Question

lets examine the relationship between educational attainment and perceived quality of health care. date for three (educational) groups are presented below. quality of health care is measured on a 5-point scale: 1=excellent, 2=very good, 3=good, 4=fair, 5=poor
less than high school | some college | college graduate
1 | 2 | 1
4 | 3 | 1
2 | 2 | 1
2 | 2 | 2
3 | 4 | 1
3 | 2 | 2
calculate anova using the five steps as presented in this chapter and match the questions with the correct answers.
the mean square between is
the df-between is
ssb (between group sum of squares) is
our calculated f-value is
if the alpha was set at .05, our f-critical would be..

Explanation:

Step 1: Organize Data

Let group 1: Less than High School (\(n_1 = 6\)): \(1, 4, 2, 2, 3, 3\)
Group 2: Some College (\(n_2 = 6\)): \(2, 3, 2, 2, 4, 2\)
Group 3: College Graduate (\(n_3 = 6\)): \(1, 1, 1, 2, 1, 2\)

Total \(N = 6 + 6 + 6 = 18\)

Step 2: Calculate Group Means (\(\bar{x}_i\))

\(\bar{x}_1 = \frac{1 + 4 + 2 + 2 + 3 + 3}{6} = \frac{15}{6} = 2.5\)
\(\bar{x}_2 = \frac{2 + 3 + 2 + 2 + 4 + 2}{6} = \frac{15}{6} = 2.5\)
\(\bar{x}_3 = \frac{1 + 1 + 1 + 2 + 1 + 2}{6} = \frac{8}{6} \approx 1.333\)

Grand Mean (\(\bar{x}_G\)):
\(\bar{x}_G = \frac{15 + 15 + 8}{18} = \frac{38}{18} \approx 2.111\)

Step 3: Calculate SSB (Between Sum of Squares)

\(SSB = \sum n_i (\bar{x}_i - \bar{x}_G)^2\)
\(SSB = 6(2.5 - 2.111)^2 + 6(2.5 - 2.111)^2 + 6(1.333 - 2.111)^2\)
\(= 6(0.389)^2 + 6(0.389)^2 + 6(-0.778)^2\)
\(= 6(0.151) + 6(0.151) + 6(0.605)\)
\(= 0.906 + 0.906 + 3.63\)
\(SSB \approx 5.442\)

Step 4: Calculate df-between

\(df_{between} = k - 1 = 3 - 1 = 2\) (where \(k = 3\) groups)

Step 5: Calculate Mean Square Between (MSB)

\(MSB = \frac{SSB}{df_{between}} = \frac{5.442}{2} \approx 2.721\)

Step 6: Calculate SSW (Within Sum of Squares)

For Group 1: \( (1 - 2.5)^2 + (4 - 2.5)^2 + (2 - 2.5)^2 + (2 - 2.5)^2 + (3 - 2.5)^2 + (3 - 2.5)^2 \)
\(= 2.25 + 2.25 + 0.25 + 0.25 + 0.25 + 0.25 = 5.5\)

Group 2: \( (2 - 2.5)^2 + (3 - 2.5)^2 + (2 - 2.5)^2 + (2 - 2.5)^2 + (4 - 2.5)^2 + (2 - 2.5)^2 \)
\(= 0.25 + 0.25 + 0.25 + 0.25 + 2.25 + 0.25 = 3.5\)

Group 3: \( (1 - 1.333)^2 + (1 - 1.333)^2 + (1 - 1.333)^2 + (2 - 1.333)^2 + (1 - 1.333)^2 + (2 - 1.333)^2 \)
\(= 0.111 + 0.111 + 0.111 + 0.444 + 0.111 + 0.444 = 1.332\)

\(SSW = 5.5 + 3.5 + 1.332 = 10.332\)

\(df_{within} = N - k = 18 - 3 = 15\)
\(MSW = \frac{SSW}{df_{within}} = \frac{10.332}{15} \approx 0.689\)

Step 7: Calculate F - value

\(F = \frac{MSB}{MSW} = \frac{2.721}{0.689} \approx 3.95\)

Step 8: F - Critical (Alpha = 0.05)

Using F - table, \(df_1 = 2\), \(df_2 = 15\), \(F_{critical} \approx 3.68\)

1. Mean Square Between:

\(MSB \approx 2.72\) (or 2.721)

2. df - between:

\(df_{between} = 2\)

3. SSB:

\(SSB \approx 5.44\) (or 5.442)

4. Calculated F - value:

\(F \approx 3.95\) (or ~4.0)

5. F - Critical (α = 0.05):

\(F_{critical} \approx 3.68\)

Answer:

• The mean square between is \(\boldsymbol{\approx 2.72}\)
• The df - between is \(\boldsymbol{2}\)
• SSB (between group sum of squares) is \(\boldsymbol{\approx 5.44}\)
• Our calculated F - value is \(\boldsymbol{\approx 3.95}\) (or ~4.0)
• If the alpha was set at.05, our F - critical would be \(\boldsymbol{\approx 3.68}\)