Question

Suppose the average lifetime of a certain type of car battery is known to be 60 months. Consider conducting a two-sided test on it based on a sample of size 25 from a normal distribution with a population standard deviation of 4 months.a) If the true average lifetime is 62 months and a =0.01, what is the probability of a type II error?b) What is the required sample size to satisfy and the type II error probability of b(62) = 0.1?

Answer 1

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 $\sigma$ = 4

true average lifetime = Sample Mean $\bar X$ = 62

We can state our null hypothesis and alternative hypothesis as follows:

Null hypothesis:

$\mathbf{H_o : \mu = 60}$

Alternative hypothesis

$\mathbf{H_1 : \mu \neq 60}$

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:

$lower \ limit = \dfrac{\bar X - \mu }{\dfrac{\sigma}{\sqrt {n}}}$

$-2.575= \dfrac{\bar x - 60 }{\dfrac{4}{\sqrt 25}}}$

$-2.575= \dfrac{\bar x - 60 }{0.8}}}$

$-2.575*0.8= {\bar x - 60 }{}}}$

$-2.06= {\bar x - 60 }{}}}$

$\bar x = 60-2.06$

$\bar x = 57.94$

The z statistics for the upper limit is:

$lower \ limit = \dfrac{\bar X - \mu }{\dfrac{\sigma}{\sqrt {n}}}$

$2.575= \dfrac{\bar x - 60 }{\dfrac{4}{\sqrt 25}}}$

$2.575= \dfrac{\bar x - 60 }{0.8}}}$

$2.575*0.8= {\bar x - 60 }{}}}$

$2.06= {\bar x - 60 }{}}}$

$\bar x = 60-(-2.06)$

$\bar x = 60+2.06$

$\bar x = 62.06$

Thus; the probability of a type II error is determined as follows:

β = P (  $57.94 < \bar x < 62.06$ )

$= P ( \dfrac{57.94 -62 }{\dfrac{4}{\sqrt{25}}}$

$= P ( \dfrac{-4.06 }{0.8}}$

= 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   $Z_{1 - \alpha/2 } = 2.575$  )

Now ;

the critical value of β is :

$Z _{1- \beta} = 1.28$

The  required sample size to satisfy and the type II error probability  is therefore determined as :

$n = [\dfrac{(Z_{1 - \alpha/2} + Z_{1 - \beta} ) \sigma }{\delta}]^2$

$n = [\dfrac{(2.575+1.28 ) 4 }{2}]^2$

$n = [\dfrac{(3.855 ) 4 }{2}]^2$

$n = [\dfrac{(15.42 ) }{2}]^2$

n = 7.71 ²

n= 59.4441

Thus; the required sample size to satisfy and the type II error probability  is 59.4441