Question

An ordinance requiring that a smoke detector be installed in all previously constructed houses has been in effect in a particular city for 1 year. The fire department is concerned that many houses remain without detectors. Let p = the true proportion of such houses having detectors, and suppose that a random sample of 20 homes is inspected. If the sample strongly indicates that fewer than 80% of all houses have a detector, the fire department will campaign for a mandatory inspection program. Because of the costliness of the program, the department prefers not to call for such inspections unless sample evidence strongly argues for their necessity. Let X denote the number of homes with detectors among the 20 sampled. Consider rejecting the claim that

p ≥ 0.8

if

x ≤ 12.

(Round your answers to three decimal places.)

(a)

What is the probability that the claim is rejected when the actual value of p is 0.8?

(b)

What is the probability of not rejecting the claim when

p = 0.7?

When

p = 0.6?

p = 0.7 p = 0.6

(c)

How do the "error probabilities" of parts (a) and (b) change if the value 12 in the decision rule is replaced by 11?

What is the probability that the claim is rejected under the new decision rule when the actual value of p is 0.8?

What is the probability of not rejecting the claim under the new decision rule when

p = 0.7?

When

p = 0.6?

p = 0.7 p = 0.6

Answer 1

Solution :

a)

P(rejected | p=0.8) =P(X<=12| p=0.8)=

∑x=0a     (nCx)px(1−p)(n-x)    =

0.032

b)

P(not rejecting | p=0.7) =P(X>=13|p=0.7)=1-P(X<=12)=

1-∑x=0x-1   (nCx)px(q)(n-x) =

0.772

P(not rejecting | p=0.6) =P(X>=13|p=0.6)=1-P(X<=12)=

1-∑x=0x-1   (nCx)px(q)(n-x) =

0.416

c)

P(rejected | p=0.8) =P(X<=11| p=0.8)=

∑x=0a     (nCx)px(1−p)(n-x)    =

0.010

P(not rejecting | p=0.7) =P(X>=13|p=0.7)=1-P(X<=12)=

1-∑x=0x-1   (nCx)px(q)(n-x) =

0.887

P(not rejecting | p=0.6) =P(X>=13|p=0.6)=1-P(X<=12)=

1-∑x=0x-1   (nCx)px(q)(n-x) =

0.596

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