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In general, without changing the sample size or the type of the test of the hypothesis, a decrease in α causes an increase in β , and a decrease in β causes an increase in α . Both probabilities α and β of the two types of errors can be decreased only by increasing the sample size or, in some way, constructing a better test of the hypothesis.

Example

If n =100 and we desire a test with significance level α =0.05, then α = P ( X ¯ c ; μ = 60 ) = 0.05 means, since X ¯ is N( μ ,100/100=1) ,

P ( X ¯ 60 1 c 60 1 ; μ = 60 ) = 0.05 and c 60 = 1.645 . Thus c =61.645. The power function is

K ( μ ) = P ( X ¯ 61.645 ; μ ) = P ( X ¯ μ 1 61.645 μ 1 ; μ ) = 1 Φ ( 61.645 μ ) .

In particular, this means that β at μ =65 is = 1 K ( μ ) = Φ ( 61.645 65 ) = Φ ( 3.355 ) 0 ; so, with n =100, both α and β have decreased from their respective original values of 0.1587 and 0.0668 when n =25. Rather than guess at the value of n , an ideal power function determines the sample size. Let us use a critical region of the form x ¯ c . Further, suppose that we want α =0.025 and, when μ =65, β =0.05. Thus, since X ¯ is N( μ ,100/n) ,

0.025 = P ( X ¯ c ; μ = 60 ) = 1 Φ ( c 60 10 / n ) and 0.05 = 1 P ( X ¯ c ; μ = 65 ) = Φ ( c 65 10 / n ) .

That is, c 60 10 / n = 1.96 and c 65 10 / n = 1.645 .

Solving these equations simultaneously for c and 10 / n , we obtain c = 60 + 1.96 5 3.605 = 62.718 ; 10 n = 5 3.605 .

Thus, n = 7.21 and n = 51.98 . Since n must be an integer, we would use n =52 and obtain α =0.025 and β =0.05, approximately.

For a number of years there has been another value associated with a statistical test, and most statistical computer programs automatically print this out; it is called the probability value or, for brevity, p -value . The p -value associated with a test is the probability that we obtain the observed value of the test statistic or a value that is more extreme in the direction of the alternative hypothesis, calculated when H 0 is true. Rather than select the critical region ahead of time, the p -value of a test can be reported and the reader then makes a decision.

Say we are testing H 0 μ =60  against H 1 μ >60 with a sample mean X ¯ based on n =52 observations. Suppose that we obtain the observed sample mean of x ¯ = 62.75 . If we compute the probability of obtaining an x ¯ of that value of 62.75 or greater when μ =60, then we obtain the p -value associated with x ¯ = 62.75 . That is,

p v a l u e = P ( X ¯ 62.75 ; μ = 60 ) = P ( X ¯ 60 10 / 52 62.75 60 10 / 52 ; μ = 60 ) = 1 Φ ( 62.75 60 10 / 52 ) = 1 Φ ( 1.983 ) = 0.0237.

If this p -value is small, we tend to reject the hypothesis H 0 μ =60  . For example, rejection of H 0 μ =60  if the p -value is less than or equal to 0.025 is exactly the same as rejection if x ¯ = 62.718 .That is, x ¯ = 62.718 has a p -value of 0.025. To help keep the definition of p -value in mind, we note that it can be thought of as that tail-end probability , under H 0 , of the distribution of the statistic, here X ¯ , beyond the observed value of the statistic. See Figure 1 for the p -value associated with x ¯ = 62.75.

The p -value associated with x ¯ = 62.75.

Suppose that in the past, a golfer’s scores have been (approximately) normally distributed with mean μ =90 and σ 2 =9. After taking some lessons, the golfer has reason to believe that the mean μ has decreased. (We assume that σ 2 is still about 9.) To test the null hypothesis H 0 μ =90  against the alternative hypothesis H 1 μ < 90  , the golfer plays 16 games, computing the sample mean x ¯ .If x ¯ is small, say x ¯ c , then H 0 is rejected and H 1 accepted; that is, it seems as if the mean μ has actually decreased after the lessons. If c =88.5, then the power function of the test is

K ( μ ) = P ( X ¯ 88.5 ; μ ) = P ( X ¯ μ 3 / 4 88.5 μ 3 / 4 ; μ ) = Φ ( 88.5 μ 3 / 4 ) .

Because 9/16 is the variance of X ¯ . In particular, α = K ( 90 ) = Φ ( 2 ) = 1 0.9772 = 0.0228.

If, in fact, the true mean is equal to μ =88 after the lessons, the power is K ( 88 ) = Φ ( 2 / 3 ) = 0.7475 . If μ =87, then K ( 87 ) = Φ ( 2 ) = 0.9772 . An observed sample mean of x ¯ = 88.25 has a

p v a l u e = P ( X ¯ 88.25 ; μ = 90 ) = Φ ( 88.25 90 3 / 4 ) = Φ ( 7 3 ) = 0.0098 ,

and this would lead to a rejection at α =0.0228 (or even α =0.01).

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research.net
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Source:  OpenStax, Introduction to statistics. OpenStax CNX. Oct 09, 2007 Download for free at http://cnx.org/content/col10343/1.3
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