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Tests for single and two population means
DC-I
Semester III
Paper : Statistical Method in Economics II
Lesson : Tests for single and two population means
Lesson : Developer : Neetu Chopra
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Tests for single and two population means
Table of contents
Chapter Outline:
1. Tests about a single population mean
Case I- A Normal population with known σ
Case II- Large Sample Tests.
Case III-A Normal population distribution
2. Tests for two population means
Case I- Normal population with known variances
Case II- Large Sample Tests.
Case III-The two Sample t-tests (Small sample test)
Case IV-Pooled t procedure.
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Tests for single and two population means
Learning objectives
In the last chapter, you learnt about the basic concepts of hypothesis testing
along with the two types of errors involved, significance levels and the p-values. In this
chapter, you will learn about the large and small sample tests for the population mean.
This will be followed by the tests on means of two different population distributions,
using the critical values as is done in single sample case. There are practice questions
given at the end of the chapter.
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Tests for single and two population means
1. Tests about a single Population Mean
Case I - A Normal population with known σ
.
We begin by looking at tests of hypothesis about a population mean for a normal
population and known σ.
The sample mean X is the tests statistic used to test Ho : µ = µ0 (null value). The
random sample of size n is taken from the normal population. X also has a normal
distribution with mean  X =  and standard deviation  X =  /
n . We first standardize
X to get Z under the assumption that Ho is true so. That  X = HO
Z
X  0
/ n
This calculated z value is the distance between X and it’s expected value (ie µ0)
when Ho is true. If this distance is quite large in a direction given by H a, Ho will be
rejected, otherwise not.
If Ho : µ > µ0, then we will reject Ho only if 𝑥 value exceeds µ0 by a significant amount.
Otherwise, we cannot eject Ho. Then, Z will be positive and large. The appropriate
rejection region will be of the form Z  - Z where Z is the cut off value chosen to
control probability of type I error ( ) at the desired level. This is the upper tailed on the
right tailed test.
On the other hand, for Ha : µ < µ0, 𝑥 will be quite less so that z becomes
negative. The appropriate rejection region will be of the form z  - z. The is the lower
tailed on left tailed test.
For the two sided alternative, Ha : µ ≠ µ0, Ho will be rejected if X is significantly
different from µ0 from any of the sides. This happens if ether z  z or Z  - Z / 2 . This
is the two tailed test, when
 is divided equally between two bat of the test z curve,
each tail giving an area of z.
The z values  z are called the critical values as they are critical in deciding
whether to reject Ho or not. For this reason, rejection region is also known as critical
region.
Test Procedure – Case I.
Null hypothesis
Ho : µ = µ0
Test statistic value : Z 
X  0
/ n
Alternative hypothesis
Rejection region
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Ho : µ > µ0
z  z (Upper tailed test)
Ha : µ < µ0
z  - Z(Lower tailed test)
Ha : µ ≠ µ0
either z  Zor Z  - Z (Two tailed test)
Question: 1. A motor car company claims that car’s average speed is 35 miles per
gallon (mpg) of petrol. In a sample of 100 cars, it was found that average is 32 mpg. If
the speed is normally distributed with standard deviation of 4 mpg, test the claim at
0.05 and 0.01 level of significance.
Solution:
1.
Ho : µ = 35
Ha : µ ≠ 35
X  0
2.
Z
3.
Reject Ho if
/ n
Z  - Z0.05 or Z ≤ –Z0.025 ( = 0.05)
Z0.05 = 1.96
If

