Given The Student Test Scores: 95, 92, 90, 90, 83, 83, 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 word summary and analysis discussing the results of your calculations. 1. State your results for the sample: the mean, median, mode, range, variance, and standard deviation 2. Explain which method is best for this data set. Why? 3. Conduct a one-sample t -test and interpret the results (use a population mean of 70). 4. In what situations would this information be useful? Prepare this assignment according to the guidelines found in the APA Style Guide

Paper For Above instruction

Introduction

Understanding descriptive and inferential statistics is essential for interpreting data accurately. This paper analyzes a set of student test scores to compute key statistical measures, evaluates the most appropriate methods for analysis, conducts a t-test to compare the sample mean to a population mean, and discusses the practical applications of these statistics. The analysis aims to provide insights into the distribution of scores and their implications within educational contexts.

Data Description and Calculations

The given test scores are: 95, 92, 90, 90, 83, 83, 83, 74, 60, and 50. These scores are used to calculate the mean, median, mode, range, variance, and standard deviation.

Mean

The mean, or average, is computed by summing all scores and dividing by the number of scores:

(95 + 92 + 90 + 90 + 83 + 83 + 83 + 74 + 60 + 50) ÷ 10 = 810 ÷ 10 = 81.

Median

The median is the middle value when scores are ordered ascendingly:

50, 60, 74, 83, 83, 83, 90, 90, 92, 95.

Since there are 10 scores, the median is the average of the 5th and 6th scores:

(83 + 83) ÷ 2 = 83.

Mode

The mode is the most frequently occurring score:

83 appears three times, more than any other score, so the mode is 83.

Range

Range is the difference between the highest and lowest scores:

95 - 50 = 45.

Variance and Standard Deviation

Variance measures the average squared deviation from the mean:

First, calculate each score's deviation from the mean, square it, then find the average of these squared deviations.

Deviations: (95-81)=14, (92-81)=11, (90-81)=9, (90-81)=9, (83-81)=2, (83-81)=2, (83-81)=2, (74-81)=−7, (60-81)=−21, (50-81)=−31.

Squared deviations: 196, 121, 81, 81, 4, 4, 4, 49, 441, 961.

Sum of squared deviations: 196+121+81+81+4+4+4+49+441+961=1946.

Variance (sample): 1946 ÷ (10−1) ≈ 216.22.

Standard deviation: √216.22 ≈ 14.71.

Summary and Analysis

The calculated mean (81) indicates that, on average, students scored relatively high, though the median (83) suggests the middle scores are slightly higher, pointing to a right-skewed distribution. The mode consistently reflects the most common score, 83, signifying its popularity among students. The range of 45 reveals considerable score variability, indicating diverse student performance.

The variance and standard deviation (≈14.71) further demonstrate the dispersion of scores around the mean, with some scores, such as 50 and 95, deviating substantially. These measures underline the heterogeneity in the student sample. In educational assessments, such variability informs teachers about the effectiveness of instruction and identifies student groups needing additional support.

Choosing the appropriate measure depends on the data's distribution. Since the data shows skewness via outliers (50 and 95), median and mode are more robust to outliers, while the mean and standard deviation provide insights into overall performance and variability.

Inferential Analysis: One-Sample T-Test

A one-sample t-test was conducted to compare the sample mean (81) with a population mean of 70. The null hypothesis (H0: μ=70) assumes no difference between the sample mean and the population mean.

Using the formula t = (x̄ - μ) / (s/√n), where x̄=81, μ=70, s≈14.71, n=10:

t = (81 - 70) / (14.71/√10) ≈ 11 / (14.71/3.16) ≈ 11 / 4.65 ≈ 2.36.

Degrees of freedom: 9.

Referring to t-distribution tables, a t-value of 2.36 with 9 degrees of freedom has a p-value less than 0.05, indicating the difference is statistically significant at the 5% level.

Thus, we reject the null hypothesis and conclude that the sample mean score significantly exceeds the population mean of 70. This suggests that the students' test performance is notably higher than the average expected score in the population.

Implications and Situations of Use

These statistical insights are valuable for educators, policymakers, and researchers. Understanding the distribution of scores helps assess teaching effectiveness, identify learning gaps, and tailor interventions. The t-test provides evidence for evaluating whether particular teaching methods have a meaningful impact compared to a standard expectation. For instance, if a new curriculum is introduced, similar analyses can determine if it improves student performance over the baseline.

In clinical research or educational policy, such statistical tools help make data-driven decisions, allocate resources effectively, and establish benchmarks for academic achievement. Moreover, statistical descriptions like variance and standard deviation inform about variability within student groups, guiding individualized instruction strategies.

Conclusion

This analysis of student test scores demonstrates how descriptive and inferential statistics can be applied to educational data to derive meaningful insights. The computed measures depict the distribution and variability of scores, while the t-test confirms a significant difference from a benchmark performance. These findings highlight the importance of statistical literacy in educational evaluation and decision-making, emphasizing the utility of robust statistical methods for assessing student achievement and informing educational strategies.

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