Respuesta :
Answer:
a) The standard deviation of the annual income σₓ = 2045
b)
The calculated value Z = 0.608 < 1.645 at 10 % level of significance
Null hypothesis is accepted
The average annual income is greater than $32,000
c)
The calculated value Z = 1.0977 < 1.96 at 5 % level of significance
Null hypothesis is accepted
The average annual income is equal to $33,000
d)
95% of confidence intervals of the Average annual income
(26 ,746.8 ,34, 763.2)
Step-by-step explanation:
Given size of the sample 'n' =100
mean of the sample x⁻ = $30,755
The Standard deviation = $20,450
a)
The standard deviation of the annual income σₓ = [tex]\frac{S.D}{\sqrt{n} }[/tex]
= [tex]\frac{20,450}{\sqrt{100} }= 2045[/tex]
b)
Given mean of the Population μ = $32,000
Given size of the sample 'n' =100
mean of the sample x⁻ = $30,755
The Standard deviation ( σ)= $20,450
Null Hypothesis:- H₀: μ > $32,000
Alternative Hypothesis:H₁: μ < $32,000
Level of significance α = 0.10
[tex]Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }[/tex]
[tex]Z = \frac{30755-32000 }{\frac{20450}{\sqrt{100} } }[/tex]
Z= |-0.608| = 0.608
The calculated value Z = 0.608 < 1.645 at 10 % level of significance
Null hypothesis is accepted
The average annual income is greater than $32,000
c)
Given mean of the Population μ = $33,000
Given size of the sample 'n' =100
mean of the sample x⁻ = $30,755
The Standard deviation ( σ)= $20,450
Null Hypothesis:- H₀: μ = $33,000
Alternative Hypothesis:H₁: μ ≠ $33,000
Level of significance α = 0.05
[tex]Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }[/tex]
[tex]Z = \frac{30755-33000 }{\frac{20450}{\sqrt{100} } }[/tex]
Z = -1.0977
|Z|= |-1.0977| = 1.0977
The 95% of z -value = 1.96
The calculated value Z = 1.0977 < 1.96 at 5 % level of significance
Null hypothesis is accepted
The average annual income is equal to $33,000
d)
95% of confidence intervals is determined by
[tex](x^{-} - 1.96 \frac{S.D}{\sqrt{n} } , x^{-} + 1.96 \frac{S.D}{\sqrt{n} })[/tex]
[tex](30755 - 1.96 \frac{20450}{\sqrt{100} } , 30755 +1.96 \frac{20450}{\sqrt{100} })[/tex]
( 30 755 - 4008.2 , 30 755 +4008.2)
95% of confidence intervals of the Average annual income
(26 ,746.8 ,34, 763.2)
