Respuesta :
Answer:
a. the probability of a type II error is 0.5319
b. the required sample size to satisfy and the type II error probability is 59.4441
Step-by-step explanation:
From the information given; we have:
sample size n = 25
Population standard deviation [tex]\sigma[/tex] = 4
true average lifetime = Sample Mean [tex]\bar X[/tex] = 62
We can state our null hypothesis and alternative hypothesis as follows:
Null hypothesis:
[tex]\mathbf{H_o : \mu = 60}[/tex]
Alternative hypothesis
[tex]\mathbf{H_1 : \mu \neq 60}[/tex]
Where ;
∝ = 0.01
From the standard normal tables at critical value ∝ = 0.01 ; the level of significance is -2.575 lower limit and 2.575 upper limit
The z statistics for the lower limit is:
[tex]lower \ limit = \dfrac{\bar X - \mu }{\dfrac{\sigma}{\sqrt {n}}}[/tex]
[tex]-2.575= \dfrac{\bar x - 60 }{\dfrac{4}{\sqrt 25}}}[/tex]
[tex]-2.575= \dfrac{\bar x - 60 }{0.8}}}[/tex]
[tex]-2.575*0.8= {\bar x - 60 }{}}}[/tex]
[tex]-2.06= {\bar x - 60 }{}}}[/tex]
[tex]\bar x = 60-2.06[/tex]
[tex]\bar x = 57.94[/tex]
The z statistics for the upper limit is:
[tex]lower \ limit = \dfrac{\bar X - \mu }{\dfrac{\sigma}{\sqrt {n}}}[/tex]
[tex]2.575= \dfrac{\bar x - 60 }{\dfrac{4}{\sqrt 25}}}[/tex]
[tex]2.575= \dfrac{\bar x - 60 }{0.8}}}[/tex]
[tex]2.575*0.8= {\bar x - 60 }{}}}[/tex]
[tex]2.06= {\bar x - 60 }{}}}[/tex]
[tex]\bar x = 60-(-2.06)[/tex]
[tex]\bar x = 60+2.06[/tex]
[tex]\bar x = 62.06[/tex]
Thus; the probability of a type II error is determined as follows:
β = P ( [tex]57.94 < \bar x < 62.06[/tex] )
[tex]= P ( \dfrac{57.94 -62 }{\dfrac{4}{\sqrt{25}}}<\dfrac{62.06 -62 }{\dfrac{4}{\sqrt{25}}})[/tex]
[tex]= P ( \dfrac{-4.06 }{0.8}}<\dfrac{2.06 }{0.8})[/tex]
= P ( -5.08 < Z < 0.08 )
= P ( Z < 0.08) - P ( Z < - 5.08)
Using Excel Function: [ (=NORMDIST (0.08)) - (=NORMDIST(-5.08)) ] ; we have:
= 0.531881372 - 0.00000001887
= 0.531881183
≅ 0.5319
b.
What is the required sample size to satisfy and the type II error probability of b(62) = 0.1
Recall that:
The critical value of ∝ = 2.575 ( i. e [tex]Z_{1 - \alpha/2 } = 2.575[/tex] )
Now ;
the critical value of β is :
[tex]Z _{1- \beta} = 1.28[/tex]
The required sample size to satisfy and the type II error probability is therefore determined as :
[tex]n = [\dfrac{(Z_{1 - \alpha/2} + Z_{1 - \beta} ) \sigma }{\delta}]^2[/tex]
[tex]n = [\dfrac{(2.575+1.28 ) 4 }{2}]^2[/tex]
[tex]n = [\dfrac{(3.855 ) 4 }{2}]^2[/tex]
[tex]n = [\dfrac{(15.42 ) }{2}]^2[/tex]
n = 7.71 ²
n= 59.4441
Thus; the required sample size to satisfy and the type II error probability is 59.4441
A)
In order to calculate the Type II Error, we proceed with stating the factors:
Hypothesized Mean is given as = μ[tex]_{0} [/tex] = 60
True Mean is given as = μa = 62
Standard Deviation is given as = σ = 4
Sample Size = n = 25
Standard error of mean = σx =σ/[tex]\sqrt{\\} [/tex]σ = 0.80
given 0.01 level and two tailed test critical value Zσ ± 2.58 or approximately 3
Acceptance region is
given as: = μ - Z∝ * σx ≤ Π ≤ μ+Z∝⇄ =57.9360 ≤x≤ 62.0640
Type II Error = probability of not rejecting β = P(57.94-μa/σx)) <Z< (62.064-μa)/σx))
= P (-5.08 <Z< 0.08)
= 0.5319-0)
= 0.532
B
Hypothesized mean = μ₀ = 60
True Mean =μₐ = 62
Standard Deviation =σ = 4
for 0.01 level and two-tailed test critical value Z∝/2 ± 2.58
for 0.01 level of type II error critical value Z[tex]\beta [/tex] = 1.28
Required sample size = n = ([tex]Z_{\alpha /2} [/tex] + [tex]Z_{\beta } [/tex])²σ²/(μ₀-μ₀)²
= 60
See the link below for more about two-sided tests:
https://brainly.com/question/8170655
