Tests of Hypotheses

1
Tests of Hypotheses
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Statistical hypothesis

• Statistical hypothesis-
• Statement about a feature of the population
  (e.g. the mean)
• Examples
• - Mean temperature of healthy adults is 98.6F
  (37c).
• - A certain medication contains a mean of 245
  ppm of a particular chemical.
• - Mean number of people that enter a certain
  restaurant in a day is 125.

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Example soft drink bottles

• A firm that produces a certain soft drink prints
  on each bottle that it contains 24 oz of drink.
  It has been suspected that the mean amount per
  bottle is less than 24 oz .
• In order to examine this claim, a sample of 100
  bottles has been taken and the mean amount per
  bottle was found to be 23.4 oz.
• Assume that the standard deviation of the
  contents of the drink in the bottles is s3 oz.
• Is this an indication that the mean amount of
  drink in a bottle is less than 24 oz?

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Set hypotheses

• Two types of hypotheses
• H0 - the null hypothesis
• Common beliefs, the claims that are assumed to
  be true up-to-date
• H0 - Mean contents of soft drink bottle, µ, is
  24 oz
• (µ24)
• H1 - the alternative hypothesis
• Alternative claims that come to challenge the
  common beliefs
• H1 mean contents of soft drink bottle, µ, is
  less than 24 oz
• (µlt24)

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Use the sample results to test the hypotheses
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Sample results
If is (i) small enough (ii) large
enough we will reject H0.
H0 µ24 H1 µlt24
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How likely are we to observe such result from a
population with mean µ24?
The distribution of
µ24 oz
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We are not very likely to observe such result
from a population with µ24 (prob0.028)
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Test statistic

• Z -2
• is an example of a test statistic
• It measures the distance of the sample results
  from what is expected if H0 is true.

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Is the value Z-2 unusual under H0?
0.0228
µ24 oz
There is only 0.0228 chance of getting values
smaller than Z-2 if H0 is true
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P-value

• The probability of getting an outcome as extreme
  or more extreme than the observed outcome.
• Extreme far from what we would expect if H0
  were true.
• The smaller the p-value, the stronger the
  evidence against H0.

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Level of significance

• a significance level.
• It is the chance we are ready to take for
  rejecting H0 while in fact H0 is true
• if p-value a, we say that we reject H0 at the a
  significance level.
• Typically, a is taken to be 0.05 or 0.01

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• In the bottles example
• If we required a significance level of a0.05
  then we would reject H0
• p-value0.0228lt0.05
• However, if we required a significance level of
  a0.01 then we would not reject H0

0.028
µ24
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Example sales of coffee

• Weekly sales of regular ground coffee at a
  supermarket have in the recent past varied
  according to a normal distribution with mean
  µ354 units per week and standard deviation s33
  units. The store reduces the price by 5. Sales
  in the next three weeks are 405, 378, and 411
  units. Is this good evidence, at the 5
  significance level, that average sales are now
  higher?
• Hypotheses
• H0
• H1
• Sample mean

µ354
µgt354
How far is from what we expect if H0 is true?
15
µ354

• Test statistic
• P-value
• p-valueprobability of getting values that are
  more extreme than the test statistic if H0 is
  true
• p(Z2.31)1-?(2.31)1-0.98960.0104
• Decision for a0.05
• P-value0.0104lt a
• We reject H0
• There is evidence that sales have increased.

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Example systolic blood pressure

• The national center for health statistics
  reports that the mean systolic blood pressure for
  males 35 to 44 years of ages is 128 and the
  standard deviation in the population is 15. The
  medical director of a large company looks at the
  records of 72 executives in this age group and
  finds that the mean systolic blood pressure in
  this sample is . Is this
  evidence that the companys executives have a
  different mean blood pressure from the general
  population?

Hypotheses H0 H1
no difference from the general population µ128
µ?128 (2 sided hypothesis)
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• Test statistic
• P-value
• p-valueprobability of getting values that are
  more extreme than the test statistic
• p(Z-1.09)0.1379
• But our H1 hypothesis is two sided we must
  also consider the probability that Z1.09
• so p-value2p(Z-1.09)2(0.1379)0.2758

0.1379
Z1.09
Z-1.09
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• Decision for a0.05
• P-value0.2758gt0.05
• we do not reject H0.
• Therefore there is no strong evidence that
  executives differ from other men in their blood
  pressure

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General rules for test of Hypotheses about the
mean

• H0 µµ0 (known s)
• Test statistic
• H1 µltµ0
• P-value p(Zz)
• H1 µgtµ0
• P-value p(Zz)
• H1 µ?µ0

Z
Z
Z
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Example Obstetrics (branch of medicine
concerned with birth of children)

• The mean birth weight in the US is 120 OZ.
• Suppose that in a sample of 100 full-term
  live-born deliveries in a hospital in a low
  socio-economic status area
• Suppose also that the standard deviation of
  birth weight is s24 OZ.
• Examine whether the birth weight in low
  socio-economic status area is lower than the rest
  of the population.
• Hypotheses
• H0
• H1
• Test statistic

µ120
µlt120
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• P-value
• Probability of getting values that are more
  extreme than the test statistic under H0.
• p-valuep(Z-2.083)0.0188
• Decision for a5
• P-value0.0188lta
• Reject H0.
• There is evidence that mean birth-weight of
  babies in the low socio-economic status area is
  smaller than mean birth weight of other babies.
• Decision for a1
• P-value0.0188gta
• Do not reject H0.

