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Topic (12) – HYPOTHESIS TESTING BASED ON A SINGLE SAMPLE
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Topic (12) – HYPOTHESIS TESTING BASED
ON A SINGLE SAMPLE
1) Testing The Population Proportion π
Let’s walk through one example, put the pieces into a
testing procedure for proportions and then use the
procedure in another example
EXAMPLE Parents of autistic children are often told
that their child is autistic around 1-2 years of age,
approximately the same age that children receive
their MMR vaccinations (mumps, measles and
rubella). As a result, some parents claim that the
vaccine caused the autism. To test this, a study was
done to compare the rate of autism in children who
receive the MMR vaccine to the known population
rate for children who do not receive the vaccine.
Among those who did not receive the vaccine, the
proportion of children with autism is 0.0021. In a
sample of 8,500 randomly selected children who did
receive the MMR vaccine, the proportion with autism
was .0028. Is this sufficient evidence to indicate that
the vaccine is related to autism?
Topic (12) – HYPOTHESIS TESTING BASED ON A SINGLE SAMPLE
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H0: π = 0.0021
HA: π > 0.0021
1.
Hypotheses:
2.
Significance level:
α = 0.03
3. If the Null Hypothesis is true, then the sampling
distribution of the sample proportion, p, obtained
from the experiment has a mean of
µ p = π = 0.0021
and a standard deviation of
σp =
π (1 − π )
= 0.00049567
n
Further the Central Limit Theorem says that p is
approximately Normally distributed if the sample
size is big enough.
The sample size is certainly big enough:
nπ = 8500(0.0021) = 17.85
n(1 − π ) = 8500 − 17.85 = 8472.15
So, let’s convert the observed sample proportion to a
z-score (we can interpret the difference more easily):
Topic (12) – HYPOTHESIS TESTING BASED ON A SINGLE SAMPLE
z=
p − µp
σp
=
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0.0028 − 0.0021
= 1.4122
0.00049567
This says that 0.0028 is 1.41 standard deviation units
above the hypothesized value of 0.0021. Is this very
likely if the null hypothesis is true? Is it supportive of
H0 or HA?
What is the probability that if we repeated the
experiment we’d see a z-score as unusual as this one,
i.e. one even more contradictory of Ho and supportive
of HA, when the null hypothesis is true.
I.e. How likely is it we’d get an even larger z-score
just by chance when the null hypothesis is true?
To do this we calculate Pr(Z>1.41).
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Pr(Z > 1.41)=1-Pr(Z ≤ 1.41)=1–0.9207 =0.0793
The probability that a random sample would yield a
sample proportion of 0.0028 or more by chance alone
when the null hypothesis is true is approximately 8%.
So, are the data sufficiently contradictory of H0 for us
to reject it?
Hold off an answer and get a few definitions first.
Defn: A TEST STATISTIC is the calculated
number on which the statistical test is based.
The P-VALUE (aka the OBSERVED SIGNIFICANCE LEVEL) of a test is the measure of how
contradictory the sample value is to the null
hypothesis while simultaneously supportive of the
alternative hypothesis. It is the probability, assuming
H0 is true, of obtaining a test statistic value at least as
inconsistent with H0 as the one actually observed.
For this example
Test statistic: z-score=1.41
P-value: Pr(z>1.41)=0.0793
Topic (12) – HYPOTHESIS TESTING BASED ON A SINGLE SAMPLE
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Determination of P-values
Each form of alternative hypothesis has a specific
p-value calculation:
1) 1-sided upper tail alternative
HA: parameter > hypothesized value
2) 1-sided lower tail alternative
HA: parameter < hypothesized value
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3) 2-sided test alternative
HA: parameter ≠ hypothesized value
To MAKE THE DECISION whether or not to
reject the null hypothesis, we use the following rule:
Reject H0 if the P-value ≤ α
Do not reject H0 if the P-value > α
In our example, the P-value = 0.0793 and α was set
to 0.03. Since 0.0793 > 0.03, we do not reject H0.
There is insufficient evidence to support the parents’
claim that the MMR vaccine is associated with
autism in children. This implies that the seemingly
higher rate is due to chance alone and not any firm
data.
Topic (12) – HYPOTHESIS TESTING BASED ON A SINGLE SAMPLE
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LARGE SAMPLE HYPOTHESIS TEST OF A
POPULATION PROPORTION:
Null hypothesis:
H0: π = πo where πo is the
hypothesized value
Alternative Hypothesis is one of three:
a) HA: π > πo
b) HA: π < πo
c) HA: π ≠ πo
p − πo
Test Statistic: z =
π o (1 − π o )
n
where n is the sample size and p is the sample
proportion
P-value: depends on the alternative hypothesis:
a) P-value = Pr( Z > z)
b) P-value = Pr( Z < z)
c) P-value = 2 Pr( Z < -|z|)
Decision Rule: reject Ho if P-value ≤ α
Assumptions:
1. n is large enough for p to be approximately
normally distributed ( nπo≥10 and n(1-πo)≥10 )
2. the sampling was random and not more than 5%
of the population.
Topic (12) – HYPOTHESIS TESTING BASED ON A SINGLE SAMPLE
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EXAMPLE The incidence rate of a certain type of
chromosome defect in adult males in the U.S. is
believed to be 1 in 80. A random sample of 1000 men
in prison revealed 20 men with defects. Is there
evidence to suggest that the rate for prisoners differs
from that in the general population? Use a
significance level of 0.05.
Null hypothesis:
H0: π = πo = 1/80 = .0125
Alternative Hypothesis: HA: π ≠ .0125
Check assumptions:
1) nπo = 1000(.0125) = 12.5 ≥10
n(1-πo)=987.5 ≥10
2) sampling was given to be random
Test Statistic:
p − πo
.02 − .0125
= 2.1347
z=
=
.0125(1 − .0125 )
π o (1 − π o )
1000
n
P-value:
2 Pr( Z < -|z|) = 2 Pr(Z<-2.13)
= 2(0.0166) = 0.0332
Conclusion: reject the null hypothesis since
0.0332<0.05=α. There is sufficient evidence based
on this sample to conclude that the population of
adult males in the U.S. penal system has a different
rate of a certain type of genetic defect than the
general adult male population.
Topic (12) – HYPOTHESIS TESTING BASED ON A SINGLE SAMPLE
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EXAMPLE Suppose a genetic crossing experiment
was performed. If independent sorting of the genes
occurs then it is expected that 25% of the offspring
would display a certain characteristic. If a particular
type of non-independent event occurs, the proportion
should be smaller. The experiment resulted in 50
plants out of 230 having the characteristic. Is this
sufficient evidence to reject independent sorting of
the genes?
Significance level: α=.10
Null hypothesis:
H0: π = πo = .25
Alternative Hypothesis: HA: π < .25
Check assumptions:
1) nπo = 230(.25) = 57.5 ≥10
n(1-πo)=172.5 ≥10
2) sampling?
Test Statistic:
p − πo
.217 − .25
z=
=
= −1.14
π o (1 − π o )
.25(1 − .25 )
230
n
P-value:
Pr( Z < z) = Pr(Z<-1.14)=0.1271
Conclusion: We fail to reject the null hypothesis
since 0.1271> α. There is insufficient evidence based
on this sample to conclude that the gene sorting is a
specific non-independent type of sort.
Topic (12) – HYPOTHESIS TESTING BASED ON A SINGLE SAMPLE
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2) Testing The Population Mean µ
To test a population mean, the test procedure is very
similar to the test for proportions except that we need
to use t-scores rather than z-scores.
Reason:
Recall that for these kinds of problems, a sample
provides a sample mean x and a sample standard
deviation s.
Further if the sample size n is sufficiently large
and sampling is random, a sample mean will be
distributed like a normal distribution.
x−µ
has a T-distribution on (n –1)
Hence, t =
s
n
degrees of freedom.
Otherwise the test procedure is quite similar.
The (obvious) other difference is in the hypotheses
themselves – we’re testing means and NOT
proportions now!
