CHAPTER 7 Hypotheses Testing About The Mean μ(mu):

CHAPTER 7
Tests for the Population Mean  - Sections: 7.1 & 7.2
1.
2.
3.
4.
-- Tests about the mean sigma given
-- Tests for the mean with sigma not given
-- Matched pairs t procedures
-- Comparing the means of two populations (independent samples)
Hypotheses Testing About The Mean (mu):
Single Population With (sigma) Given
NULL HYPOTHESIS
Ho : o
ONE SAMPLE ZSTATISTIC
ALTERNATIVE
HYPOTHESIS
REJECT Ho AT LEVEL
 IF:
P-VALUE
Ha : o
zobs z*
P[Zzobs]
Ha : o
zobs z*
P[Z  - zobs]
Ha : 
zobs  z* OR
zobs z*
2P[Z |zobs|]
Notation and Conditions
1. The Z-statistic has the Standard Normal distribution, provided the sample is taken from a
normal population OR n 30.
2. zobs is the observed value of the statistic computed from the sample data.
3. z* or z*/2 is the critical z-value based on the given  (one-sided alternative) or /2
(two-sided alternative).
TEST ABOUT THE MEAN  OF A SINGLE POPULATION ( Known)
EXAMPLE
A company that produces snack foods uses a machine to package 454 oz bags of peanuts. We
will assume that the net weights are normally distributed with mean 454 oz and that the standard
deviation of all such weights is 7.8 oz. To test whether the machine is working properly, we
check the weights of 25 bags. The weights are as follows:
465
468
447
456
433
456
438
454
456
454
463
435
447
450
450
449
446
447
442
447
444
449
456
446
452
At the 10% level of significance, does the sample provide sufficient evidence to conclude that
the packaging machine is not working properly?
SOLUTION:
Step 1.
Population parameter under investigation:
 = True mean weight of a bag of peanuts packaged by the machine.
Null Hypothesis
Ho : = 454 oz
Alternative Hypothesis
Ha : 454
Step 2.
Test statistic:
Here we can use the one-sample Z-statistic because the population standard deviation is given.
Step 3.
Decision Rule:
Reject Ho with a = 0.10 if:
zobs z = 1.645 OR
zobs - z = - 1.645
Step 4.
Drawing the conclusion based on the available data.
For the 25 bags whose weights we have, the sample mean
= 450 therefore,
Since -2.56 falls inside the rejection region, the test rejects the null hypothesis. Our conclusion
should be that the true mean weight of a bag of peanuts packaged by the machine is not equal to
454 oz. Therefore the packaging machine is not working properly.
Step 5
Computing the P-value
P-Value = 2*P[Z > |-2.56|] = 2*(.0052) = 0.0105 or 1.05%
Using the TI-83 to carry out the test
Enter the data into L1 > STAT > TESTS > 1:Z-Test
Hypotheses Testing About The Mean (mu):
Single Population With (sigma) Not Given
NULL HYPOTHESIS
Ho : o
ONE SAMPLE TSTATISTIC
ALTERNATIVE
HYPOTHESIS
REJECT Ho AT LEVEL
 IF:
P-VALUE
Ha : o
tobs t*
P[Ttobs]
Ha : o
tobs t*
P[T  - tobs]
Ha : 
tobs  t* OR tobs t* 2P[T |tobs|]
Notation and Conditions
1. The T-statistic has the student t-distribution with df = n - 1 provided random sample is
from a normal population OR n 
2. tobs is the observed value of the statistic, which is computed from the sample data.
3. t* or t*/2 is the critical t-value based on the given  (one-sided alternative) or /2 (twosided alternative).
TEST ABOUT THE MEAN  OF A SINGLE POPULATION (Not known)
EXAMPLE
The following figures show the amount of coffee (in ounces) filled by a machine in six randomly
picked jars.
15.7 15.9 16.3 16.2 15.7 15.9
At the 2.5% level of significance, is the true mean amount of coffee filled in a jar less than 16
ounces? Assume a normal distribution for the amount of coffee filled in a jar.
SOLUTION:
Step 1
Population parameter under investigation:
 = True mean amount of coffee filled in a jar by the machine.
Null Hypothesis
Ho :  = 16 oz
Alternative Hypothesis
Ha : 16 oz
Step 2.
Test statistic:
Because the population standard deviation is not given, we use the one-sample T-statistic.
Step 3.
Decision Rule:
Reject Ho with  = 0 .025 if: tobs  t = -2.571
To find twe use table D with df = 5)
Step 4.
Drawing the conclusion based on the available data.
For the 6 jars whose weights we have, the sample mean
deviation s = 0.251 oz.
= 15.95 oz and the sample standard
Therefore,
Since -0.4879 falls outside the rejection region, the test does not reject the null hypothesis. Our
conclusion should be that the true mean amount of coffee filled in a jar by the machine is not less
than 16 oz.
Step 5.
Computing the P-value
(a) Using Table D
In the row with df = 5 of Table D, look for the two t-values the bracket the absolute value of the
test statistic (i.e., tobs ). Hera that value is |-0.4879| or 0.4879.
Because the value 0.4879 of tobs is smaller than the lowest t-value of 0.727 we find in row 5 of
the table, the p-value = P[T < -0.4879] = P[T > 0.4879] must be larger than 0.25 (or 25%), the
Upper-tail probability p for t = 0.727.
(b) Using the TI-83
Use the tcdf( ) function found on the 2nd|DISTR screen of the TI-83. This function works like
the normalcdf ( ) function for computing normal probabilities. To compute a probability using
tcdf( a, b, df) = P[ a < T < b] where T follows the t-distribution. df = the degrees of freedom.
In our case, tcdf(-100, -0.4879, 5) = 0.3231 or 32.31%.
Using the TI-83 to carry out the test
Enter the data into L1 > STAT > TESTS > 2:T-Test
Comparing Two Treatments:
Matched Pairs T-Procedures (Before & After)
NULL HYPOTHESIS
Ho : D
TEST STATISTIC
ALTERNATIVE
HYPOTHESIS
REJECT Ho AT LEVEL  P-VALUE
IF:
Ha : D
tobs t*
P[Ttobs]
Ha : D
tobs t*
P[T  - tobs]
Ha : D
tobs  t* OR tobs t*
2P[T |tobs|]
Notation and Conditions
1. The T-statistic has the student t-distribution with df = n - 1 provided random sample is
from a normal population OR n .
2. tobs is the observed value of the statistic, which is computed from the sample data.
3. t* or t*/2 is the critical t-value based on the given  (one-sided alternative) or /2 (twosided alternative).
Difference in population means using the paired t-Test
(Dependent samples)
EXAMPLE.
Is it safe for an individual to drive after drinking two beers? To test whether consuming alcohol
impairs driving ability, the DMV conducted the following experiment. Seven volunteers were
asked to drive through an obstacle course prior to consuming any alcohol and again after
drinking two beers. Driving ability was measured by recording the total reaction time (in
seconds) of a driver in avoiding the various obstacles on the course. The data are given below.
Driver
Reaction time w/o alcohol
Reaction time w alcohol
Difference (After - Before)
1
8.82
10.61
1.79
2
6.62
8.18
1.56
3
9.84
12.24
2.4
4
10.81
13.45
2.64
5
9.20
8.88
-0.32
6
11.67
10.88
-0.79
7
4.42
5.38
0.96
At the 5% level of significance, is it true that consumption of two beers impairs driving ability?
Assume differences between reaction times before & after are normally distributed.
SOLUTION:
In this type of test we compute the difference D = after - before (we could compute the
difference "before - after" if we wanted) and test the null hypothesis
Ho: D = 0
against one of the alternatives:
Ha: D  0
Ha: D  0
Ha: D ≠ 0
Step 1.
Population parameter under investigation:
D = True mean difference between "after - before".
Null Hypothesis
Ho : D = 0
Alternative Hypothesis
Ha : D0 (i.e., mean reaction time "after" is larger than mean reaction time "before")
If we can reject the null hypothesis, we would prove that the true mean of all differences "after before" is greater than zero. i.e., that the reaction time "after" is greater than the reaction time
"before".
