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?

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 = 7.71 ²

n= 59.4441

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

3-4

step-by-step explanation:

the answer is 3-4 because both 45 and 60 has a multiple of 15.

45/3=15

60/4=15

therefore the odds against selecting a red is 4-3

$6.42

step-by-step explanation: