hypothesis testing skills set
TRANSCRIPT
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Skills Set
Chapter 7: Hypothesis Testing
No. Skills Examples of questions involving the skills
1 To identify
the null and
alternativehypothesis
(If you
determine it’s
a one tail test:
look at sample
mean)
Lecture example 1 In the following situations, state suitable null and alternative hypotheses.
(a) The mean factory assembly time for a particular electronic component is
84 seconds. The factory manager wants to test whether the introduction
of a new assembly process results in a different assembly time.
(b) In a report, it was stated that the average age of all hospital patients was
53 years. The hospital director wants to test whether this figure is an
underestimate.
Solution
(a) Let be the mean factory assembly time (in seconds) after the
introduction of the new assembly process.
H0: 84 against H1: 84
(b) Let be the average age (in years) of the hospital patients.
H0 : 53 against H1 : 53 Tutorial Q4
A bank has branches in two cities A and B in the same country. The manager ofthe bank claims that the waiting time for a customer to open a new account
with the bank is not more than 5 minutes.
0H : 5 against 1H : 5
8235 486666667
150
ww
n
.
Note: 5w corresponds with 1H : 5
2 Using G.C (p-value) to
conduct
hypothesis
tests when no
unknowns are
present
Lecture example 2
A machine packs sand into bags. The mass of a filled bag follows anormal distribution with mean 1500 g and variance 0.16 g2. A sample of
10 filled bags is randomly chosen and found to have a mean mass of
1499.8 g.
(a) Test whether the sample provides significant evidence, at the 1%level of significance, that the machine packs bags that are
underweight.
Solution
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(a) Let X be the mass (in grams) of a filled bag.
Let be the mean mass (in grams) of the bags produced by the
machine.
0
1
H : 1500
H : 1500
n = 10, x = 1499.8, 0.16 0.4 .
Under H0,0.16
~ N 1500,10
X
.
Test statistic, Z 1500
~ N(0, 1)0.4
10
X under H0.
Using z -test, 1.58113883 z . p-value P( 1.58113883) Z = 0.0569.
Conclusion: Since p-value = 0.0569 > 0.01 , we do not reject H0 at
the 1% level of significance. There is insufficient evidence to saythat the machine packs bags that are underweight.
3 Using Test-Statistic and
Critical Region
when there are
unknowns in
the question.
Tutorial Q10b
A random sample of 50 packets of ‘Nilo Bar’ snacks is weighed and the mass,
x grams, is recorded. The results are summarised by
( x – 150) = –250, ( x – 150)2 = 3000.
The population mean mass is grams. A test is carried out at the 1%
significance level with the following hypotheses:
0 : H 0
:1 H 0
Given that 0 H is rejected in favour of H1, find the set of possible values of
0 .
Solution
x = 150
150 x
n
250 150 145
50
2 s =
2
2 1501150
1
x x
n n
2
25013000
49 50
35.714
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0Test statistic: ~ 0, 135.714
50
X Z N
Since0 H
is rejected in favour of H1,
Critical region: 2.576 or 2.576 z z
0
0
1452.576
35.714
50
147.18
0
0
1452.576
35.714
50
142.82
0 0 0: 142.82 or 147.18
4 Identify theneed to use
Central Limit
Theorem for
distribution of
sample mean.
(n large and
population not
known to be
normal)
Lecture example 5Clara, who has diabetes, has to monitor her blood glucose levels which vary
throughout a week. The results from a sample of 75 readings, x , taken at
random times over a week, are summarized by
511.5 x and 2 4027.89 x .(ii) State, giving a reason, whether we can conclude that the alternative
hypothesis0
is to be accepted if the blood glucose level cannot be
assumed to have a normal distribution.
Solution
(ii) Since the sample size, n, is sufficiently large, by the Central Limit
Theorem, the mean blood glucose level is approximately normally
distributed, so the alternative hypothesis0 can still be accepted.
6 conduct t-testsfor a
population
mean based on
a sample taken
from a normal
population with
unknown
variance and a
small sample
size.
Assumption
Lecture example 7An automatic coffee dispensing machine is set to dispense 440 cm
3 of coffee
into a mug. A random sample of 12 filled mugs was taken and the mugs were
found to contain the following amounts of coffee, in cm3.
420 446 372 430 436 441
433 422 445 422 438 442
Assuming no spillage and that the amount of coffee dispensed follows a normal
distribution, test at the 5% level of significance, whether the machine is
dispensing too little coffee.
Solution
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phrasing.
Using data lists
vs stats values.
Let be the mean amount of coffee in a mug (in cm3).
H0 : = 440
H1 : < 440
n = 12, x = 428.92, s = 20.07.
Under H0, test statistic, T = 440 X S
n
t (11).
Using t -test, t = –1.912418829.
p-value P( –1.912418829)T 0.0411.
Conclusion: Since p-value = 0.0411 < 0.05, we reject H0 at the 5% level of
significance. There is sufficient evidence to conclude that the machine is
dispensing too little coffee.
7 Comparing thedifferences
between 1 tail
and 2 tail test,
z test and t-
tests
Tutorial Question 7In a test at the 5% significant level it is found that there is significant evidence
that the population mean talk-time is less than 5 hours. Using only this
information, and giving a reason in each case, state whether each of the
following statements is
(i)
necessarily true, (ii) necessarily false,
(ii)
or (iii) neither necessarily true nor necessarily false.
(a) There is significant evidence at the 10% significance level that the
population mean talk-time is less than 5 hours.
(b) There is significant evidence at the 5% significance level that the
population mean talk-time is not 5 hours.
Solution
(a) Test at 10% level of significance: the p-value remains the same; hence if p-
value 0.05 , it is definitely 0.1 , and we reject 0H . This statement is
necessarily true.
(b) Test0H : 5 vs 1H : 5 at 5% level of significance: new p -value is
twice old p -value; new p-value may or may not be 0.05 . This statement is
neither necessarily true nor necessarily false.
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8 Comparing thedifferences
between z- test
and t-tests
Tutorial Question 8(b)After the implementation of another teaching pedagogy, the SAT score of 10
students are collected. To investigate the effectiveness of this pedagogy, a
hypothesis test is carried out at the %α level of significance.
(ii) If the null hypothesis is not rejected when a t -test is carried out, explain
whether it is necessarily true that the same conclusion is obtained when
a z-test is carried out at the same significance level.
Let andt z p p be the p-values obtained using a t -test and z-test
respectively. Note that for the same test statistic, >t z p p .
Since H0 is not rejected for a t -test, pt >100
. There are two possible
conclusions for a z-test.
OR
Z
N(0,1)
t( )
Test statistic
t p
z p
N(0,1)
t( )
z
Zt
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Reason:
For a fixed level of significance, , z
< t
(refer to graph above).
When H0 is not rejected for a t -test, there are two possible conclusions
for a z-test:
1) z
< test statistic < t
: H0 will be rejected if z-test is used.
2) test statistic < z
< t
: H0 will not be rejected if z-test is used.
9 CommonAssumptions
1. For t-test, we may need to assume population is normallydistributed if sample size is given to be small and population
variance is unknown.
2. For z-test, we do not need to assume population is normally
distributed if sample size is large.3. Sample are independently chosen.