Title: Tests of Hypotheses about the mean of 1 population: 1' s is known 2' s is unknown
1Tests of Hypothesesabout the mean of 1
population1. s is known2. s is unknown
2Statistical 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.
3Example 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?
4Set 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)
5Use the sample results to test the hypotheses
6Sample results
If is small enough we will reject H0.
7How likely are we to observe such result from a
population with mean µ24?
The distribution of
µ24 oz
8(No Transcript)
9Test 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
10P-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.
11Level 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
13Summary 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.
14Example 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?
15µ354
- 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.
16Example 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
19General 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)
211. 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.
232. 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
253. 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
274. 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
29Testing 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
32practice
- 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)
33Common mistakes What is wrong with the
following sets of Hypotheses?
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.
34Testing 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
36What 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
37t-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)
38back to question1
back to question2
ta
39Practice 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
40Practice 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
44A confidence interval to the mean when s is
unknown
When s is known
When s is unknown
45Testing hypotheses about the mean using
confidence interval
- If µ0 is outside the CI ? reject H0
-
-
Confidence interval
46T-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