(a)
\( H_0: \mu_c = \mu_s \), \( H_1: \mu_c > \mu_s \)

Define: \( \mu_c \) = mean weight of chinchilla rabbits, \( \mu_s \) = mean weight of sable rabbits
Null hypothesis: No difference in mean weights, \( \mu_c = \mu_s \)
Alternative hypothesis: Chinchilla rabbits are heavier, \( \mu_c > \mu_s \)
Test type: One-tailed

Result: \( H_0: \mu_c = \mu_s \), \( H_1: \mu_c > \mu_s \) [2]

(b)
0.041

Test: Two-sample unpaired t-test, assuming equal variances
Data: Chinchilla weights (4.9, 4.2, 4.1, 4.4, 4.3, 4.6, 4.0, 4.7, 4.5, 4.4), \( n_c = 10 \)
Sable weights (4.2, 4.1, 4.1, 4.2, 4.5, 4.4, 4.5, 3.9, 4.2, 4.0), \( n_s = 10 \)
Chinchilla mean: \( \bar{x}_c = \frac{4.9 + 4.2 + 4.1 + 4.4 + 4.3 + 4.6 + 4.0 + 4.7 + 4.5 + 4.4}{10} = 4.41 \)
Sable mean: \( \bar{x}_s = \frac{4.2 + 4.1 + 4.1 + 4.2 + 4.5 + 4.4 + 4.5 + 3.9 + 4.2 + 4.0}{10} = 4.21 \)
Chinchilla variance: \( s_c^2 \approx \frac{(4.9 – 4.41)^2 + \cdots + (4.4 – 4.41)^2}{9} \approx 0.0765333333 \)
Sable variance: \( s_s^2 \approx \frac{(4.2 – 4.21)^2 + \cdots + (4.0 – 4.21)^2}{9} \approx 0.0409888889 \)
Pooled variance: \( s_p^2 = \frac{(10 – 1) \times 0.0765333333 + (10 – 1) \times 0.0409888889}{10 + 10 – 2} \approx 0.0587611111 \)
T-statistic: \( t = \frac{4.41 – 4.21}{\sqrt{\frac{0.0587611111}{10} + \frac{0.0587611111}{10}}} \approx 1.844463643 \)
Degrees of freedom: \( df = 10 + 10 – 2 = 18 \)
P-value: \( P(T > 1.844463643, df = 18) \approx 0.0408 \)
Round to three decimal places: 0.041

Result: 0.041 [2]

(c)
Reject \( H_0 \). There is sufficient evidence to suggest that chinchilla rabbits are generally heavier than sable rabbits

Significance level: \( \alpha = 0.05 \)
P-value: 0.041
Compare: \( 0.041 < 0.05 \)
Decision: Reject \( H_0 \)
Reason: P-value is less than the significance level, indicating sufficient evidence for \( H_1 \)

Result: Reject \( H_0 \). There is sufficient evidence to suggest that chinchilla rabbits are generally heavier than sable rabbits [2]