Answer:
a) The standard deviation of the annual income σₓ = 2045
b)
<em>The calculated value Z = 0.608 < 1.645 at 10 % level of significance</em>
<em>Null hypothesis is accepted </em>
<em>The average annual income is greater than $32,000</em>
c)
<em>The calculated value Z = 1.0977 < 1.96 at 5 % level of significance</em>
<em>Null hypothesis is accepted </em>
<em>The average annual income is equal to $33,000</em>
<em>d) </em>
<em>95% of confidence intervals of the Average annual income</em>
<em>(26 ,746.8 ,34, 763.2)</em>
<u>Step-by-step explanation:</u>
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 =
=
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
<u><em>Null Hypothesis:- H₀</em></u>: μ > $32,000
<u><em>Alternative Hypothesis</em></u>:H₁: μ < $32,000
Level of significance α = 0.10
Z= |-0.608| = 0.608
<em>The calculated value Z = 0.608 < 1.645 at 10 % level of significance</em>
<em>Null hypothesis is accepted </em>
<em>The average annual income is greater than $32,000</em>
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
<u><em>Null Hypothesis:- H₀</em></u>: μ = $33,000
<u><em>Alternative Hypothesis</em></u>:H₁: μ ≠ $33,000
Level of significance α = 0.05
Z = -1.0977
|Z|= |-1.0977| = 1.0977
The 95% of z -value = 1.96
<em>The calculated value Z = 1.0977 < 1.96 at 5 % level of significance</em>
<em>Null hypothesis is accepted </em>
<em>The average annual income is equal to $33,000</em>
<em>d) </em>
<em>95% of confidence intervals is determined by</em>
<em></em><em></em>
<em></em><em></em>
<em>( 30 755 - 4008.2 , 30 755 +4008.2)</em>
<em>95% of confidence intervals of the Average annual income</em>
<em>(26 ,746.8 ,34, 763.2)</em>
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