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Example Nicotine

• The nicotine content in cigarettes of a certain
  brand is normally distributed with mean ( in
  milligrams) µ and standard deviation s0.1. The
  brand advertises that the mean nicotine content
  of its cigarettes is 1.5, but measurements on a
  random sample of 100 cigarettes of this brand
  gave a mean . Is this evidence ,at
  the 1 significance level, that the mean nicotine
  content is actually higher than advertised?
• Hypotheses
• H0
• H1
• Test statistic

µ1.5
µgt1.5
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• P-value
• Probability of getting values that are more
  extreme than the test statistic under H0.
• p(Z3)1-p(Zlt3)1-0.99870.0013
• Decision for a1
• P-value0.0013lt a
• Reject H0.
• There is evidence that the mean nicotine content
  is higher than advertised

P-value
24
Example body temperature

• - Mean temperature of healthy adults98.6F
  (37C)
• (found by Carl Wunderlich, German physician,
  1868)
• - In 1992, a random sample of n50 gave
• - Assume s0.67
• Is there evidence, at the 0.01 significance
  level, to suspect that the mean temperature
  differ from 98.6F?
• Hypotheses
• H0
• H1

µ98.6
µ?98.6
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• Test statistic
• P-value
• 2p(Zgt-3.9)2(1-?(3.9))2(0.00012)0.00024
• Decision for a0.01
• P-value0.00024lt a
• We reject H0
• The is evidence to suspect that the mean
  temperature differ from 98.6

-3.9
3.9
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What is wrong with the following sets of
Hypotheses?

• H0
• H1

Answer the hypotheses should be about µ !
H0 µlt5 H1 µ5
Answer the equal sign hypothesis should be in H0.
H0 µ?5 H1 µ5
Answer the equal sign hypothesis should be in H0.
H0 µ5 H1 µlt5
Answer nothing is wrong
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Testing hypotheses using a confidence interval

• Example
• A certain maintenance medication is supposed to
  contain a mean of 245 ppm of a particular
  chemical. If the concentration is too low, the
  medication may not be effective if it is too
  high, there may be serious side effects. The
  manufacturer takes a random sample of 25 portions
  and finds the mean to be 247 ppm. Assume
  concentrations to be normal with a standard
  deviation of 5 ppm. Is there evidence that
  concentrations differ significantly (a5) from
  the target level of 245 ppm?
• Hypotheses
• H0 µ245
• H1 µ?245

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First, lets examine the Z test statistic

• Test statistic
• P-value
• 2P(Zgt2)2(0.0228)0.0456
• Decision at 5 significance level
• P-valuegta ? reject H0
• The concentration differs from 245

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Now, examine the hypotheses using a confidence
interval

• a 0.05 ? confidence level is 1- a 95
• 95 CI
• 245.04 , 248.96
• We are 95 certain that the mean concentration
  is between 245.04 and 248.96.
• Since 245 is outside this CI - reject H0.
• The concentration differs from 245

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Examine the hypotheses using a confidence interval

• H0 µµ0
• H1 µ?µ0
• If µ0 is outside the confidence interval, then
  we reject the null hypothesis at the a
  significance level.
• Note this method is good for testing two-sided
  hypotheses only
• confidence interval

µ0
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Example

• Suppose a claim is made that the mean weight µ
  for a population of male runners is 57.5 kg. A
  random sample of size 24 yields . s
  is known to be 5 kg.
• Based on this, test the following hypotheses
• H0 µ57.5
• H1 µ?57.5
• Answer using
• a) A Z test statistic
• b) A confidence interval

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• a)
• Test statistic
• P-value
• 2P(Zgt2.45)2(1-.9929)2(.0071).0142
• Decision at 5 significance level
• P-valuelta ? reject H0
• Conclusion
• Mean weight differs from 57.5

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• b)
• a 0.05 ? confidence level is 1- a 95
• 95 CI
• 58 , 62
• 57.5 is outside this CI - reject H0.
• Mean weight differs from 57.5
• question? Would you reject H0 µ59 vs. H1
  µ?59?
• No, because 59 is in the interval 58, 62

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Example

• In a certain university, the average grade in
  statistics courses is 80, and s11.
• A teacher at that university wanted to examine
  whether her students received higher grades than
  the rest of the stat classes. She took a sample
  of 30 students and recorded their grades
• hypotheses
• H0µ80
• H1µgt80
• data are
• mean

95 100 82 76 75 83 75 96 75 98 79 80 79 75 100 91
81 78 100 72 94 80 87 100 97 91 70 89 99 54
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• Test statistic
• P-value
• P(Zgt2.51)1-0.99400.006
• Decision at 5 significance level
• P-valuelta ? reject H0
• conclusion
• The grades are higher than 80

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• Testing hypotheses about the mean when s is
  unknown

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What happens when s is unknown?

