A dry cleaning establishment claims that a new spot remover will remove more than 70% of the spots to which it is applied. To check this claim, the spot remover will be used on 15 spots chosen at random. If fewer than 11 of the spots are removed, we shall not reject thenull hypothesis that p= 0.7; otherwise, we conclude that p > 0.7. Required:
a. Evaluate α, assuming that p = 0 .7.
b. Evaluate β for the alternative p = 0.9.

α = 1 - 0,915 = 0.085

β = 0.314

Step-by-step explanation:

Dry cleasning clain that spot remover will remove the 70% of spot. The claim support the spot and chose at random.

Null hypothesis:

H0 : p = 0.7

If 11 of spot are removed then hypothesis rejected.

If 11 spots are not removed then hypothesis accepted

H1 : p >0/7

X is the number of spot and binomial random variable.

For a)

The probability of committing the type 1 error use alpha such as

α = P ( type I error)

where X > 10

and p = 0.7

for the given values where n = 12 and r = 10 and p - 0.7

we get

α = 1 - 0,915 = 0.085

b)

The nonrejection of the hypothesis where type II error is false .

The hypothesis for this is 0.9

This is Type II error

β = P ( type II error) = P ( X <= 10 where p = 0.9 )

Where n = 12 , r = 10 and p = 0.9

then

β = 0.314