Question

A manufacturer of sprinkler systems used for fire protection in office buildings claims that the true average system-activation temperature is 130oF. A sample of n = 9 systems, when tested, yields a sample average activation temperature of 131.08oF. Suppose the distribution of activation temperature is normal with standard deviation 3.0 oF. Denote the true average system-activation temperature by µ ( oF). Consider testing H0 : µ = 130 versus Ha : µ 6= 130 at the significance level α = 0.05.

(a) (7 points) What is β(131)?

(b) (7 points) For this level 0.05 test, in order to control the Type II error probability at 0.01, that is to make β(131) ≤ 0.01, how many sprinkler systems should be used?

Answer 1

a.
Given that,
Standard deviation, σ =3
Sample Mean, X =131.08
Null, H0: μ<130
Alternate, H1: μ>130
Level of significance, α = 0.05
From Standard normal table, Z α/2 =1.6449
Since our test is right-tailed
Reject Ho, if Zo < -1.6449 OR if Zo > 1.6449
Reject Ho if (x-130)/3/√(n) < -1.6449 OR if (x-130)/3/√(n) > 1.6449
Reject Ho if x < 130-4.9347/√(n) OR if x > 130-4.9347/√(n)
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Suppose the size of the sample is n = 9 then the critical region
becomes,
Reject Ho if x < 130-4.9347/√(9) OR if x > 130+4.9347/√(9)
Reject Ho if x < 128.3551 OR if x > 131.6449
Implies, don't reject Ho if 128.3551≤ x ≤ 131.6449
Suppose the true mean is 131
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(128.3551 ≤ x ≤ 131.6449 | μ1 = 131)
= P(128.3551-131/3/√(9) ≤ x - μ / σ/√n ≤ 131.6449-131/3/√(9)
= P(-2.6449 ≤ Z ≤0.6449 )
= P( Z ≤0.6449) - P( Z ≤-2.6449)
= 0.7405 - 0.0041 [ Using Z Table ]
= 0.7364
For n =9 the probability of Type II error is 0.7364

b.
type 2 error is 0.01
level of significance=0.05
sample size is needed,
true mean 131
n= ((σ(Zalpha +Zbeta))/(U-Uo))^2
n =((9*(Z0.05+Z0.01)/(131-130))^2
n= ((9*(1.64+2.32))/(131-130))^2
n =1270.2096
n= 1271