Reject Ho if
Z ≥ Z0.005 or Z ≤ –Z0.005
Z0.005 = 2.575
4.
Z
32  35
4 / 100
  7.5
Since – 7.5 < -1.96, Ho will be rejected at 0.05 level of significance.
Also, –7.5 < -2.575. Hence, Ho will be again rejected at 0.01 level of significance. The
data is significant enough to reject Ho at both levels of significance.
Question: 2. In the previous question, test the claim that the performance of the car
has gone down at 0.05 and 0.01 levels of significance.
Solution :
1.
Ho : µ = 35
Ho : µ < 35
2.
Z=
x – 0
/ n
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3.
Reject Ho if
Z ≤ -Z0.05 (at
 = 0.05)
Z0.05 = 1.645
Reject Ho if
Z ≤ -Z0.05 (at
 = 0.01)
Z0.01 = 2.33
4.
Z=
32 –35
4/ 100
= –7.5
Since –7.5 < –1.645 and –7.5 < –2.33, Ho will be rejected at 0.05 and 0.01 levels of
significance respectively.
Case II : Large Sample Test
When the sample size is large enough, we can use the z test from case I. In this
case, neither of the conditions of normal population or known σ are required.
If the sample size (n) is large, standardized variable
Z
X 
S/ n
has approximately standard normal distribution. If Ho is true, the test statistic
Z
X 
S/ n
Will have approximately standard normal distribution. The test procedure remains
same as in the previous case A large sample is denoted by a sample size of more than
40. The intuition behind this is that central limit theorem can be used to justify using the
test for normal population. Even if σ2 is unknown, we can substitute it with S2 to
compute the test statistic.
Question: 3. Light bulbs of a certain type are advertised as having an average lifetime
of 750 hours. A potential consumer has decided to go ahead with a purchase
arrangement unless it can be conclusively demonstrated that the true average lifetime is
smaller than what is advertised. A random sample of 50 bulbs was selected that gave
mean as 738.44 hours and standard deviation of 38.20 hours. What are the appropriate
hypotheses for this test? What conclusion would be appropriate for a significance level of
0.05? Does your conclusion change at 0.01 level of significance ? What is the P-value of
the test?
Solution:
1.
Ho : µ = 750
Ha : µ < 750
x – 0
2.
Test statistic z =
3.
Reject Ho if z  -z
s/ n
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 = 0.05, z0.05 = 1.645
At  = 0.01, z0.01 = 2.33
At
4.
z=
7.8.44 – 750
= –2.14
38.20 / 50
Since – 2.14 < - 1.645, H0 is rejected at 0.05 level but not at 0.01 level as -2.14 > 2.33
Hence, consumer will purchase the bundle at 0.01 level of significance but not at 0.05
level.
P-value is the area under the Z curve beyond – 2.14
P-value = 0.0162
Question: 4. Manager of a certain city hotel reports that mean number of rooms rented
out per night is at least 212. A sample of 150 night gives a mean value of 201.3 rooms
and standard deviation of 45.5 rooms. At 0.01 level of significance cheek the manager’s
claim?
Solution:
1.
Ho : µ = 212
Ha : µ < 212
x – 0
2.
Z
3.
Reject Ho : if z  - z0.01
s/ n
z0.01 = 2.33
4.
z
201.3 – 212
45.5 / 150
 – 2.88
Since – 2.88  - 2.33, Ho is rejected at 0.01 level. Manager’s claim of at least 212 rooms
being rented out per night is an overstatement.
Case III – A Normal Population Distribution
When n < 40 and  2 is unknown, the z test for testing the population mean cannot be
used. This is primarily because the central limit theorem cannot be invoked to justify the
use of large sample test. As a result, we have to use a t-test.
If X,, X2-----------Xn is a random sample from a normal distribution, the standardized
variable
T 
X 
S/ n
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Tests for single and two population means
has a t distribution with n-1 degrees of freedom. (This has been discussed in the
previous chapters).
When Ho : µ = µ0 is tested against Ha : µ < µ0, the appropriate test statistic in this case
is T 
X 
S/ n
. Here, we use estimated standard deviation of X , which is s/ 𝑛 rather than
 / n . If Ho is true, the appropriate rejection region for which type I error probability is
controlled at the desired level is given by t  t, n-1. Here - t, n-1, is the lower tail t critical
value. Also,
P(type I error) = P (Ho is rejected when t is true)
= P (T  - t, n-1 when T has t distribution with n-1 df)
=

