15. ANOVA

Math 115

Author

Yurk

Example 1: Sleep Quality and Study Location

A campus wellness researcher wants to know whether students’ sleep quality differs depending on where they study. A random sample of 156 college students was surveyed. Each student reported their primary study location (Library, Dorm Room, Coffee Shop, or Student Center) and their average sleep quality score over the past month (measured on a validated 0-100 scale, where higher scores indicate better sleep quality).

The summary statistics are:

Study Location n Mean SD
Library 42 69.1 8.9
Dorm Room 48 61.4 16.4
Coffee Shop 31 61.7 10.6
Student Center 35 67.7 13.7

The ANOVA output from statistical software is:

Source df Sum Sq Mean Sq F p-value
Location 3 1943.2 647.72 3.83 0.011
Residuals 152 25692.9 169.03

Conduct a hypothesis test at the \(\alpha = 0.05\) significance level to determine whether mean sleep quality differs across the four study locations.

Step 1: State the hypotheses

Let \(\mu_1\), \(\mu_2\), \(\mu_3\), \(\mu_4\) represent the population mean sleep quality scores for students who primarily study at the Library, Dorm Room, Coffee Shop, and Student Center, respectively.

  • \(H_0: \mu_1 = \mu_2 = \mu_3 = \mu_4\) (all population means are equal)
  • \(H_A:\) At least one population mean is different from the others

Step 2: Check conditions

  1. Independence:
    • The sample is random
    • Observations are independent within and between groups
    • Condition met ✓
  2. Normality:
    • Sample sizes range from 31 to 48, all reasonably large
    • No extreme outliers mentioned
    • Condition met ✓
  3. Equal Variance:
    • Standard deviations range from 8.9 to 16.4
    • These are similar (largest is less than twice the smallest)
    • Condition met ✓

Step 3: Identify the test statistic

From the ANOVA table, the F-statistic is \(F = 3.83\) with \(df_1 = 3\) and \(df_2 = 152\).

Step 4: Find the p-value

From the software output, \(p = 0.011\)

This is the area to the right of \(F = 3.83\) under the F-distribution with \(df_1 = 3\) and \(df_2 = 152\).

Step 5: Make a decision

Since \(p = 0.011 < 0.05 = \alpha\), we reject \(H_0\).

Step 6: State the conclusion in context

There is convincing evidence that mean sleep quality differs across the four study locations. At least one study location is associated with a different average sleep quality score than the others.

Note: This was an observational study, so we cannot conclude that study location causes differences in sleep quality. There may be confounding variables (e.g., students who study at coffee shops may also consume more caffeine, which could affect sleep).

Example 2: Background Audio and Concentration

A psychology researcher wants to know whether the type of background audio affects students’ concentration during studying. She randomly assigns 120 college students to one of four conditions: Silence, Classical Music, Lo-fi Beats, or Pop Music (with lyrics). Each student completes a 30-minute reading comprehension task while listening to their assigned audio (or silence), and their concentration is measured using eye-tracking software (scored 0-100, with higher scores indicating better focus).

The researcher reports the following partial ANOVA results:

Source df Sum of Squares Mean Square F
Audio Type ? 1845 ? ?
Residuals ? 16530 142.5

Use this information to answer the following questions.

Part A

Calculate the degrees of freedom for Audio Type (\(df_1\)) and for Residuals (\(df_2\)).

Degrees of freedom for Audio Type (between groups):

\[df_1 = k - 1 = 4 - 1 = 3\]

Degrees of freedom for Residuals (within groups):

\[df_2 = n - k = 120 - 4 = 116\]

Part B

Using the Sum of Squares for Audio Type (SSG = 1845) and your answer from Part A, calculate the Mean Square for Audio Type (MSG).

\[MSG = \frac{SSG}{df_1} = \frac{1845}{3} = 615\]

Part C

Using the Mean Square Error (MSE = 142.5) and your answer from Part B, calculate the F-statistic.

\[F = \frac{MSG}{MSE} = \frac{615}{142.5} = 4.32\]

Part D

The completed ANOVA table is shown below. Based on the p-value, what conclusion should the researcher draw at the \(\alpha = 0.05\) level?

Source df Sum of Squares Mean Square F p-value
Audio Type 3 1845 615 4.32 0.006
Residuals 116 16530 142.5

Since \(p = 0.006 < 0.05 = \alpha\), we reject \(H_0\).

Conclusion: There is convincing evidence that mean concentration scores differ across the four audio conditions. At least one type of background audio is associated with a different level of concentration than the others.

Note: Because this was a randomized experiment (students were randomly assigned to audio conditions), we can conclude that the type of background audio causes differences in concentration.

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