Chapter 7
Hypothesis Tests for a Population Mean (7.1-7.3)
Hypothesis Testing is an attempted proof by
statistical contradiction where the preconceived idea
about the value of a parameter is called a hypothesis.
General Statements:
(1) The hypothesis to be tested is called the null
hypothesis and is denoted by H .
0
(2) The null hypothesis is always tested against
some ALTERNATIVE HYPOTHESIS which
his denoted by H a .
(3) The null hypothesis is rejected whenever the
evidence provided by the data is sufficient to
reject the null hypothesis. If we reject the null
hypothesis, then we are (in some sense)
accepting the alternative hypothesis.
(4) When do we fail to reject the null hypothesis?
There is not sufficient evidence to reject the null
hypothesis. This does not necessarily mean that
we accept the null hypothesis, it just means that
based on the available data, we cannot reject it.
Procedure in brief:
Assume that the null hypothesis is true.
Calculate p = P[observed data | H is true]
0
Reject H if p is a very small number. That is, reject
H if the data we observed would be very unlikely to
occur (assuming that H is true).
0
0
0
Type I and Type II Errors
There are two types of incorrect decisions – rejecting
of a true null hypothesis and not rejecting a false null
hypothesis.
Type I Error: Rejecting the null hypothesis when
it is true.
Type II Error: Not rejecting the null hypothesis
when it is false.
Notations:
 = P[Type I Error]
= P[Reject H | H is true]
0
0
 = P[Type II Error]
= P[Not Rejecting H | H is false]
0
0
The probability of making a Type I error is called
the significance level, , of a hypothesis test. In our
testing, we will be given a significance level, . In
general, for our purposes, , the probability of Type
II error will be unknown.
Example 1: The maker of a certain car model claims
that its cars average 31 miles per gallon. A random
sample of 36 cars showed a mean of 29.6
miles/gallon with a sd of 4 miles. At the 0.05 level
of significance, can we conclude that the
manufacturer's cars average less than 31 miles per
gallon?
Soln:
Step1: State the null and alternative hypotheses.
H :  = 31
H :  < 31
0
a
Step2: Write the given information using proper
symbols.
n = 36, x = 29.6, s = 4,
X ~ N(, / n ) = N(31, 4/6) = N(31, 2/3)
Step3: Find the test statistic.
Z=
X 
S .d ( X )
= -2.1
Step4: Find the p-value
p-value = P(Z < -2.1) = 0.0179
Step5: Conclusion
Since the p-value<, we reject the null hypothesis.
Yes, the manufacturer's cars average less than 31
miles per gallon.
Example 2. Proline fishing line has a breaking
strength that is normally distributed with a mean of
30 pounds and a sd of 3 pounds. The engineers have
developed a new, less expensive manufacturing
process, but are worried that the mean breaking
strength may be less than 30 pounds. They assume
that the breaking strength is unchanged, unless there
is a convincing evidence to the contrary. A sample
of 100 of these new lines was tested and their
average breaking strength came out to 28.4 pounds.
Assume that the standard deviation is unchanged. Is
there a compelling evidence that the new lines are
weaker? Use 0.05 as the level of significance.
Soln:
Step1: State the null and alternative hypotheses.
H :  = 30
H :  < 30
0
a
Step2: Write the given information using proper
symbols.
n = 100, x = 28.4,  = 3,  = 0.05
X ~ N(, / n ) = N(30, 3/10)
= N(30, 0.3)
Step3: Find the test statistic.
Z=
X 
S .d ( X )
= -5.33
Step4: Find the p-value.
p-value = P(Z < -5.33) = 0
Step5: Conclusion
Since the p-value<, we reject the null
hypothesis.
Yes, the new lines are weaker.
Example 3. In justifying their demands for higher
wages, the employees in the shipping department of
a large mail order house report that, on the average,
the department completes an order within 13
minutes. The general manager took a random
sample of 400 orders and found the average
completion time of 14 minutes with a sd of 10
minutes. Test the claim of the employees. Use 0.03
as the level of significance.
Soln.
Step1: State the null and alternative hypotheses.
H :  = 13
H :  > 13
0
a
Step2: Write the given information using proper
symbols.
n = 400, x = 14, s = 10,  = 0.03
~ N(, / n ) = N(13, 10/20)
= N(30, 0.5)
Step3: Find the test statistic.
Z = SX.d( X) = 2 (called the z-value)
X
Step4: Find the p-value
p-value = P(Z > 2) = 0.0227
Step5: Conclusion
Since the p-value<, we reject the null
hypothesis.
The average completion time per order is more
than 13 minutes.
Example 4. In a report prepared by the Economics
Research Department of a major bank, the
department manager maintains that the average
annual family income in L.A. is $22,432. Test the
above claim. A random sample of 400 families
shows an average income of $22,574 with a sd of
$2,000. Use 0.05 as the level of significance.