The One Sample t – test
Null hypothesis
Ho : µ = µ0
Test statistic value :
t
x – 0
s/ n
Alternative hypothesis
Rejection Region for a level
Ha : µ > µ0
t  t, n-1 (upper tailed)
Ha : µ < µ0
t  - t, n-1 (lower tailed)
Ha : µ ≠ µ0
either t  t, n-1
 Test
or t  -t, n-1 (two tailed)
Question: 5. The specifications for a certain kind of ribbon call for a mean breaking
strength of 180 pounds. If five pieces of the ribbon (randomly selected from different
rolls) have a mean breaking strength of 169.5 pounds with a standard deviation of 5.7
pounds, test at 0.01 level of significance that the mean breaking strength is lower than
assumed. Assume that the population distribution is normal.
Solution:
1.
Ho : µ = 180
Ha : µ < 180
x – 0
2.
t
3.
Reject Ho if t  -t, n-1
s/ n
t, n-1 = t0.01,4 = 3.747
n-1 = 5-1 = 4 degrees of freedom
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4.
t
169.5 –180
5.7 / 5
 – 4.12
Since – 4.12 < –3.747, Ho is rejected at
 = 0.01. This implies that the mean breaking
strength is lower than specified.
Question: 6. A random sample of 6 steel beams has a mean compressive strength of
58, 392 psi (pounds per square inch) with σ of 648 psi. Use this information and the
level of significance
 = 0.05 to test whether the true average compressive strength of
the steel from which this sample came is 58,000 psi. Assume normality.
Solution:
1.
Ho : µ = 58000
Ha : µ ≠ 58000
x – 0
2.
t
3.
Reject Ho if t  t, n-1
s/ n
or t  - t, n-1
t, n-1 = t0.025,5 = 2.571
4.
t
58392  58000
648 / 5
 1.48
Since -2.571 < 1.48 < 2.571, Ho will not be rejected. The data is not significant enough
to reject the claim that the mean compressive strength of the steel is 58000 psi.
Question: 7. A census of retail establishments in a particular month revealed that mean
monthly turnover of food stores was $ 2500. A random sample of 16 such stores showed
a mean monthly turnover of $ 2660 and σ of $ 480. Could you conclude that mean
monthly turnover had changed since census? Use
Solution:
1.
 = 0.05 and assume normality.
Ho : µ = 2500
Ha : µ ≠ 2500
x – 0
2.
t
3.
Reject Ho if t  t0.025,15 or t  -t0.025,15
s/ n
t0.025,15 = 2.131
4.
t
2660  2500
480 / 16
 1.33
Since -2.131 < 1.33 < 2.131, the data is not significant enough to reject Ho.
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Tests for single and two population means
2.
Tests for two population means
When we are comparing two different populations, we can either conduct an
interval estimate of difference between two population means or we can test the
hypothesis that the two population means are equal. For instance, we might wish to
investigate the difference between the efficacy of two drugs. As another example, we
might want to investigate if girls perform better than boys in exams. Let µ, be the
average marks obtained by girls and µ2 be the average marks obtained by boys. Hence,
the appropriate null hypothesis in this case could be Ho : µ1 - µ2 = 0 (ie µ1 = µ2) against
the alternative hypothesis Ha : µ1 - µ2 > 0 (ie µ1 > µ2).
Difference between two population means
The difference between two population means is denoted by µ1 - µ2. In this
section, we are going to discuss inferences on this difference, i.e µ1 - µ2. Often, we are
interested in testing whether the two population mean are equal, so that their difference
is zero. This implies µ1 - µ2 = 0 or µ1 = µ2. In the more general case, an investigator
might want to test µ1 - µ2 =  0 , where  0 is some constant.
Basic assumptions
1. Let there be two normal populations with mean µ1 and µ2 and the known
variances  12 and  2 2 .
2. Let X1, X2 ----------------Xm be a random sample of size m from the first
population.
3. Let Y1, Y2----------------Yn be a random sample of size n from the second
population.
4. The sample X and Y are mutually independent.
The estimator of µ1 - µ2 is X - Y , is difference between the corresponding sample
means. To get the test statistic for the test, we standardize the estimator.
Expected Value of 𝑋 – 𝑌
E ( X - Y ) = E ( X ) – E (Y )
= µ1 - µ2
Hence, X - Y is an unbiased estimator of µ1 - µ2
Standard deviation of 𝑋 – 𝑌
Standard deviation of 𝑋 – 𝑌
V(X – Y)  V(X)  V(Y)  2Cov (X – Y) 
12
m

 22
n
(The Covariance term is equal to zero)
S.D. ( X - Y ) =  X Y 
 12
m

 22
n
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Test Procedures for difference in means
Case I : Normal populations with known variances
In this case, both the population distributions are assumed to be normal and their
variances,  12 and  2 2 are known. As a result, both X and Y will be normally
distributed. This implies, the estimator. X - Y will have a normal distribution, with mean
(µ1 - µ2) and standard deviation  X  Y .
When we standardize X - Y , we get a standard normal variable.
Z
( X  Y )  ( 1  2 )
 12
m

 22
n
When we do testing of hypothesis, we denote the null value by  0 , which is usually zero,
so that Ho :
µ1 - µ2 =  0 Hence, we replace µ1 - µ2 by  0 in the expression for test
statistic.
If the alternative hypothesis is Ha : µ1 - µ2 >  0 , a value 𝑥 - 𝑦 significantly greater than
 0 will provide evidence against Ho, in which case we reject it. Such a 𝑥 - 𝑦 value implies
a positive and large value of z. Therefore, Ho is rejected if z is greater than or equal to a
chosen critical value, which is z (Since Z has a standard normal distribution). Here,
is
the level of significance or type I error probability.
Test procedure
Null hypothesis
Ho : µ1 - µ2 =  0
Test Statistic Value : Z
x – y – 0
 12  22
m

n
Alternative hypothesis
Rejection region for level
Ha : µ1 - µ2 >  0
z  z (Upper tailed)
Ha : µ1 - µ2 <  0
Z  -z (Lower tailed)
Ha : µ1 - µ2 ≠  0
either z  z or z  - z
 Test
(Two tailed)
Note : P-value is calculated in the same manner as z test for a single population mean.
Question: 8 An experiment was performed to compare the toughness of high
purity steel with commercial steel. The 32 specimens of high purity steel yielded
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the sample average toughness of 65.6, where as 38 specimens of commercial
steel gave the corresponding average of 59.8. Since the high purity steel is more
expensive, it’s use for a certain application can be justified only if its toughness
exceeds that of commerce steel by more than 5. Assuming both toughness
distributions to be normal, σ1 = 1.2 and σ2 = 1.1, test the relevant hypothesis at
0.01
Solution:
1.
Ho : µ1 - µ2 = 5
Ha : µ1 - µ2 > 5
2.
Z
x – y – 0
 12  22
m
3.

n
Reject Ho if z  z
z0.01 = 2.33
4.
Z
(65.6 – 59.8) – 5
(1.2) 2 (1.1) 2

32
38

0.8
0.0768
 2.88
Since 2.886 > 2.33, we will reject Ho The data is significant and does not justify the use
of high purity steel.
Question: 9. To estimate the average nicotine content of two brands of cigarettes, 50
cigarettes of the first brand were taken that gave average nicotine content of 2.61 mg.
The second sample of 40 cigarettes of the second brand gave average nicotine content of
2.38 mg. Assume nicotine content to be normally distributed for both the brands. Also,
σ1 = 0.12mg and σ2 = 0.14mg. Test the Ho : µ1 - µ2 = 0.2 against µ1 - µ2 ≠ 0.2 using
 = 0.05
Solution:
1.
Ho : µ1 - µ2 = 0.2
Ha : µ1 - µ2 ≠ 0.2
2.
Z
x – y – 0
 12  22
m
3.

