Math Problem Statement

Several magazines were grouped into three categories based on what level of education of their readers the magazines are geared towards: high, medium, or low level. Then random samples of the magazines were selected to determine the number of three-plus-syllable words were in the advertising copy, and the data is in the table. Is there enough evidence to show that the mean number of three-plus-syllable words in advertising copy is different for at least one of the education levels? Test at the 5% level.

Number of Three Plus Syllable Words in Advertising Copy

High Education Medium Education Low Education 30 20 7 22 18 9 32 26 9 21 22 9 19 23 7 15 0 8 6 12 9 32 20 9 26 20 8 23 21 8 25 0 8 34 29 9 36 4 8 36 8 9 34 11 8 14 18 9 39 20 7 20 7 9 State the hypotheses. H0: Select an answer

Ha: Select an answer

Calculate the F test statistic. Round to two decimal places. F =

Calculate the p-value. Round to four decimal places. p-value =

State your decision. Since p-value ? 0.05, Select an answer H0.

Interpret the results. At the 5% significance level, there Select an answer evidence to show that Select an answer .

Solution

To test whether there is a difference in the mean number of three-plus-syllable words in advertising copy for at least one education level, we perform a One-Way ANOVA test.

1. Hypotheses

  • Null hypothesis (H₀): The mean number of three-plus-syllable words in advertising copy is the same across all education levels. H0:μHigh=μMedium=μLowH_0: \mu_{\text{High}} = \mu_{\text{Medium}} = \mu_{\text{Low}}
  • Alternative hypothesis (Hₐ): At least one mean is different. Ha:At least one μ is different.H_a: \text{At least one } \mu \text{ is different.}

2. Data Table

HighMediumLow
30207
22189
32269
21229
19237
1508
6129
32209
26208
23218
2508
34299
3648
3689
34118
14189
39207
2079

3. Calculate the F-test statistic

We compute the test statistic using the formula for ANOVA: F=Between-group variabilityWithin-group variabilityF = \frac{\text{Between-group variability}}{\text{Within-group variability}} I'll calculate the required values. pandas as pd import scipy.stats as stats

Data

data = { "High": [30, 22, 32, 21, 19, 15, 6, 32, 26, 23, 25, 34, 36, 36, 34, 14, 39, 20], "Medium": [20, 18, 26, 22, 23, 0, 12, 20, 20, 21, 0, 29, 4, 8, 11, 18, 20, 7], "Low": [7, 9, 9, 9, 7, 8, 9, 9, 8, 8, 8, 9, 9, 9, 8, 9, 7, 9] }

Create DataFrame

df = pd.DataFrame(data)

Perform ANOVA

f_statistic, p_value = stats.f_oneway(df["High"], df["Medium"], df["Low"])

Display results

f_statistic, p_value### 4. Results

  • F-test statistic (F): 26.14 (rounded to two decimal places)
  • p-value: 1.5319×1081.5319 \times 10^{-8} (rounded to four decimal places, 0.0000)

5. Decision

Since the p-value (0.00000.0000) is less than 0.05, we reject the null hypothesis (H₀).


6. Interpretation

At the 5% significance level, there is strong evidence to show that the mean number of three-plus-syllable words in advertising copy is different for at least one education level.


Would you like me to provide a detailed breakdown of calculations, or conduct further post hoc tests to determine which groups differ?
Here are 5 follow-up questions and one tip:

  1. What does rejecting the null hypothesis mean in practical terms for advertisers?
  2. How can we calculate confidence intervals for the group means?
  3. Should a post hoc test (e.g., Tukey's HSD) be conducted to identify the differences?
  4. What are the assumptions of ANOVA, and were they satisfied?
  5. How does sample size affect the results of ANOVA?

Tip: Always verify the assumptions of normality and equal variance when performing ANOVA.

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Math Problem Analysis

Mathematical Concepts

Statistics
Analysis of Variance (ANOVA)

Formulas

F = Between-group variability / Within-group variability
p-value calculation for ANOVA

Theorems

F-distribution in ANOVA

Suitable Grade Level

Undergraduate Level or Advanced High School Statistics