Respuesta :
Answer:
The test statistic t = 3.61270
The tabulated value t₀.₀₅ = 3.61270 is greater than the tabulated value 2.4185 at 0.05 level of significance.
The null hypothesis is rejected at 0.05 level of significance.
the population means are not different
Step-by-step explanation:
Step(i):-
Given first sample size 'n₁' = 19
Given the mean annual family income for 19 people making inquiries at the first development is $148,000, with a standard deviation of $41,000
The mean of first sample 'x₁⁻ ' = $148,000
The standard deviation of first sample S₁ = $41,000
Given data a corresponding sample of 25 people at the second development had a mean of $186,000, with a standard deviation of $27,000
The second sample size n₂ = 25
The mean of second sample 'x₂⁻ = $186,000
The standard deviation of first sample S₂ = $27,000
Step(ii) :-
Null hypothesis : H₀ : μ₁= μ₂
Alternative hypothesis :H₁: μ₁≠μ₂
Level of significance ∝ = 0.05
Degrees of freedom : ν = n₁+n₂-2 = 19+25-2 = 42
Test of hypothesis
[tex]t=\frac{x^{-} _{1}-x^{-} _{2} }{\sqrt{S^2(\frac{1}{n_{1} } +\frac{1}{n_{2} } } )}[/tex]
where
[tex]S^{2} = \frac{n_{1}S_{1} ^{2} +n_{2} S^{2} _{2} }{n_{1}+n_{2} -2 }[/tex]
on calculation , we get
[tex]S^{2} = \frac{19(41,000) ^{2} +25(27,000)^2 }{19+25-2 } = 1,194,380,952.381[/tex]
[tex]t=\frac{1,48,000 -1,86,000 }{\sqrt{1,194,380,952.381(\frac{1}{19 } +\frac{1}{25 } } )}[/tex]
on calculation , we get
[tex]t = \frac{-38,000}{10,518.4311} =-3.61270[/tex]
Taking modulus
|t| = |-3.61270|
t = 3.61270
Step(iii):-
The degrees of freedom ν = n₁+n₂-2 = 19+25-2 = 42
The tabulated value t₀.₀₅ = 3.61270 is greater than the tabulated value 2.4185 at 0.05 level of significance.
Conclusion:-
The null hypothesis is rejected at 0.05 level of significance.
we accepted alternative hypothesis that is H1:μ1 ≠ μ2
the population means are not different
