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Statistics 305A
Section 6.2. Large-Sample Significance Tests for a
Population Mean
The significance test is used to decide whether the evidence provided by a random sample of size
n indicates that µ is most likely a given value # or that µ is most likely in a region not including
#. The structure features two hypotheses, a null hypothesis H 0 : µ = # , and an alternative
hypothesis Ha. Three different forms of Ha are defined.
Forms of Hypotheses
1. H0 : µ = #, Ha : µ < #
2. H0 : µ = #, Ha : µ > #
3. H0 : µ = #, Ha : µ ≠ #
If, based on the sample information, it appears most likely that µ < # when the test has form 1,
then we reject H0 and say the evidence favors Ha : µ < #.
The significance test is carried out in the following 5 steps for tests about µ.
Step 1. State the null hypothesis H0 : µ = #.
Step 2. State the alternative hypothesis Ha : (one of 3 forms).
Step 3. State that the test statistic is z = ( x − # ) / ( s / n ) , the test distribution is N(0, 1), and
that an extreme value of z (specify meaning for the form of Ha used) will constitute
evidence against the null hypothesis.
Step 4. Give the sample mean x0 , sample standard deviation s 0 and compute the value of the
test statistic
z0 =
( x0 − # )
s0 / n
Step 5. Compute and report the p-value, and give your conclusion (either reject H0 or fail to
reject H0).
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The p-value of the test is the probability P ( Z ≤ z 0 ) for form 1, P ( Z ≥ z 0 ) for form 2, and
P ( | Z | ≥ | z 0 | ) for form 3, where Z ~ N (0, 1) . We reject H0 if the p-value is small. (Note:
P ( | Z | ≥ | z 0 | ) = P ( Z > z 0 ) + P ( Z < − z 0 ) when z 0 ≥ 0 ).
Example:
Suppose we wish to test
H 0 : µ = 10
H a : µ < 10
versus
and the sample of n = 25 yields x0 = 8 and s 0 = 5 . The 5-step test procedure is:
Step 1.
H 0 : µ = 10
Step 2.
H a : µ < 10
Step 3. The test statistic is z = ( x − 10) / ( s / n ) . The distribution is Z ~ N (0, 1) . A value of
z substantially less than zero will be evidence against H 0 .
Step 4. n = 25, x0 = 8 , s 0 = 5
z0 =
8 − 10
5 / 25
= −2
Step 5.
(The value of the test statistic).
P ( Z ≤ −2) = 0.0228 is the p-value. The p-value is small so reject H 0 . The evidence
indicates µ < 10 .
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What is the basis for this test procedure? What is the logic behind it? Consider the following
for the test form H 0 : µ = # versus H a : µ < # , (form #1).
We expect that a sample mean x will be “rather close” to the population mean µ, and that
z = ( x − µ ) / ( s / n ) will therefore be “rather close” to zero. Let us assume µ = # . Then we
expect z = ( x − # ) / ( s / n ) will be “rather close” to zero when we get a sample mean x and
sample standard deviation s. The random variable Z = ( X − # ) / ( S / n ) is approximately
distributed N(0, 1), for large n, when µ = # . Suppose the sample yields x0 and s 0 so that
z 0 = ( x0 − # ) / ( s 0 / n ) .
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Then we reason as follows based on the p-value P ( Z ≤ z 0 ) :
(a)
If P ( Z ≤ z 0 ) is small, then a very unusual event has occurred if indeed µ = # . But if
µ = µ a << # then P ( Z ≤ ( x0 − µ a ) /( s0 / n ) ) is not so small and a very unusual
event will not have occurred. Small p-value is evidence that our assumption that
µ = # is not good, so we reject H0, and conclude the evidence favors Ha .
(b)
If P ( Z ≤ z 0 ) is not small, then z0 is about what we expect from a random sample when
µ ≥ # and our assumption that H0 is true (i.e., that µ = # ) led us to a usual event. We
fail to reject H0 because there is no evidence to support rejecting it. This does not
mean we believe H 0 is true. Rather, we believe that our sample is not large enough to
detect a difference between µ and #.
Thus, generally, the logic is that we assume H0 is true and see if the sample yields an outcome
that has small probability (reject H0) or not so small probability (fail to reject H0). We are at
liberty to decide what is a small p-value. People tend to call 0.05 or less a small value.
Suppose a different test form is considered (i.e., not form #1). If the form is H 0 : µ = # vs.
H a : µ > # then in Step 5 we would compute P ( Z ≥ z 0 ) as the p-value. If the form is
H 0 : µ = # vs. H a : µ ≠ # then the p-value is P ( | Z | ≥ | z 0 | ) . Always, a small p-value causes
rejection of H 0 : µ = # and the conclusion that the evidence supports H a . If the p-value is
not small, we do not reject H 0 and conclude that the evidence does not support H a .
Finally, consider a hypothetical example to illustrate the logic of significance tests generally.
Suppose you decide to drive your car from Ames to Des Moines. You head down the highway,
proceeding under the null hypothesis
H0: I am heading toward Des Moines.
Your alternative is
Ha: I am not heading toward Des Moines.
You see a sign on the highway and read MINNESOTA STATE LINE 50 MILES STRAIGHT
AHEAD. This permits you to test your null hypothesis. You say to yourself “Self, the
probability of seeing that sign when my null hypothesis is true is very small. The evidence
suggests that I should reject H0 in favor of Ha.” The sample information here is provided by the
text of the sign.
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