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:

Alternative hypothesis

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:







The z statistics for the upper limit is:








Thus; the probability of a type II error is determined as follows:
β = P (
)


= 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 :

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