Question

Is the mean bacteria colonies count in the west basin of Lake Macatawa less than 10? Suppose 34 samples of water were taken from the west basin. The mean colonies count in these samples was 11.95 with the variance 153.76. What is the power of the test if the true mean count is 7.4?

a) 0.41

b) 0.62

c) 0.06

d) 0.84

e) none of the answers provided here

Answer 1

Given that,
Standard deviation, σ =12.4
Sample Mean, X =11.95
Null, H0: μ=10
Alternate, H1: μ<10
Level of significance, α = 0.05
From Standard normal table, Z α/2 =1.6449
Since our test is left-tailed
Reject Ho, if Zo < -1.6449 OR if Zo > 1.6449
Reject Ho if (x-10)/12.4/√(n) < -1.6449 OR if (x-10)/12.4/√(n) > 1.6449
Reject Ho if x < 10-20.4/√(n) OR if x > 10-20.4/√(n)
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Suppose the size of the sample is n = 34 then the critical region
becomes,
Reject Ho if x < 10-20.4/√(34) OR if x > 10+20.4/√(34)
Reject Ho if x < 6.5 OR if x > 13.5
Implies, don't reject Ho if 6.5≤ x ≤ 13.5
Suppose the true mean is 7.4
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(6.5 ≤ x ≤ 13.5 | μ1 = 7.4)
= P(6.5-7.4/12.4/√(34) ≤ x - μ / σ/√n ≤ 13.5-7.4/12.4/√(34)
= P(-0.42 ≤ Z ≤2.87 )
= P( Z ≤2.87) - P( Z ≤-0.42)
= 0.9979 - 0.3372 [ Using Z Table ]
= 0.66
For n =34 the probability of Type II error is 0.66
power = 1- type 2 error
power = 1-0.66
power = 0.34

Answer:
option:e
None of the above answer
power = 0.34