Answer:
p-value: 0.453
Step-by-step explanation:
Hello!
You have two study variables
X~N(μₓ;σₓ²)
nₓ= 4
X[bar]ₓ= 149.50
Sₓ= 10.66
Y~N(μ[tex]_{y}[/tex];σ[tex]_{y}[/tex]²)
[tex]n_{y}[/tex]= 6
X[bar][tex]_{y}[/tex]= 158.17
S[tex]_{y}[/tex]= 19.87
The hypothesis is:
H₀: μₓ = μ[tex]_{y}[/tex]
H₁: μₓ ≠ μ[tex]_{y}[/tex]
The statistic to use is a pooled t for independent samples with unknown but equal population variances
t= [(X[bar]ₓ - X[bar][tex]_{y}[/tex]) - (μₓ - μ[tex]_{y}[/tex])]/[Sa/√(1/nₓ + 1/[tex]n_{y}[/tex])]
The value of the statistic is:
t= -0.789
For this statistic value the corresponding p-value is:
P(t ≤ -0.789) + (1 - P(t ≤ 0.789) = 0.453
I hope it helps!