n
Reject Ho if z  z or z  -z
z = z0.025 = 1.96
4.
Z
(2.61  2.38)  0.2
(0.12) 2 (0.14)2

50
40
1.08
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Since, -1.96 < 1.08 < 1.96, Ho cannot be rejected. This implies that difference between
𝑥 - 𝑦 = 2.61-2.38 = 0.23 and the hypothesized null value (  0 = 0.20) is not significant.
This difference can be attributed to chance.
Case II : Large Sample Tests
When we are dealing with independent random sample from the two populations (may or
may not be normal) having unknown variances, we can still use the Z test from case I
provided both the samples are large enough to invoke central limit theorem. In this
situation, we substitute S1 for σ1 and S2 for σ2. Also, central limit theorem assures that
X - Y will have approximately normal distribution regardless of the population
distribution. The standardization of X - Y gives the standard normal variable:
Z
X  Y  ( 1  2 )
S12
S2
 2
m
n
This test is known as large sample test with approximate significance levels
. It use the
test statistic value
𝑍=
𝑥 – 𝑦 – ∆0
𝑠12
𝑠2
+ 2
𝑚
𝑛
These tests are based on z critical values as calculated in the previous case. These tests
are appropriate if both m > 40 and n > 40.
Question: 10. There are two training programmes for the newly hired telephone
marketing representatives. To test the relative effectiveness of the two programmes, a
proficiency test was taken 45 representatives from the first training programme gave a
mean score of 76 points with standard derivation of 13.5, whereas 40 representatives
from the second training programme gave a mean score of 77.97 points with a standard
deviation of 9.05. The manager wants to test whether the difference in the mean scores
is significant. Use
Solution:
1.
 = 0.05
Ho : µ1 - µ2 = 0
Ha : µ1 - µ2 ≠ 0
𝑥 –𝑦 – ∆0
2.
𝑍=
3.
Reject Ho if z  z0.025 or z  -z0.025
2
𝑠2
1 + 𝑠2
𝑚
𝑛
z0.025 = 1.96
4.
Z
(76  77.97)  0
(13.5)
45
2

(9.05)
40
2

1.97
= –0.797
2.469
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Since, -1.96 < -0.797 < 1.96, Ho cannot be rejected. Hence, there is no significant
difference between the two training programmes as they are equally efficient.
Question: 11. A company claims that it’s light bulbs are superior to those of its’ main
competitor. If a study showed that a sample of 40 of its bulbs has a mean lifetime of 647
hours of continuous use with a standard deviation of 27 hours, while a sample of 40
bulbs made by its main competitor had a mean lifetime of 638 hours of continuous use
with standard deviation of 31 hours, does this substantiate the claim at the 0.05 level of
significance?
Solution:
1.
Ho : µ1 - µ2 = 0
Ha : µ1 - µ2 ≠ 0
𝑥 –𝑦 – ∆0
2.
𝑍=
3.
Reject Ho if z  z0.05
2
𝑠2
1 + 𝑠2
𝑚
𝑛
Z0.05 = 1.645
Z
4.
647  638
27 2
312

40
40
1.38
Since 1.38 < 1.645, Ho cannot be rejected. Therefore, the observed difference between
the two sample means is not significant.
Case III : The Two Sample t-test (Small sample tests)
We usually do not know the value of the underlying population variances. The Central
Limit thereon cannot be invoked if either of the sample sizes is small and population
variances are unknown. In this case, we have to make certain assumptions about the
underlying population distributions.
Assumptions:
1. Both populations are normal.
2. Both the samples X and Y are random and independent of one another.
Now, when we want to test for the difference in the population means, we standardize, 𝑥
- 𝑦. Since the sample sizes are small, the standardized variable will not have az
distribution. Rather, it will have a t distribution.
Theorem
When the population distributions are both normal, the standardized variable
T 
X  Y  ( 1  2 )
S12 S 2 2

