because it is less affected by the outlier at 50 and provides a more accurate reflection of a typical student's score. The mode is also valuable for understanding common performances. For variability, the standard deviation effectively measures how spread out scores are around the mean, which is particularly relevant when considering educational interventions or assessing score consistency.
Inferential Statistics: One-Sample T-Test
A one-sample t-test compares the sample mean to a population mean—in this case, 70—to assess whether the observed scores significantly differ from the hypothesized population average.
Null hypothesis (\(H_0\)): The true population mean equals 70.
Alternative hypothesis (\(H_A\)): The true population mean is not equal to 70.
The test statistic is calculated as:
\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
\]
Where:
- \(\bar{x} = 85\)
- \(\mu_0 = 70\)
- \(s = 15.65\)
- \(n = 10\)
Plugging in the numbers:
\[
t = \frac{85 - 70}{15.65 / \sqrt{10}} = \frac{15}{15.65 / 3.162} \approx \frac{15}{4.954} \approx 3.026
\]
Degrees of freedom = \(n - 1 = 9\).
Using a t-distribution table or calculator, the critical t-value at \(\alpha = 0.05\) (two-tailed) for 9 degrees of freedom is approximately ±2.262. Since 3.026 > 2.262, the null hypothesis is rejected.
Interpretation: There is a statistically significant difference between the sample mean of 85 and the population mean of 70, indicating that the students performed significantly better than the average.
Practical Significance and Applications
These statistical measures are useful in various educational settings. For example, teachers and administrators can identify whether students are generally performing above or below expectations. The variability measures—variance and standard deviation—offer insights into score consistency, aiding decisions about targeted interventions. The t-test provides evidence of whether observed differences are statistically meaningful, informing curriculum adjustments and resource allocation.
Such analyses are invaluable in identifying disparities, tracking the impact of educational programs, and setting realistic learning goals. They can also be applied in accreditation processes, standardized testing evaluations, and in the development of benchmarks for student achievement.
Conclusion
Analyzing test scores through descriptive and inferential statistics offers a comprehensive understanding of student performance. The median emerges as the most representative measure of central tendency in this data set due to outliers, while the standard deviation quantifies score variability. The significant t-test result confirms that students scored notably higher than the population mean of 70, highlighting the effectiveness of instructional strategies or inherent student ability levels. Such statistical analyses are vital tools for educators aiming to improve educational outcomes and tailor interventions effectively.
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