(a)
\( H_0 \): The die is fair (\( P = \frac{1}{6} \))

\( H_1 \): The die is not fair

Each number has probability \( \frac{1}{6} \) if fair

Result: \( H_0 \): Probabilities are equal, \( H_1 \): Probabilities are not equal [2]

(b)
5

\( df = k – 1 \), where \( k = 6 \) (categories)

\( df = 6 – 1 = 5 \)

Result: 5 [1]

(c)
10

Expected frequency: \( E = n \times p \)

\( n = 60 \), \( p = \frac{1}{6} \)

\( E = 60 \times \frac{1}{6} = 10 \)

Result: 10 [1]

(d)
0.287

\( \chi^2 = \sum \frac{(O – E)^2}{E} \), \( E = 10 \)

Observed: \( [8, 7, 6, 15, 12, 12] \)

For 1: \( \frac{(8 – 10)^2}{10} = 0.4 \)

For 2: \( \frac{(7 – 10)^2}{10} = 0.9 \)

For 3: \( \frac{(6 – 10)^2}{10} = 1.6 \)

For 4: \( \frac{(15 – 10)^2}{10} = 2.5 \)

For 5: \( \frac{(12 – 10)^2}{10} = 0.4 \)

For 6: \( \frac{(12 – 10)^2}{10} = 0.4 \)

\( \chi^2 = 0.4 + 0.9 + 1.6 + 2.5 + 0.4 + 0.4 = 6.2 \)

\( df = 5 \), p-value \( \approx 0.287 \)

Result: 0.287 [2]

(e)
Fail to reject \( H_0 \)

\( p = 0.287 > 0.05 \)

Insufficient evidence to reject the null hypothesis

Conclusion: No evidence at 5% level that die is unfair

Result: Fail to reject \( H_0 \), p-value > 0.05 [2]