Answer:
sorry I don't know the answer
The answer is C- Midsegment
Answer: the value for the associated test statistic is 1.2653
Step-by-step explanation:
Given that;
sample size one n₁ = 10
mean one x"₁ = 6.4
standard deviation one S₁ = 1.1
sample size two n₁ = 11
mean two x"₂ = 5.6
standard deviation one S₁ = 1.7
H₀ : μ₁ = μ₂
H₁ : μ₁ ≠ μ₂
Pooled Variance
sp = √( { [(n₁ - 1) × s₁² + (n₂ - 1) × s₂²] / (n₁ + n₂ - 2)} × (1/n₁ + 1/n₂))
we substitute
= √( { [(10 - 1) × (1.1)² + (11 - 1) × (1.7)²] / (10 + 11 - 2)} × (1/10 + 1/11))
= √( { [(9) × 1.21 + (10) × 2.89] / (19) } × (0.1909))
= √({[ 39.79 ] / 19} × (0.1909))
= √( 2.0942 × 0.1909)
= √( 0.39978 )
= 0.63228
Now Test Statistics will be;
t = ( x"₁ - x"₂) / sp
we substitute
t = ( 6.4 - 5.6) / 0.63228
t = 0.8 / 0.63228
t = 1.2653
Therefore the value for the associated test statistic is 1.2653
5 x 3 = 15 + 1 = 16 feet
16 x 8 = 128 feet
128 / 3 = 42 yards 2 feet
Hope this helps!! :)
