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11.06.2020 •
Mathematics
4. The average annual income of 100 randomly chosen residents of Santa Cruz is $30,755 with a standard deviation of $20,450. a) What is the standard deviation of the annual income? b) Test the hypothesis that the average annual income is $32,000 against the alternative that it is less than $32,000 at the 10% level. c) Test the hypothesis that the average annual income is equal to $33,000 against the alternative that it is not at the 5% level. d) What is the 95% confidence interval of the average annual income?
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Ответ:
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 =![\frac{S.D}{\sqrt{n} }](/tpl/images/0683/4665/fc1b3.png)
=![\frac{20,450}{\sqrt{100} }= 2045](/tpl/images/0683/4665/e33b1.png)
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
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
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
( 30 755 - 4008.2 , 30 755 +4008.2)
95% of confidence intervals of the Average annual income
(26 ,746.8 ,34, 763.2)
Ответ: