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
= 4
true average lifetime = Sample Mean
= 62
We can state our null hypothesis and alternative hypothesis as follows:
Null hypothesis:
![\mathbf{H_o : \mu = 60}](https://tex.z-dn.net/?f=%5Cmathbf%7BH_o%20%3A%20%5Cmu%20%20%3D%2060%7D)
Alternative hypothesis
![\mathbf{H_1 : \mu \neq 60}](https://tex.z-dn.net/?f=%5Cmathbf%7BH_1%20%3A%20%5Cmu%20%20%5Cneq%2060%7D)
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:
![lower \ limit = \dfrac{\bar X - \mu }{\dfrac{\sigma}{\sqrt {n}}}](https://tex.z-dn.net/?f=lower%20%5C%20limit%20%3D%20%5Cdfrac%7B%5Cbar%20X%20-%20%5Cmu%20%7D%7B%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%20%7Bn%7D%7D%7D)
![-2.575= \dfrac{\bar x - 60 }{\dfrac{4}{\sqrt 25}}}](https://tex.z-dn.net/?f=-2.575%3D%20%5Cdfrac%7B%5Cbar%20x%20-%2060%20%7D%7B%5Cdfrac%7B4%7D%7B%5Csqrt%2025%7D%7D%7D)
![-2.575= \dfrac{\bar x - 60 }{0.8}}}](https://tex.z-dn.net/?f=-2.575%3D%20%5Cdfrac%7B%5Cbar%20x%20-%2060%20%7D%7B0.8%7D%7D%7D)
![-2.575*0.8= {\bar x - 60 }{}}}](https://tex.z-dn.net/?f=-2.575%2A0.8%3D%20%7B%5Cbar%20x%20-%2060%20%7D%7B%7D%7D%7D)
![-2.06= {\bar x - 60 }{}}}](https://tex.z-dn.net/?f=-2.06%3D%20%7B%5Cbar%20x%20-%2060%20%7D%7B%7D%7D%7D)
![\bar x = 60-2.06](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20%2060-2.06)
![\bar x = 57.94](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%2057.94)
The z statistics for the upper limit is:
![lower \ limit = \dfrac{\bar X - \mu }{\dfrac{\sigma}{\sqrt {n}}}](https://tex.z-dn.net/?f=lower%20%5C%20limit%20%3D%20%5Cdfrac%7B%5Cbar%20X%20-%20%5Cmu%20%7D%7B%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%20%7Bn%7D%7D%7D)
![2.575= \dfrac{\bar x - 60 }{\dfrac{4}{\sqrt 25}}}](https://tex.z-dn.net/?f=2.575%3D%20%5Cdfrac%7B%5Cbar%20x%20-%2060%20%7D%7B%5Cdfrac%7B4%7D%7B%5Csqrt%2025%7D%7D%7D)
![2.575= \dfrac{\bar x - 60 }{0.8}}}](https://tex.z-dn.net/?f=2.575%3D%20%5Cdfrac%7B%5Cbar%20x%20-%2060%20%7D%7B0.8%7D%7D%7D)
![2.575*0.8= {\bar x - 60 }{}}}](https://tex.z-dn.net/?f=2.575%2A0.8%3D%20%7B%5Cbar%20x%20-%2060%20%7D%7B%7D%7D%7D)
![2.06= {\bar x - 60 }{}}}](https://tex.z-dn.net/?f=2.06%3D%20%7B%5Cbar%20x%20-%2060%20%7D%7B%7D%7D%7D)
![\bar x = 60-(-2.06)](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20%2060-%28-2.06%29)
![\bar x = 60+2.06](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20%2060%2B2.06)
![\bar x = 62.06](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20%2062.06)
Thus; the probability of a type II error is determined as follows:
β = P (
)
![= P ( \dfrac{57.94 -62 }{\dfrac{4}{\sqrt{25}}}](https://tex.z-dn.net/?f=%3D%20P%20%28%20%5Cdfrac%7B57.94%20-62%20%7D%7B%5Cdfrac%7B4%7D%7B%5Csqrt%7B25%7D%7D%7D%3C%5Cdfrac%7B62.06%20-62%20%7D%7B%5Cdfrac%7B4%7D%7B%5Csqrt%7B25%7D%7D%7D%29)
![= P ( \dfrac{-4.06 }{0.8}}](https://tex.z-dn.net/?f=%3D%20P%20%28%20%5Cdfrac%7B-4.06%20%7D%7B0.8%7D%7D%3C%5Cdfrac%7B2.06%20%20%7D%7B0.8%7D%29)
= 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
)
Now ;
the critical value of β is :
![Z _{1- \beta} = 1.28](https://tex.z-dn.net/?f=Z%20_%7B1-%20%5Cbeta%7D%20%3D%201.28)
The required sample size to satisfy and the type II error probability is therefore determined as :
![n = [\dfrac{(Z_{1 - \alpha/2} + Z_{1 - \beta} ) \sigma }{\delta}]^2](https://tex.z-dn.net/?f=n%20%3D%20%5B%5Cdfrac%7B%28Z_%7B1%20-%20%5Calpha%2F2%7D%20%2B%20Z_%7B1%20-%20%5Cbeta%7D%20%29%20%5Csigma%20%7D%7B%5Cdelta%7D%5D%5E2)
![n = [\dfrac{(2.575+1.28 ) 4 }{2}]^2](https://tex.z-dn.net/?f=n%20%3D%20%5B%5Cdfrac%7B%282.575%2B1.28%20%29%204%20%7D%7B2%7D%5D%5E2)
![n = [\dfrac{(3.855 ) 4 }{2}]^2](https://tex.z-dn.net/?f=n%20%3D%20%5B%5Cdfrac%7B%283.855%20%29%204%20%7D%7B2%7D%5D%5E2)
![n = [\dfrac{(15.42 ) }{2}]^2](https://tex.z-dn.net/?f=n%20%3D%20%5B%5Cdfrac%7B%2815.42%20%29%20%7D%7B2%7D%5D%5E2)
n = 7.71 ²
n= 59.4441
Thus; the required sample size to satisfy and the type II error probability is 59.4441