m
n
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has approximately t distribution with df
υ estimated from the data by
2
 s12 s22 
  
m n
υ= 2  2
( s1 / m) ( s22 / n) 2

m –1
n –1
We replace µ1 - µ2 by  0 to get the test statistic value.
Two Sample t test procedure
Null hypothesis
Ho : µ1 - µ2 =  0
Test Statistic Value :
𝑡=
𝑥 –𝑦 – ∆0
2
𝑠2
1 +𝑠 2
𝑚
𝑛
Alternative hypothesis
Rejection Region for Approximate Level  Test
Ha : 1 -  2 >  0
t  t, υ
(upper tailed)
Ha : 1 -  2 <  0
t  -t, υ
(lower tailed)
Ha : 1 -  2 #  0
either t  t, υ or t  -t, υ (two tailed)
Question: 12. A. A toy manufacturer sells two types of rubber baby buggy bumpers.
Tests for durability reveal that 13 bumpers of type I lasted for 11.3 weeks with s1 = 3.5
weeks while 10 bumpers of types II lasted for 7.5 weeks with s2 = 2.7 weeks. Type I
costs more to manufacturer and company does not want to produce it until on an
average, it lasts for at least 8 weeks longer than type II. Find at 0.01 level of
significance, what will the manager do?
Solution:
1.
Ho : 1 -  2 = 8
Ha : 1 -  2 < 8
𝑥 –𝑦 – ∆0
2.
𝑡=
3.
Reject Ho if t  t0.01, V
4.
t
2
𝑠2
1 +𝑠 2
𝑚
𝑛
(11.3  7.5)  8
2
(3.5) (2.7)

13
10
2

4.2
  3.25
1.29
Also,
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(
V=
3.52 2.7 2 2

)
13
10
=
(3.52 /13) 2 (27 2 /10) 2

12
9
(0.94  0.729) 2
(0.94) 2 (0.729)2

12
9
2.785
 21
0.132
=
So, the t test will be based on 21 df.
t,
V
= t0.01,21 = 2.518
Since, -3.25 < -2.518, Ho will be rejected. Since the difference between the two types of
bumpers is not at least 8 weeks, the manager will not use type I bumpers.
Question: 13. Let µ1 and µ2 denote true average densities for two different types of
bricks. Assuming normality of the two density distributions, test Ho : µ1 - µ2 = 0 versus
Ha : µ1 - µ2 ≠ o using the following data : m = 6, 𝑥 = 22.73, s1 = 0.164, n = 5, 𝑦 =
21.95, and s2 = 0.24. Use
Solution:
1.
 = 0.01
Ho : µ1 - µ2 = 0
Ha : µ1 - µ2 ≠ 0
𝑥 –𝑦 – ∆0
2.
𝑡=
3.
Reject Ho if t  t0.005, V or t  - t0.005,
4.
t
2
𝑠2
1 +𝑠 2
𝑚
𝑛
22.73  21.95
2
(0.164) (0.24)

6
5
2

V
0.78
 6.190
0.126
Also,
(
V =
(0.164) 2 (0.24) 2 2

)
6
5
((0.164) 2 / 6) 2 ((0.24) 2 / 5) 2

6
4
=
(0.0045  0.0115) 2
(0.0045) 2 (0.0115) 2

5
4
=
(0.016) 2
0.000004  0.000033
=
6.91
The test will be based on 6 df.
t0.005,
6
= 3.707
Since 6.190 > 3.707, Ho will be rejected. There is a significant difference between the
true average densities for two different types of brick.
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Tests for single and two population means
Case IV : Pooled t procedures
This case assumes the two normal population distributions have equal variances, ie  12 =
 2 2 =  2 . The underlying population distributions have equal spreads but their means
are different, which implies that they are centered at different point.
We begin by standardizing X - Y , the estimation of 1 -  2
Z
X  Y  ( 1  2 )

