type of hypothesis test being evaluated. This is why many statisticians always recommend a two
tailed alternative, because you have made a stronger case for rejecting H
0.
Probability Values (PValues) and Hypothesis Testing
The level of significance chosen is rather arbitrary. For that reason, most statisticians use the
concept of probability values (
pvalues)
to report the outcome of a hypothesis test. The pvalue
for a
hypothesis test is the smallest value of α for which the data indicates rejection of H
0
. Thus,
to reject the null hypothesis the pvalue must be <=
α.
If it is >
α, we fail to reject the
H
0
.
H
a
: μ < μ
0
H
0:
μ = μ
0
H
a
: μ
≠
μ
0
𝑥
̅
−
μ
0
 >= 1.96σ/
√𝑛
5 
P a g e
The pvalue may also be interpreted as the probability of observing a value of the test statistic at
least as extreme as the observed value of the test statistic if H
0
is true
. In other words, it’s the
probability of incorrectly rejecting the null hypothesis.
If we let
𝑿
̅
represent the random variable for the sample mean under H
0
and x be the observed
value of
𝑥̅
, then the pvalue for the one sample Ztest is computed as follows:
PValues for One Sample ZTest
All probabilities are computed under the assumption that H
0
is true.
In our Cooley High School example, the pvalue for the twotailed test is:
2*Prob(
𝑿
̅
75>=4) = 2*Prob(
𝑿
̅
>=79),
which can be computed as the following in Excel:
2*(1NORM.DIST(79,75,15/
√49
), True) = 2*(0.030974) = 0.061948.
Because our pvalue of 0.06 > our alpha value of .05, we fail to reject H
0
.
The pvalue for a onetailed test is:
Prob(
𝑿
̅
>=79) = 0.030974.
Because our pvalue of 0.03 < than our alpha value of .05, we reject H
0
.
One Sample Hypothesis for Mean: Small Sample, Normal Population,
Variance Unknown
Suppose we are interested in testing a hypothesis about the mean of a normal population where
the population variance is unknown, and the sample size n is <30. Then, it can be shown that
(
𝑥̅

µ
)/(s/√n) follows a t
distribution with n1 degrees of freedom. Here s = sample standard
deviation.
Like the standard normal distribution, the tdistribution has a density symmetric around 0. As
shown below, the tdistribution has fatter tails than the standard normal density, and as n
increases the tdistribution approaches the standard normal density.
Hypotheses
PValue
H
0:
μ <= μ
0
H
a
: μ > μ
0
Prob(
𝑿
̅
>= x)
H
0:
μ >= μ
0
H
a
: μ < μ
0
Prob(
𝑿
̅
<= x)
H
0:
μ = μ
0
H
a
: μ
≠
μ
0
Prob(
𝑿
̅
− µ
0
 ≥ 𝑥
)
6 
P a g e
As shown in the
Hypothesis Testing.xlsx
spreadsheet, T random variable worksheet shows the
percentiles of the tdistribution and how they can be computed using the T.INV function. We let
t
(α,n
1)
represent the α percentile of a t
distribution with n1 degrees of freedom. Below we find,
for example, t
(.025,28)
= 2.04841.
Basically, one sample ttests look just like one sample z
tests with s replacing σ and the t
percentiles replacing the z percentiles.
Critical Region for One Sample ttests
If we let
T
n1
stand for a tdistribution with n1 df and t represent the observed value of (
𝑥̅

µ
)/(s/√n)
, then the pvalues for a one sample ttest may be computed as follows:
PValues for One Sample tTest
0
0.1
0.2
0.3
0.4
0.5
4
3
2
1
0
1
2
3
4
T and Normal Densities
Normal
T 5 df
T 15 df
T 30 df
6
7
8
9
D
E
F
G
H
2.5 %ile 28 df
2.04841 =T.INV(0.025,28)
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 Summer '16
 naveen rathi
 Normal Distribution, Variance, Null hypothesis, Hypothesis testing, Statistical hypothesis testing