Respuesta :
Answer:
Step-by-step explanation:
Hello!
The historical data suggests that the life expectancy in Argentina is equal to the life expectancy in Bolivia.
With the objective of testing if that it hasn't changed, the records of recently deceased people from Argentina and Bolivia were selected at random:
Group 1: Argentina
X₁: Years of life of a recently deceased Argentinian.
n₁= 265
X[bar]₁= 74.8 years
S₁= 4.1 years
Group 2: Bolivia
X₂: Years of life of a recently deceased Bolivian.
n₂= 300
X[bar]₂= 75.4 years
S₂= 4.3 years
The parameters of interest are the population means of the years of life of people in both countries:
H₀: μ₁ = μ₂
H₁: μ₁ ≠ μ₂
α: 0.05
The statistic to use is an approximate standard deviation, and since both samples are quite large, it is valid to use the sample standard deviations in place of the population standard deviations:
[tex]Z= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{\sqrt{\frac{S^2_1}{n_1} +\frac{S_2^2}{n_2} } }[/tex]≈N(0;1)
[tex]Z_{H_0}= \frac{(74.8-75.4)-(0)}{\sqrt{\frac{16.81}{265} +\frac{18.49}{300} } }= -2.54[/tex]
The critical values for this test are:
[tex]Z_{\alpha /2}= Z_{0.025}= -1.96[/tex]
[tex]Z_{1-\alpha /2}= Z_{0.0975}= 1.96[/tex]
Using the critical value approach, the decision rule is:
If [tex]Z_{H_0}[/tex] ≤ -1.96 or if [tex]Z_{H_0}[/tex] ≥ 1.96, reject the null hypothesis.
If -1.96 < [tex]Z_{H_0}[/tex] < 1.96, do not reject the null hypothesis.
[tex]Z_{H_0}[/tex] ≤ -1.96 so the decision is to reject the null hypothesis.
So with a 5% significance level, there is enough evidence to reject the null hypothesis, you can conclude that the life expectancy in Argentina is different from the life expectancy in Bolivia.
I hope this helps!