91Ó°ÊÓ

Clifford Adelman, a researcher with the Department of Education, followed a cohort of students who graduated from high school in \(1992,\) monitoring the progress the students made toward completing a bachelor's degree. One aspect of his research was to compare students who first attended a community college to those who immediately attended and remained at a four-year institution. The sample standard deviation time to complete a bachelor's degree of the 268 students who transferred to a four-year school after attending community college was \(1.162 .\) The sample standard deviation time to complete a bachelor's degree of the 1145 students who immediately attended and remained at a four-year institution was \(1.015 .\) Assuming the time to earn a bachelor's degree is normally distributed, does the evidence suggest the standard deviation time to earn a bachelor's degree is different between the two groups? Use the \(\alpha=0.05\) level of significance.

Step by step solution

01

Define the hypotheses

H_1: σ_1 ≠ σ_2

02

Identify the sample statistics

Note the given sample statistics. For community college transfer students (Group 1):n_1 = 268s_1 = 1.162For students who went straight to a four-year school (Group 2): n_2 = 1145s_2 = 1.015

03

Calculate the test statistic

Use the formula for the ratio of two sample variances: F = (s_1^2)/(s_2^2).Substitute the given values: F = (1.162^2)/(1.015^2)

04

Find the critical value

Determine the degrees of freedom for the numerator (df_1) and the denominator (df_2). For Group 1: df_1 = n_1 - 1 = 268 - 1 = 267. For Group 2, df_2 = n_2 - 1 = 1144. Using the F-distribution table, find the critical value for a two-tailed test at α = 0.05: F_{critical}.

05

Compare the test statistic and critical value

Compare the calculated F-value to the critical values. If the F-value is greater than F_{critical} for df_1 = 267 and df_2 = 1144, or less than F_{critical} found using the inverse F-distribution table, reject the null hypothesis (significant difference). Otherwise, fail to reject the null hypothesis (no significant difference).

06

Conclusion

Based on the comparison in Step 5, state the conclusion. If H_0 was rejected, there is evidence to suggest the standard deviation of time to complete a degree is different between the two groups. If H_0 was not rejected, there is no evidence to suggest a difference.

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Key Concepts

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

sample standard deviation

The sample standard deviation is a measure of the variation or dispersion of a set of values. In hypothesis testing, it's used to compare the spread of different samples. The formula for the sample standard deviation is denoted as:
\[ s = \sqrt{\frac{\sum{(x_i - \bar{x})^2}}{n-1}} \], where:

• \(x_i\) is each value in the sample
• \(\bar{x}\) is the mean of the sample
• \(n\) is the sample size

In our exercise, we have two groups of students: those who transferred from a community college and those who directly joined a four-year institution.
The standard deviation for the first group is \(1.162\) and for the second group is \(1.015\). These values tell us how much the time taken to complete a bachelor's degree varies within each group. Comparing these standard deviations will help us understand if there's a significant difference in the time taken to complete degrees between the two groups.

F-distribution

The F-distribution is critical in comparing two sample variances. It is a right-skewed distribution used in the analysis of variance and other statistical tests. In our context, the ratio of the variances follows an F-distribution under the null hypothesis.

The test statistic, known as the F-value, is calculated as: \[ F = \frac{s_1^2}{s_2^2} \]
In this formula:

• \(s_1^2\) and \(s_2^2\) are the sample variances from the two groups
• \(df_1\) and \(df_2\) represent the degrees of freedom for the two groups

For our exercise, substituting the given standard deviations, we get:
\[ F = \frac{(1.162)^2}{(1.015)^2} \]
This calculated F-value helps us determine whether the sample variances are significantly different by comparing it with a critical value from the F-distribution table, based on a significance level of \(\alpha = 0.05\).

If the calculated F-value is greater than the critical value or less than its reciprocal, we reject the null hypothesis, indicating a significant difference in variances.

two-sample variance test

A two-sample variance test, also known as an F-test, is used to compare the variability of two different samples. This test helps determine if there is a significant difference between the variances of two populations. The main steps involved are:

• Formulate the null and alternative hypotheses.
• Calculate the sample statistics, mainly the standard deviations.
• Compute the test statistic (F-value).
• Determine the degrees of freedom for both samples.
• Find the critical value from the F-distribution table.
• Compare the calculated F-value and the critical value.
• Make a conclusion based on the comparison.

In our exercise, we established the null hypothesis \(H_0: \sigma_1 = \sigma_2\) and the alternative hypothesis \(H_1: \sigma_1 e \sigma_2 \).
After calculating the F-value and comparing it with the critical value, we decide whether the difference in variance is statistically significant. If we find a significant difference, it indicates that the time taken to complete a bachelor's degree varies more in one group than the other. Otherwise, we conclude there's no substantial difference in the time taken between the two groups. This method provides a systematic approach to understanding differences in sample variability, crucial for many statistical analyses.