Topic (12) – HYPOTHESIS TESTING BASED ON A SINGLE SAMPLE
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LARGE SAMPLE HYPOTHESIS TEST OF A
POPULATION MEAN:
Null hypothesis:
H0: µ = µ o where µ o is the
hypothesized value
Alternative Hypothesis is one of three:
a) HA: µ > µ o
b) HA: µ < µ o
c) HA: µ ≠ µ o
x − µo
Test Statistic: t =
s
n
where n is the sample size and x is the sample mean
and s is the sample standard deviation
P-value: depends on the alternative hypothesis:
a) P-value = Pr( T > t)
b) P-value = Pr( T < t)
c) P-value = 2 Pr( T > |t|)
Decision Rule: reject Ho if P-value ≤ α
Assumptions:
1. n is large enough for x to be approximately
normally distributed ( n≥30)
2. the sampling was random and not more than 5%
of the population.
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Alternative Approach to a Decision Rule:
We have seen the p-value approach where we
compare the p-value of a test with α. We can instead
define an interval (or two) of values of the test
statistic that lead us to reject with a type I error rate
of α.
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Call the t-value that corresponds to t* the critical or
cutoff value for df = n-1 and the type I error rate = α.
To indicate that it’s value depends on the sample size
(through the df) and the choice of α, we denote the
cutoff value by
t*(α, n-1)
So, an alternative decision rule is:
Decision Rule: reject Ho if the test statistic, t,
a) t > t*(1-α, n-1)
b) t < - t*(1-α, n-1)
c) |t| > t*(1-α/2, n-1)
where t* is that value that makes the statement
Pt(T>t*) = α.
Example values:
t*(0.95, 9) = 1.833
- t*(0.90, 14) = - 1.345
t*(0.975, 21) = 2.0796
Topic (12) – HYPOTHESIS TESTING BASED ON A SINGLE SAMPLE
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EXAMPLE In an experiment with rats, a behavioral
scientist used an auditory signal to indicate that food
was available through an open door in the cage. The
scientist believed that it would take the rats less than
18 trials on average to learn to recognize the signal.
He ran the experiment with 23 rats and obtained the
following results: x = 16.261 trials to learn to go to
the open door with a standard deviation of s = 3.441
trials. Is there sufficient evidence to support his
contention?
1) state the hypotheses and choose an α:
Ho: µ = 18
α=0.10
HA: µ < 18
2) review the assumptions:
Each rat tested independently of the other rats?
The rats constitute a random sample of rats from the
population of interest?
Is a sample of n=23 sufficiently large to invoke the
CLT and use a T-test?
Let’s look at the dot plot for this data before
deciding
Topic (12) – HYPOTHESIS TESTING BASED ON A SINGLE SAMPLE
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9
8
7
Count
6
5
4
3
2
1
0
5
10
15
20
NUMTRIALS
25
3) do the calculations:
Here are the results as given by SYSTAT v. 9
Mean
t
= 16.261
= -2.424
SD
df
= 3.441
= 22
2-sided P-value = 0.024
Let’s check these numbers:
df = n-1 = 23-1 = 22
ts =
x − µ o 16.26 − 18
=
= −2.424
s
3.44
n
23
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The software prints out the 2-sided P-value but we
have a one-sided test so we need to divide the given
P-value by 2 to get the right P-value for our test:
our p-value =0.024 / 2 = 0.012 < 0.10 = α.
4) Draw a conclusion
Since our P-value = 0.012 < 0.10 = α, we reject Ho
and conclude that the data provide sufficient evidence
that the average number of trials required for a rat to
recognize an auditory signal that food is available is
less than 18.
The alternative approach for a decision is to use the
critical value method:
Need – t*(0.90, 22) = - 1.322 .
Since t = - 2.424 < - 1.322, we reject the null
hypothesis.
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EXAMPLE Wastewater discharged from a dairy
processing plant is high in bicarbonates and calcium.
As a result it is thought that irrigating fields of kidney
bean plants with the discharge will promote growth.
An experiment was done to test this. 40 plants were
irrigated with a 50% mix of discharge and water,
resulting in a mean root length of 5.46 cm and a
standard deviation of 0.55 cm. It is known that
average root growth with regular feeding is 5.20 cm.
Is there sufficient evidence to suggest that irrigation
with the discharge effects average root length? Use a
0.025 level test.
Significance level:
α=.025
Null hypothesis:
Ho: µ = 5.20
Alternative Hypothesis: HA: µ > 5.20
Check assumptions:
1) n large enough?