Soln.
Step1: State the null and alternative hypotheses.
H :  = 22,432
H :   22,432
0
a
Step2: Write the given information using proper
symbols.
n = 400, x = 22574, s = 2000,
 = 0.05, X ~ N(, / n )
=N(22432, 2000/20) = N(22432, 100)
Step3: Find the test statistic.
Z=
X 
S .d ( X )
= 1.42 (called the z-value)
Step4: Find the p-value
p-value = P(Z > 1.42 or Z < -.142) = 0.1556
Step5: Conclusion
Since the p-value>=, do not reject the null
hypothesis.
HW: 31-41 (odd) pp. 346-348
7.3 (Small Samples) Use t-distribution
Example 5: A California wine producer states that
the volume of wine in its standard size bottles
averages 750 ml. An independent agency examines
a random sample of 17 bottles and finds the average
volume for the sample to be 721 ml with a sd of 48
ml. Is the average volume per bottle less than 750
ml? Use  = 0.01. Assume that the volume of wine
is normally distributed.
Soln.
Step1: State the null and alternative hypotheses.
H :  = 750 ml
H :  < 750 ml
0
a
Step2: Write the given information using proper
symbols.
n = 17, x = 721, s = 48,  = 0.01, df = 16
Step3: Find the test statistic.
t=
X 
s/ n
= -2.491 (called the t-value)
Step4: Find the p-value
p-value = 0.0121 (From TI-83)
Step5: Conclusion
Since the p-value >=, do not reject the null
hypothesis.
No, the average volume per bottle is not less than
750 ml.
Example 6. A recent newspaper article claimed that
the average fuel economy figure for the autos
traveling at 70 mph was 25.7 mpg. Two neighbors,
one with a small car and one with a large car debated
this claim and decided to test it at the 0.05 level of
significance. From a random sample of 11 drivers
who did their driving on highways, averaging 70
mph, they found that these 11 drivers averaged 23.6
mpg with a sd of 3.5 mpg. Were the neighbors
correct in challenging the newspaper's claim?
Assume that "distance traveled per gallon" is a
normal random variable.
Soln.
Step1: State the null and alternative hypotheses.
H :  = 25.7
0
H :   25.7
a
Step2: Write the given information using proper
symbols.
n = 11, x = 23.6, s = 3.5,  = 0.05, df = 10
Step3: Find the test statistic.
t = Xs /  n = -1.99 (called the t-value)
Step4: Find the p-value
p-value = 0.0746 (From TI-83)
Step5: Conclusion
Since the p-value >= , do not reject the null
hypothesis.
No, the neighbors were not correct in challenging
the newspaper's claim.
HW: 29, 31, 32, 37 and 38 (pp. 358-359)
NOTE: Study Section 6.3 before moving to the
next section.
7.4 Hypothesis Testing – Population Proportion
Example 8. Mr. Dixon, a Republican, claims that he
has the support of 55% of all voters in the 23rd U.S.
Congressional District. Can the Central Committee
conclude that less than 55% of all voters support Mr.
Dixon, if, out of a random sample of 500 registered
voters, only 245 expressed their preference for Mr.
Dixon? Use  = 0.01 as the level of significance.
Solution
Step1: State the null and alternative hypotheses.
H : p = 0.55
H : p < 0.55
Step 2: Write the given information using proper
symbols.
n = 500, p = 245/500 = 0.49,  = 0.01
SD( p ) = p(1n p) = 0.02225
0
a
Step 3: Find the test statistic.
Z = (VariableSD Mean)
= (0.49 – 0.55)/0.02225 = -2.69662
Step 4: Find the p-value
p-value = P[Z < -2.69662] = 0.0035
Step 5: Conclusion
Since the p-value < , reject the null hypothesis.
Yes, the Central Committee can conclude that less
than 55% of all voters support Mr. Dixon
Example 9. The sponsor of a weekly television show
believes that the studio audience consists of an equal
number of men and women. Out of 400 persons
attending the show on a given night, 220 are men.
Using a level of significance of 0.03, can we
conclude that more than 50% of the people attending
the show are men?
Solution Let p = proportion of male population
attending the show.
Step1: State the null and alternative hypotheses.
H : p = 0.5
H : p > 0.5
0
a
Step 2: Write the given information using proper
symbols.
n = 400,
SD( p ) =
p
= 220/400 = 0.55,  = 0.03
p(1  p)
= 0.025
n
Step 3: Find the test statistic.
Z = (VariableSD Mean)
= (0.55 – 0.5)/0.025 = 2
Step 4: Find the p-value
p-value = P[Z > 2] = 0.0228
Step 5: Conclusion
Since the p-value < , reject the null hypothesis.
Yes, more than 50% of the people attending the
show are men?
HW: 9, 11, 13 and 15 pp. 364-365