Step 2.
Test statistic:
Because the differences are normally distributed, the sample statistic has the student tdistribution with degrees of freedom df = 6.
Step 3.
Decision Rule:
Reject Ho with  = 0 .05 if: tobs t
We find tusing table D with df = 7 - 1 = 6
Step 4.
Drawing the conclusion based on the available data.
For the 7 volunteers who took part in this experiment, the sample mean difference "after before" and sample standard deviation of these differences were: = 1.177143 sec.
and s = 1.31122 sec.
Therefore,
Since 2.375 falls inside the rejection region, our test rejects the null hypothesis. Our conclusion
should be that the true mean difference in reaction times "after - before" is greater than 0.
Therefore the test, at this level, has shown that after drinking two beers drivers do take longer to
react.
Step 5
Computing the P-value
(a) Using table D
The two t-values in table D the bracket tobs = 2.375 in row 6 are 1.943 & 2.447. The Upper-tale
probabilities that correspond to these two vales are .05 & .025.
Since 1.943 < 2.375 < 2.447 ==> .025 < P[t >2.375] < .05
(b) Using the tcdf( ) function of TI-83
P[t >2.375] = tcdf(2.375, 100, 6) = 0.0276 or 2.76%
Using the TI-83 to carry out the test
Enter the data (differences) into L1 > STAT > TESTS > 2:T-Test
Comparing The Means of Two Populations: Two Independent
Samples
NULL HYPOTHESIS
ALTERNATIVE
HYPOTHESIS
Ho : 
TWO SAMPLE T-STATISTIC Ha : 
REJECT Ho AT
LEVEL  IF:
P-VALUE
tobs t*
P[Ttobs]
Ha : 
tobs t*
P[T  - tobs]
Ha : 
tobs  t* OR
tobs t*
2P[T |tobs|]
Notation and Conditions
1. The T-statistic has the student t-distribution with df* = the smaller of n1 - 1 or n2 - 1,
provided the samples come from two normal populations (or that eache sample size is
greater than 30) normal population OR n 
2. tobs is the observed value of the statistic, which is computed from the sample data.
3. t* or t*/2 is the critical t-value based on the given  (one-sided alternative) or /2 (twosided alternative).
(*) When the sample sizes are large, the two sample t-procedures are fairly accurate with df = the
smaller of n1-1 or n2-1. With small sample sizes it is recommended that t*be computed using
the t-distribution with degrees of freedom calculated using the formula:
COMPARING THE MEANS  AND  OF TWO POPULATIONS
(INDEPENDENT LARGE SAMPLES)
EXAMPLE.
When 100 raw commercial apples of Variety A and 150 of Variety B were tested for their iron
content (in milligrams), the following information was obtained.
Sample Mean
(in milligrams)
Sample Standard
Deviation
Variety A
0.32
0.08
Variety B
0.29
0.05
At the 5% level of significance, do the two varieties of apples differ in their iron content?
SOLUTION:
Step 1.
Population parameters under investigation:
 = True mean iron content of apples of variety A.
 = True mean iron content of apples of variety B.
Null Hypothesis
Ho :  = 
Alternative Hypothesis
Ha : 
Step 2.
Test statistic:
Because both sample sizes are large, the standard normal distribution Z is a good approximation
for the distribution of the T-statistic. i.e., The distribution of T is approximately normal with
mean 0 & standard deviation 1.
Step 3.
Decision Rule:
Reject Ho with  = 0 .05 if:
tobs z = 1.960 OR tobs - z = - 1.960
Step 4.
Drawing the conclusion based on the available data.
Since 3.34 falls inside the rejection region our test rejects the null hypothesis. The mean iron
content in the two varieties of apples is different.
Step 5.
Computing the p-value
P-value = 2*P[Z > 3.34] = 0.0004 or 0.04%
Using the TI-83 to carry out the test
STAT > TESTS > 4:2-SampTTest...