Download Chapter 5 - hypothesis

WHY WE HAVE TO DO THE HYPOTHESIS?
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To make decisions about populations based on the sample
information.
Example :- we wish to know whether a medicine is really effective
to cure a disease. So we use a sample of patients and take their
data in effect of the medicine and make decisions.
To reach the decisions, it is useful to make assumptions about
the populations. Such assumptions maybe true or not and called
the statistical hypothesis.
Definitions
Hypothesis Test:
It is a process of using sample data and statistical procedures to
decide whether to reject or not to reject the hypothesis (statement)
about a population parameter value (or about its distribution
characteristics).
Null Hypothesis,
:
Generally this is a statement that a population has a specific value.
The null hypothesis is initially assumed to be true. Therefore, it is
the hypothesis to be tested.
Alternative Hypothesis,
:
It is a statement about the same population parameter that is used
in the null hypothesis and generally this is a statement that
specifies that the population parameter has a value different in
some way, from the value given in the null hypothesis. The
rejection of the null hypothesis will imply the acceptance of this
alternative hypothesis.
Test Statistic:
It is a function of the sample data on which the decision is to be
based.
Critical/ Rejection region:
It is a set of values of the test statistics for which the null
hypothesis will be rejected.
Critical point:
It is the first (or boundary) value in the critical region.
P-value:
The probability calculated using the test statistic. The smaller the
p-value is, the more contradictory is the data to
.
Procedure for hypothesis testing
1.
Define the question to be tested and formulate a hypothesis for
a stating the problem.
H o :  a or   a or   a
H1:  a or   a or  > a
2. Choose the appropriate test statistic and calculate the sample
statistic value. The choice of test statistics is dependent upon
the probability distribution of the random variable involved in
the hypothesis.
3. Establish the test criterion by determining the critical value and
critical region.
4. Draw conclusions, whether to accept or to reject the null
hypothesis.
Example 4:
The average monthly earnings for women in managerial and
professional positions is RM 2400. Do men in the same positions
have average monthly earnings that are higher than those for
women? A random sample of n = 40 men in managerial and
professional positions showed X = RM3600 and s = RM 400. Test
the appropriate hypothesis using
 = 0.01.
Solution:
The hypothesis to be tested are:
H 0 :  2400
H1:  2400
We use normal distribution n > 30, as n = 40
Rejection region:
Z  z
z  z0.01  2.33 (from normal distribution table)
Test statistic:
Z
x   3600  2400

 18.97
s
400
n
40
Conclusion :
Since 18.97 > 2.33, falls in the rejection region, we reject H 0 and conclude
that average monthly earnings for men in managerial and professional
positions are significantly higher than those for women.
Example 5:
Aisyah makes “kerepek ubi” and sell them in packets of 100g each. 12
randomly selected packets of “kerepek ubi” are taken and their weights in g
are recorded as follows:
98
102
98
100
96
91
97
97
100
94
101
97
Perform the required hypothesis test at 5% significance level to check
whether the mean weight per packet if “kerepek ubi” is not equal to 100g.
Solution:
The hypothesis to be tested are:
H 0 :   100
H1:   100
We use t distribution,
Two-tailed test
 unknown, n = 12 < 30

 0.025  t0.025 ,11  2.201 and t0.025 ,11  2.201
2
Test Statistic:
t
x   97.5833  100

 2.737
s
3.0588
n
12
ttest  2.737  2.201  t0.025,11
Cocnlusion:
Since – 2.737 < -2.201, falls in the rejection region, we reject H 0 and
conclude that weight per packet of “kerepek ubi” is not equal to 100g.
Exercise 3:
A teacher claims that the student in Class A put in more hours studying
compared to other students. The mean numbers of hours spent studying per
week is 25hours with a standard deviation of 3 hours per week. A sample of
27 Class A students was selected at random and the mean number of hours
spent studying per week was found to be 26hours. Can the teacher’s claim
be accepted at 5% significance level?
Answer: Z = 1.7321, reject H 0
Example 6:
When working properly, a machine that is used to make chips for calculators
produce 4% defective chips. Whenever the machine produces more than
4% defective chips it needs an adjustment.To check if the machine is
working properly, the quality control department at the company often takes
sample of chips and inspects them to determine if they are good or defective.
One such random sample of 200 chips taken recently from the production line
contained 14 defective chips. Test at the 5% significance level whether or not
the machine needs an adjustment.
Exercise 4:
A manufacturer of a detergent claimed that his detergent is at least 95%
effective is removing though stains. In a sample of 300 people who had used the
Detergent and 279 people claimed that they were satisfied with the result.
Determine whether the manufacturer’s claim is true at 1% significance level.
Answer: Fail to Reject H 0
Exercise 5
A new concrete mix being designed to provide adequate compressive
strength for concrete blocks. The specification for a particular application
calls for the blocks to have a mean compressive strength  greater than
1350kPa. A sample of 100 blocks is produced and tested. Their mean
compressive strength is 1366 kPa and their standard deviation is 70 kPa.
Test hypothesis using  = 0.05. Answer : Fail to reject