2
m


2
n
= Z
X – Y – ( 1 – 2 )
 2(
1 1
 )
m n
which has a standard normal distribution. The common variance, σ2 must be estimated
from the sample data, this can be done in variance ways. One way is that we use
estimator S12, which is the variant of m observations from the first sample. Alternatively,
we can also use S22, the variance of the n observations from the second sample. A better
way is to combine or pool both these sample variances. Hence, we use the weighted
average of the two, where the corresponding sample sizes are the weights. By using the
weighted average, we make sure that the bigger sample that contains more information
about σ2 should be given a larger weight.
Pooled/Combined estimator of σ2
𝑠𝑝2 = t 
𝑠𝑝2 =
512  492
1 1
28.609

4 4
 0.99
𝑚 –1
𝑛 –1
𝑠21 +
𝑠2
𝑚 + 𝑛 –2
𝑚+ 𝑛 –2 2
This estimator has m+n-2 df, where m-1 df are contributed by the first sample in
estimation of σ2 and n-1 df are contributed by the second sample in the estimation of σ2.
If we replace σ2 by 𝑠𝑝2 in the standardized variable, then it will have a t distribution with
m+n-2 df.
Note: We should use two sample t procedure for testing the differences in population me
unless there is a compelling reason to believe that the two population variances are
some, in which case we use the pooled t test.
Question: 14. In comparing two kinds of paints, a testing agency finds that 4 1-gallon
cans of one brand cover on the average 512 square feet with s1 = 31 sq. feet while 4 1gallon cans of the other brand cover on an average 492 sq. feet with s2 = 26 sq. feet.
Test if the difference between the average area covered by the two brands is significant
at 0.05 level of significance. Assume that the two populations sampled are normal and
have equal variances.
Solution:
1.
Ho : µ1 - µ2 = 0
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Tests for single and two population means
Ha : µ1 - µ2 ≠ 0
2.
𝑡=
𝑥 –𝑦 – ∆0
2
𝑠2
1 +𝑠 2
𝑚
𝑛
Where 𝑆𝑃2 =
3.
m –1
n –1
. s12 
.s 22
mn – 2
mn –2
Reject Ho if t  t, V or t  - t, V
υ = m+n-2
Where
=4+4-2=6
t0.025,
4.
6
3(31)2  3(26)2
 28.609
6
𝑆𝑃2 =
t
= 2.447
512  492
28.609
1 1

4 4
 0.99
Since, -2.447 < 0.99 < 2.447, Ho cannot be rejected. Sample results are not strong
enough to reject Ho.
Question: 15. To test the effectiveness of a given fertilizer on the wheat production, 24
plots of equal area were chosen, half of them where treated with fertilizers and half were
not (this was the control group). Mean yield of the treated plots was 51 bushels with s1 =
0.36 bushels and the mean yield of the untreated plots was 4.8 bushels with s2 = 0.40
bushels. Can we conclude that there is a significant improvement in wheat production
because of the fertilizer treatment if (a)
Solution:
1.
= 0.01 and (b) = 0.05
Ho : µ1 - µ2 = 0
Ha : µ1 - µ2 > 0
2.
𝑡=
𝑥 –𝑦 – ∆0
𝑆21 (
1
1
+ )
𝑚
𝑛
Where 𝑆𝑃2 =
3.
(m –1) s12  (n –1) s22
mn – 2
Reject Ho if t  t, υ
Where
υ = m+n-2
= 12+12-2 = 22
t0.01,22 = 2.51
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Tests for single and two population means
and t0.05,22 = 1.72
𝑆𝑃2 =
4.
t
At
11(0.4)2  11(0.36)2
 0.1448 s p  0.1448  0.38
22
5.1  4.8
1 1
0.38

