She takes a sample of three elementary school students, four high school students, eight college students, and four adults not in school and records the average number of hours each person sleeps.

Assume that the conditions for a one-way ANOVA are met. Determine the critical value for this one-way ANOVA test using a [tex]$5 \%$[/tex] significance level based on the table of critical values of the [tex]$F$[/tex]-distribution shown below.

\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
\multirow{2}{*}{
\begin{tabular}{l}
Degrees of Freedom \\
for the Denominator
\end{tabular}} & \multicolumn{6}{|c|}{ Degrees of Freedom for the Numerator } \\
\hline & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline 1 & 161.45 & 199.50 & 215.71 & 224.58 & 230.16 & 233.99 \\
\hline 2 & 18.51 & 19.00 & 19.16 & 19.25 & 19.30 & 19.33 \\
\hline 3 & 10.13 & 9.55 & 9.28 & 9.12 & 9.01 & 8.94 \\
\hline 4 & 7.71 & 6.94 & 6.59 & 6.39 & 6.26 & 6.16 \\
\hline 5 & 6.61 & 5.79 & 5.41 & 5.19 & 5.05 & 4.95 \\
\hline 6 & 5.99 & 5.14 & 4.76 & 4.53 & 4.39 & 4.28 \\
\hline 7 & 5.59 & 4.74 & 4.35 & 4.12 & 3.97 & 3.87 \\
\hline 8 & 5.32 & 4.46 & 4.07 & 3.84 & 3.69 & 3.58 \\
\hline 9 & 5.12 & 4.26 & 3.86 & 3.63 & 3.48 & 3.37 \\
\hline 10 & 4.96 & 4.10 & 3.71 & 3.48 & 3.34 & 3.24 \\
\hline
\end{tabular}



Answer :

We are tasked with determining the critical value for a one-way ANOVA test using a 5% significance level based on the provided F-distribution table. Here are the detailed steps to solve this problem:

### Step 1: Determine the number of groups
There are four groups in the sample:
1. Elementary school students
2. High school students
3. College students
4. Adults not in school

Thus, [tex]\(k = 4\)[/tex].

### Step 2: Calculate the Total Sample Size
We have the following sample sizes for each group:
- Elementary school students: 3
- High school students: 4
- College students: 8
- Adults not in school: 4

The total sample size [tex]\(N\)[/tex] is:
[tex]\[ N = 3 + 4 + 8 + 4 = 19 \][/tex]

### Step 3: Calculate the Degrees of Freedom
For a one-way ANOVA, we need to calculate the degrees of freedom for the numerator (between groups) and the denominator (within groups).

Degrees of Freedom for the Numerator (Between Groups):
[tex]\[ \text{df}_{\text{between}} = k - 1 = 4 - 1 = 3 \][/tex]

Degrees of Freedom for the Denominator (Within Groups):
[tex]\[ \text{df}_{\text{within}} = N - k = 19 - 4 = 15 \][/tex]

### Step 4: Use the F-Distribution Table
Using the F-distribution table, we need to find the critical value corresponding to:
- Degrees of freedom for the numerator: 3
- Degrees of freedom for the denominator: 15
- 5% significance level ([tex]\(\alpha = 0.05\)[/tex])

Here is the relevant portion of the provided table:
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|c|} \hline \multirow{2}{*}{\begin{tabular}{l} Degrees of Freedom \\ for the Denominator \end{tabular}} & \multicolumn{6}{|c|}{ Degrees of Freedom for the Numerator } \\ \hline & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & 161.45 & 199.50 & 215.71 & 224.58 & 230.16 & 233.99 \\ \hline 2 & 18.51 & 19.00 & 19.16 & 19.25 & 19.30 & 19.33 \\ \hline 3 & 10.13 & 9.55 & 9.28 & 9.12 & 9.01 & 8.94 \\ \hline 4 & 7.71 & 6.94 & 6.59 & 6.39 & 6.26 & 6.16 \\ \hline 5 & 6.61 & 5.79 & 5.41 & 5.19 & 5.05 & 4.95 \\ \hline 6 & 5.99 & 5.14 & 4.76 & 4.53 & 4.39 & 4.28 \\ \hline 7 & 5.59 & 4.74 & 4.35 & 4.12 & 3.97 & 3.87 \\ \hline 8 & 5.32 & 4.46 & 4.07 & 3.84 & 3.69 & 3.58 \\ \hline 9 & 5.12 & 4.26 & 3.86 & 3.63 & 3.48 & 3.37 \\ \hline 10 & 4.96 & 4.10 & 3.71 & 3.48 & 3.33 & 3.23 \\ \hline \end{tabular} \][/tex]

From this table, we can see that for 15 degrees of freedom for the denominator (the row for 15 is not given in the table, so we conservatively pick the closest available, which is row for 10 since typical F-tables provided will go beyond):

- For [tex]\(\text{df}_{\text{between}} = 3\)[/tex]: the critical value when [tex]\(\text{df}_{\text{within}} = 15\)[/tex] is approximately closest approximations available:

Corresponding value may need reference to standard F-distribution complete tables (typically around 2.76 for higher degrees).

The critical value for this one-way ANOVA test, considering closest possible find may be required from standard complete tables is [2.74-2.76] rounded effectively.