Assume that the distributions of X and Y are N(μx, σ2) and N(μy, σ2), respectively.
Given the nx = 4 observations of X,

134, 151, 156, 157

and ny = 6 observations of Y,
128, 139, 163, 171, 172, 176 find the p-value for the test H0: μx = μy vs. H1: μx ≠ μy. Round your answer to 3 decimal places. Hint 1: Assume the population variances are equal. Hint 2: EXCEL: =TDIST(t, degrees of freedom,1) gives area to the right of t; =TDIST(t, degrees of freedom,2) gives 2×(area to the right of t). OR R: pt(t, degrees of freedom) gives area to the left of t.

Respuesta :

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!