12 12

0.3
1
0.38
6
 1.93
 = 0.01,
1.93 < 2.51 ie t < t0.01, υ
Hence, Ho will not rejected
At
 = 0.05,
1.93 > 1.72 ie t > t0.05, υ
We will reject Ho
Practice Questions
Q. 1.
A random sample of 25 employees has a mean weekly wage of $ 130. Could
this sample have been drawn from a population normally distributed about a mean of $
120 with a standard deviation of $ 10 ?
Q. 2.
A trucking firm is suspicious of the claim that the average lifetime of certain
tires is at least 28,000 miles. To check the claim, the firm puts 40 of these tires on its
trucks and gets a mean lifetime of 27,463 miles with a standard deviation of 1348 miles.
What can it conclude if the probability of type I error is to be at most 0.01 ?
Q. 3.
The specifications for a certain kind of ribbon call for a mean breaking strength
of 185 pounds. If five pieces randomly selected from different rolle have breaking
strengths of 171.6, 191.8, 178.3, 184.9 and 189.1 pounds, test the null hypothesis µ =
185 pounds against the alternative hypothesis µ < 185 pounds at the 0.05 level of
significance.
Q. 4.
The security department of a factory wants to know whether the true average
time required by the night guard to walk his round is 30 minutes. If, in a random sample
of 32 rounds, the night guard averaged 30.8 minutes with s = 1.5 minutes, determine
whether there is sufficient evidence to reject the null hypothesis µ = 30 in favor of the
alternative hypothesis µ ≠ 30 minutes. Use
 = 0.01
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Tests for single and two population means
Q. 5.
The electric bulbs of manufacturer A have a mean lifetime of 1400 hours with a
standard deviation of 200 hours and those of B have a mean of 1200 hours with a
standard deviation of 100 hours. If a sample of 125 bulbs each is tested, what is the
probability that. A has a mean lifetime that is (a) less than 160 hours than B’s bulbs? (b)
250 hours more than B's bulbs ?
Q. 6.
Two samples were drawn independently from two normal populations. The
relevant results are:
m = 15
x = 60.2
m
 (x – x)
2
1
n = 19
y = 61.8
m
( y – y)
2
 83.27
1
Assuming that the population variance are equal, test the hypothesis that the population
means are equal.
Q. 7.
A random sample of 40 male employees is taken at the end of an year and the
mean number of hours of absenteeism for the year is found to be 63 hours. A similar
sample of 50 female employees has a mean of 66 hours. Could these samples have been
drawn from a population with the same mean and standard deviation of 10 hours?
Q. 8.
The management of a plant wishes to test the effectiveness of a new technique
in assembling 9 certain device. Two groups of a employees each were selected. The
group using the old technique gave the mean length of time taken to assemble the
device to be 52.8 minutes and the group using the new technique gave the mean to be
56mins. An estimate of the variance obtained by pooling the sample results was found to
be 22.1. What conclusions can be drawn about the effectiveness of the two techniques ?
Q. 9.
To test the claim that the resistance of an electric wire can be reduced by more
than 0.05 ohm. by alloying 32 valves obtained for standard wire yielded x = 0.136 ohm
and s1 = 0.004 ohm, and 32 valves obtained for alloyed wire yielded y = 0.083 ohm
and s2 = 0.005 ohm. At the 0.05 level of significance, does this support the claim?
Q. 10. The following random samples are measurements of the heat producing capacity
(in millions of calories per ton) of specimens of coal from two mines:
Mine 1 : 8260 8130 8350 8070 8340
Mine 2 : 7950 7890 7900 8140 7920 7840
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Tests for single and two population means
Use the 0.01 level of significance to test whether the difference between the means of
these two samples is significant Assume equal variances for the two underlying
population distributions.
Symbols:
 X  0  X
Z     1
Z / 2
/ n
m
(X  X )
1
 12
2
 56.2 t  2
t
X  0
S/ n
Y  2 2 1 -  2 X - Y
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t / 2
0  2