• We can estimate it from the sample
• When the standard deviation is estimated from
  the sample, the test statistic is not Z

Or
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t-distribution

• Symmetric around zero
• Bell-shaped
• Has wider tails than those of Z

Z
t(n-1)
0
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t-distribution becomes more similar to Z as n
increases
t(9)
Z
t(2)
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Example

• During a recent water shortage in a southern
  city, the water company randomly sampled
  residential water consumption on a daily basis. A
  random sample of 20 residents revealed
• , S24.3 gallons.
• Suppose the mean water consumption before the
  water shortage was 250 gallons. Test, at the 5
  significance level, whether there was a decrease
  in the mean daily consumption.
• Hypotheses
• H0
• H1

µ250
µlt250
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• Test statistic
• P-value
• P(t(19)lt-4.84) t table
• P(t(19)gt4.84)
• p-value lt 0.0025
• Decision for a0.05
• P-valuelt0.05 ? Reject the null hypothesis

-4.84
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Example

• The mean age of all CEOs for major corporations
  in the U.S was 48 years in 1991. A random sample
  of 25 CEOs taken recently from major
  corporations showed that years,
  s5 years. Assume that the age of CEOs of major
  corporations have an approximate normal
  distribution. Would you conclude, at the 5
  significance level, that the current mean age of
  all CEOs of major corporations is not equal to
  48?

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• Hypotheses
• H0
• H1
• Test statistic
• P-value
• 2P(t(24)lt-2) 2P(t(24)gt2) t table
• 2(0.025 to 0.05)
• 0.05 to 0.1
• Decision for a0.05
• P-valuegt0.05 ? Do not reject the null
  hypothesis
• Conclusion
• The mean age of CEOs of major companies is not
  different from 48

µ48
µ?48
2
-2
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Example

• The police department will be eligible for a new
  hire if they can produce convincing evidence that
  their response times to non-emergency crime call
  average more than 15 minutes. A random sample of
  41 calls averaged 17 minutes, with standard
  deviation 6 minutes. Carry out a test and decide
  if they are eligible for the new hire.

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• Hypotheses
• H0
• H1
• Test statistic
• P-value
• P(t(40)gt2.13)
• ( 0.01 to 0.025)
• 0.01ltp-valuelt0.025
• Decision for a0.05
• P-valuelt0.05 ? Reject H0
• Conclusion
• The mean response time to non-emergency calls is
  greater than 15 they are eligible for the new
  hire.

µ15
µgt15
2.13
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back to question1
back to question2
ta
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Practice the t-table

• P(t(20)gt1.325)
• P(t(10)gt2.870)
• P(t(10)lt-2.780)
• P(t(10)gt3.6)
• P(t(10)lt-3.6)
• P(t(25)gt2.51)?
• P(t(9)lt-2)?
• P(t(19)lt-4.84)

ta
0.1
0.0083
0.0083
lt0.0025
lt0.0025
0.0083lt P(t(25)gt2.51) lt 0.01
0.025lt P(t(9)lt-2) lt 0.05
lt0.0025
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A confidence interval to the mean when s is
unknown
When s is known
When s is unknown
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Testing hypotheses about the mean using
confidence interval

• If µ0 is outside the CI ? reject H0

Confidence interval
50
T-test with Minitab

• Example
• Most people believe that the mean age at
  which babies start to walk is one year. A A
  researcher thought that the mean age is higher.
  She took a sample of 10 babies and documented the
  age (in months) at which they started to walk.
• The data are
• Examine the researchers claim (a5).
• mean
• SD1.633

12 11 13 14 15 13 12 11 16 13
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• hypotheses
• H0µ12
• H1µgt12
• Test statistic
• P-value
• P(t(9)gt1.94)
• 0.025ltp.vlt0.05
• Decision at 5 significance level
• P-value lt 0.05 ? reject H0
• conclusion
• The mean age at which babies start to walk is
  higher than 12 months

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Flow diagram for testing hypotheses about the mean
s known?
Yes
No
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Flow diagram for testing hypotheses about the mean
s known?
Yes
No
Is the sample VERY large?
Yes
No