2) sampling done appropriately?
Topic (12) – HYPOTHESIS TESTING BASED ON A SINGLE SAMPLE
Test Statistic:
ts =
P-value:
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x − µo 5.46 − 5.20
=
= 2.99
s
0.55
40
n
Pr( T> t) = Pr(T>3)=0.002
Conclusion: We reject the null hypothesis since
0.002< α=0.025. There is strong evidence to suggest
that irrigating kidney bean plants with 50% mix of
discharge from the dairy processing plant increases
average root lengths.
3) Testing the Population Variance σ 2
To test a population variance, the test procedure is
very similar to the test for the mean except that we
need to use chi-square scores rather than t-scores.
Recall that χ 2 =
(n − 1)s 2
σ
2
has a Chi-Square
distribution on (n – 1) degrees of freedom when the
population from which we sampled has a Normal
Topic (12) – HYPOTHESIS TESTING BASED ON A SINGLE SAMPLE
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distribution with a mean µ and variance σ 2 . We can
use this to test hypotheses about that σ 2 .
HYPOTHESIS TEST OF A POPULATION
VARIANCE:
Null hypothesis:
H0: σ 2 = σ 02 where σ 02 is the
hypothesized value
Alternative Hypothesis is one of three:
a) HA: σ 2 > σ 02
b)
HA: σ 2 < σ 02
c)
HA: σ 2 ≠ σ 02
2
Test Statistic: χ =
( n − 1) s 2
σ 02
where n is the sample size and s 2 is the sample
variance
Decision Rules:
a) Reject H0 if χ 2 > χU2 , the upper-tail value for
chosen α and df = n – 1. and df = n – 1.
b) Reject H0 if χ 2 < χ L2 , the lower-tail value for 1 – α
and df = n – 1.
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c) Reject H0 if χ 2 > χU2 , the upper-tail value for α/2
or if χ 2 < χ L2 , the lower-tail value for 1 – α/2 and df
= n – 1.
Assumptions:
The population from which the sample was taken is
approximately Normally distributed and the sampling
was done randomly.
EXAMPLE The USGS maintains historical records
of water flow into various bodies of water in the
United States.
Of interest is whether climate change is affecting the
variability in water flow in recent years, i.e. the
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effects of warming may be in the variability of events
rather than through a shift in the mean rate.
So, let’s test the hypothesis that the flow rate
variance has increased above the historical level.
2
2
versus HA: σ 2 > σ historical
H0: σ 2 = σ historical
Historically (prior to 1990), the variability in flow
2
= 277x106 cfs.
rates was σ historical
Recent data from 1993 to 2003 are given in the
following table:
Year
1993
1994
1995
1996
1997
1998
1999
2000
2001
2002
2003
CFS
95600
97200
59400
137800
63000
92600
45700
72100
39500
47200
135300
Topic (12) – HYPOTHESIS TESTING BASED ON A SINGLE SAMPLE
CFS
25000
Moments
Mean
Std Dev
Std Err Mean
upper 95% Mean
lower 95% Mean
N
Sum Wgt
Sum
Variance
Skewness
Kurtosis
CV
N Missing
75000100000
150000
80490.909
34296.631
10340.823
103531.7
57450.119
11
11
885400
1.17626e9
0.6045854
-0.751232
42.609323
0
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Topic (12) – HYPOTHESIS TESTING BASED ON A SINGLE SAMPLE
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Test Standard Deviation=value
Hypothesized Value
16643.3 = sqrt( 277x106)
Actual Estimate
34296.6
df
10
Test Statistic
Prob > |ChiSq|
Prob < ChiSq
Prob > ChiSq
ChiSquare
42.4643
<.0001
1.0000
<.0001
The software provides
the p-values so that
we do not need to find
the critical values for
making decisions. We
can use the p-values
directly!
2
⎛ 34296.6 ⎞
2
Check the calculations: χ = 10⎜
⎟ = 42.464 .
⎝ 16643.3 ⎠
So, for an upper tail test, the p-value is < 0.0001 which is
smaller than any reasonable α we might choose, so we
reject the null hypothesis and conclude that variance in
flow rates has increased based on the data for the last ten
years.