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.

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;σ²)

= 6

X[bar]= 158.17

S= 19.87

The hypothesis is:

H₀: μₓ = μ

H₁: μₓ ≠ μ

The statistic to use is a pooled t for independent samples with unknown but equal population variances

t= [(X[bar]ₓ - X[bar]) - (μₓ - μ)]/[Sa/√(1/nₓ + 1/)]

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!

yes because there is exactly one y value for every x value

for function there can't be more than one value for every x value

m∠5 + m∠3 = m∠1   is true.

m∠5 + m∠6 = 180 is true

m∠2 +m∠3 = m∠6   is true

m∠2 +m∠3 +m∠5 = 180   is true