Answer:
Step-by-step explanation:
Hello!
You want to compare two populations to see if there is a difference in the average annual income of the studied families.
H₀:μ₁ = μ₂
H₁:μ₁ ≠ μ₂
α: 0.10
Since the two samples are less than 30, I'll make the propper asumptions and use a Student t statistic to test the hypothesis.
t = (x₁[bar]-x₂[bar]) - (μ₁ - μ₂)
[tex]\sqrt{S²_{1}/n_{1} + S²_{2}/n_{2}}[/tex]
This distribution has [tex]t_{n1+n2-2}[/tex] degrees of freedom.
The rejection region for this hypothesis set is two-tailed, this means, you will reject the null hypothesis if the statistic takes small values or big values. There are two critical values.
[tex]t_{13+28-2;\0.10/2 }[/tex] = [tex]t_{39;\0.05 }[/tex] = -1.6849 ≅ -1.68
[tex]t_{13+28-2;\1-(0.10/2) }[/tex] = [tex]t_{39;\0.95}[/tex] = 1.6849 ≅ 1.68
So you'll rejec the null hypothesis if the calculated statistic is t ≤ -1.68 or t ≥ 1.68.
Sample 1
n₁ = 13
Mean (x₁[bar] = $151,000
S₁ = $41,000
Sample 2
n₂ = 28
Mean (x₂[bar] = $183,000
S₂ = $28,000
t = (151,000-183,000) - (0)
[tex]\sqrt{(41,000)²/13 + (28,000)²/28}[/tex]
t = -32,000 = -2.551
12542.24
Since -2.551 is less than -1.68, you have significant evidence to reject the null hypothesis. The average annual income of the population 1 is different than the average annual income of the the population 2.
