Page 389 - DECO504_STATISTICAL_METHODS_IN_ECONOMICS_ENGLISH

Statistical Methods in Economics

Notes The complement −1 β of β , i.e. the probability of Type-II error, is called the power of a statistical test
because it measures the probability of rejecting H when it is true.
0
For example, suppose null and alternative hypotheses are stated as.
H μ = 80 and H μ = 80
0: 1:
Power of a test: The ability (probability) of a test to reject the null hypothesis when it is false.
Often, when the null hypothesis is false, another alternative value of the population mean, μ is
unknown. So for each of the possible values of the population mean μ , the probability of committing
Type II error for several possible values of μ is required to be calculated.
Suppose a sample of size n = 50 is drawn from the given population to compute the probability of
committing a Type II error for a specific alternative value of the population mean, μ . Let sample
mean so obtained be x = 71 with a standard deviation, s = 21. For significance level, α = 0.05 and a
two-tailed test, the table value of z = ± 1.96. But the deserved value from sample data is
0.05
x − μ 71 − 80
z = σ x = 21/ 50 = – 3.03

Since z = – 3.03 value falls in the rejection region, the null hypothesis H is rejected. The rejection of
cal
0
null hypothesis, leads to either make a correct decision or commit a Type II error. If the population
mean is actually 75 instead of 80, then the probability of commiting a Type II error is determined by
computing a critical region for the mean x . This value is used as the cutoff point between the area
c
of acceptance and the area of rejection. If for any sample mean so obtained is less than (or greater
than for right-tail rejection region), x , then the null hypothesis is rejected. Solving for the critical
c
value of mean gives
x c − μ x c − 80
z = or ± 1.96 =
c σ x 21/ 50
x = 80 ± 5.82 or 74.18 to 85.82
c
If μ = 75, then probability of accepting the false null hypothesis H : μ = 80 when critical value is
0
falling in the range x = 74.18 to 85.82 is calculated as follows:
c
74.18 − 75
z = = – 0.276
1 21/ 50
The area under normal curve for z = – 0.276 is 0.1064.
1
85.82 − 75
z = = 3.643
2 21 50
The area under normal carve for z = 3.643 is 0.4995
2
β
Thus the probability of committing a Type II error () falls in the region:
x
β = P (74.18 < c < 85.82) = 0.1064 + 0.4995 = 0.6059
The total probability 0.6059 of committing a Type II error () is the area to the right of x = 74.18 in
β
c
the distribution. Hence the power of the test is −1 β = 1 – 0.6059 = 0.3941 as shown in Figure 1 (b).

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