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Tests of Hypotheses about the mean of 1 population: 1' s is known 2' s is unknown

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Statement about a feature of the population (e.g. the mean) Examples: - Mean temperature ... Obstetrics (branch of medicine concerned with birth of children) ... – PowerPoint PPT presentation

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Title: Tests of Hypotheses about the mean of 1 population: 1' s is known 2' s is unknown


1
Tests of Hypothesesabout the mean of 1
population1. s is known2. s is unknown
2
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.

3
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?

4
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)

5
Use the sample results to test the hypotheses
6
Sample results
If is small enough we will reject H0.
7
How likely are we to observe such result from a
population with mean µ24?
The distribution of
µ24 oz
8
(No Transcript)
9
Test statistic
  • measures the distance of the sample results from
    what is expected if H0 is true.
  • Z -2 is an example of a test statistic

10
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.

11
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

12
  • 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
13
Summary soft drink bottles example
  • Hypotheses
  • H0 µ24
  • H1 µlt24
  • Test statistic
  • P-value P(Z-2)0.0228
  • Decision at the a.05 significance level
  • Reject H0 p-value0.0228lta0.05
  • Conclusion
  • Contents of soft drink bottles is less than
    24 oz.

14
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(Z2.31)1-?(2.31)1-0.98960.0104
  • Decision for a0.05
  • P-value0.0104lt a
  • We reject H0.
  • Conclusion
  • There is evidence that sales have increased.

16
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)
17
  • Test statistic
  • P-value
  • 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
18
  • Decision for a0.05
  • P-value0.2758gt0.05
  • we do not reject H0.
  • Conclusion
  • Therefore there is no strong evidence that
    executives differ from other men in their mean
    blood pressure

19
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
20
  • Practice
  • Tests of hypotheses about the mean (s known)

21
1. 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
22
  • P-value
  • 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.

23
2. 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
24
  • P-value
  • p(Z3)1-p(Zlt3)1-0.99870.00
    13
  • Decision for a1
  • P-value0.0013lt a
  • Reject H0.
  • There is evidence that the mean nicotine content
    is higher than advertised

P-value
25
3. 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
26
  • Test statistic
  • P-value
  • 2p(Z-3.9)2(0.00012)0.00024
  • Decision for a0.01
  • P-value0.00024lt a
  • We reject H0
  • Conclusion
  • There is evidence to suspect that the mean
    temperature differs from 98.6

27
4. Grades
  • 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

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
28
  • Test statistic
  • mean
  • P-value
  • P(Z2.51)1-0.99400.006
  • Decision at 5 significance level
  • P-valuelta ? reject H0
  • conclusion
  • The mean grade is higher than 80

29
Testing 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
30
  • 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
  • H1

µ245
µ?245
31
  • A 95 CI for µ
  • a 0.05 ? confidence level is 1- a 95
  • 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

32
practice
  • A 95 confidence interval for the mean time
    (in hours) to complete an audit task is
    7.04,7.76.
  • Use the relation between confidence intervals
    and two-sided tests to examine the following sets
    of hypotheses
  • (a) H0 µ7.5 H1 µ?7.5 (a.05)
  • (b) H0 µ7 H1 µ?7 (a.05)

33
Common mistakes 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.
34
Testing hypotheses using Minitab
  • List values of your variable in a column.
  • 2. Stat gt Basic Statistics gt1-Sample Z.
  • 3. Pick your column of values in variables.
  • 4. For tests of hypotheses Choose options and
    pick the alternative hypothesis.
  • For confidence intervals Choose options
    and pick not equal for the alternative.
  • Use Minitab to do the gradesexample
  • grades

35
  • Testing hypotheses about the mean when s is
    unknown

36
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

37
t-distribution
  • Symmetric around zero
  • Bell-shaped
  • Has wider tails than those of Z
  • t-distribution becomes more similar to Z as n
    increases

t(16)
t(9)
Z
t(2)
38
back to question1
back to question2
ta
39
Practice the t-table
  • P(t(20)gt1.325)
  • P(t(10)gt2.764)
  • P(t(10)lt-2.764)
  • 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
.01
MTB.exe
0.01
0.001ltplt0.005
0.001ltplt0.005
0.005lt P(t(25)gt2.51) lt 0.01
0.025lt P(t(9)lt-2) lt 0.05
lt0.001
40
Practice t-test
  • Use the survey data to test whether the mean GPA
    of students is higher than 3.
  • Hypotheses
  • H0
  • H1
  • Test statistic

Survey 0200morning.MPJ Survey 0200evening.MPJ
41
  • P-value
  • Decision for a0.05
  • Conclusion

42
  • 2. Most people believe that the mean age at which
    babies start to walk is one year. 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
43
  • hypotheses
  • H0µ12
  • H1µgt12
  • Test statistic
  • P-value
  • P(t(9)1.94)
  • 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

44
A confidence interval to the mean when s is
unknown
When s is known
When s is unknown
45
Testing hypotheses about the mean using
confidence interval
  • If µ0 is outside the CI ? reject H0

Confidence interval
46
T-test with Minitab
1. List values of your variable in a column 2.
Stat gt Basic Statistics gt1-Sample t 3. Pick your
column of values in variables 4. For tests of
hypotheses Choose options and pick the
alternative hypothesis For confidence
intervals Choose options and pick not equal
for the alternative
Practice t confidence interval A survey of 100
married couples was conducted to find out how
many months they dated before getting married.
The sample mean was 11.41 with a sample
deviation of Sx8.12. Find a 95 confidence
interval for the true average number of months
dated among all married couples
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