This document discusses hypothesis testing for a single mean. It explains the steps for conducting a hypothesis test on a single mean: 1) State the null and alternative hypotheses, 2) Choose a significance level, 3) Select the appropriate test statistic (z-test or t-test), 4) Identify the critical region, 5) Calculate the test statistic, and 6) Make a conclusion about whether to reject or fail to reject the null hypothesis based on where the test statistic falls relative to the critical region. It then provides examples demonstrating how to conduct hypothesis tests for single means.

1. HYPOTHESIS TESTING PART-II
SINGLE MEAN
NADEEM UDDIN
ASSOCIATE PROFESSOR
OF STATISTICS

2. Hypothesis Test for single mean :
Statisticians follow a formal process to determine whether to
reject a null hypothesis, based on sample data. This process
called hypothesis testing.
1. State the hypotheses:
This involves stating the null and alternative hypotheses. The
hypotheses are stated in such a way that they are mutually
exclusive. That is, if one is true, the other must be false.
Set Null hypothesis Alternative
hypothesis Number of tails
1 μ = μ ≠ 2
2 μ > μ < 1
3 μ < μ > 1
o
o
o
o
o
o

3. 2. Level of significance:
α = 0.01, 0.05 or any given value
3. Test statistic:
2
1. when known
2
2. when unknown and n 30
2
3. t= when unknown and n<30
X
z
X
z
s
n
X
s
n
n










 


4. 4. Critical Region:
The set of values outside the region of
acceptance is called the region of rejection. If
the test statistic falls within the region of
rejection, the null hypothesis is rejected. In such
cases, we say that the hypothesis has been
rejected at the α level of significance. The
following steps are use to find the critical region.

5. For Test statistic(1) and (2)
Z > Zα/2 and Z< - Zα/2 when H1: μ ≠
Z > Zα when H1: μ >
Z< - Zα when H1: μ <
o
o
o

6. For Test statistic(3)
t > tα/2,n-1 and t < - tα/2,n-1 when H1: μ ≠
t > tα, n-1 when H1: μ >
t < - tα, n-1 when H1: μ <
o
o
o

7. 5. Computation:
Find the value of the test statistic
6. Conclusion:
If the calculated value of test statistic falls in
the area of rejection ,we reject the null
hypothesis otherwise accept it.

8. Test Concerning Single Means
Example-1:
American Theaters knows that a certain hit movie ran an average
of 84 days in each city, and the corresponding standard deviation
was 10 days. The manager of the southeastern district was
interested in compairing the movie’s popularity in his region with
that in all of American’s other theaters. He randomly chose 75
theaters in his region and found that they ran the movie an average
81.5days. State appropriate hypotheses for testing whether there
was a significant difference in the length of the picture’s run
between theaters in the southeastern district and all American’s
other theaters at a 1 percent significance level.

9. Solution:
n =75, σ =10, 𝑥 = 81.5, 𝛼 = 0.01, 𝜇 = 84
1. Hypothesis
H0: 𝜇 = 84
H1: 𝜇 ≠ 84
2. Level of significance 𝛼 = 0.01
3. Test statistic 𝑍 =
𝑥−𝜇
σ
𝑛

10. 4. Critical Region
In case of two tail test i.e. H1 𝑖𝑠 ≠.
Reject H0, if 𝑍 𝑐𝑎𝑙 ≤ −𝑍𝑡𝑎𝑏 or 𝑍 𝑐𝑎𝑙 ≥ 𝑍𝑡𝑎𝑏.
Where 𝑍𝑡𝑎𝑏 = 𝑍 𝛼
2
= 𝑍0.01
2
= 𝑍0.005 = 2.58
𝑍 𝑐𝑎𝑙 ≤ −2.58 or 𝑍 𝑐𝑎𝑙 ≥ 2.58.
(Using inverse area of normal table)

11. 5. Computation
𝑍 =
𝑥−𝜇
σ
𝑛
𝑍 =
81.5−84
10
75
= −2.17
6. Conclusion:
The value of Zcal lies in the area of acceptance
therefore we accept H0.The length of run in the
southeast is not significantly different from the
length of run in other regions.

12. Example-2:
Hinton press Hypothesizes that the average life of its
largest web press is 14500 hours. They know that the
standard deviation of press life 2100 hours. From a
sample of 25 presses, the company finds a sample mean
of 13000 hours. At a 0.01 significance level, should the
company conclude that the average life of the presses is
less than the hypothesized 14500 hours?

13. Solution:
n=25, σ =2100, 𝑥=13000, 𝛼=0.01, 𝜇=14500
1. Hypothesis
H0: 𝜇 = 14500
H1: 𝜇 < 14500
2. Level of significance 𝛼 = 0.01
3. Test statistic 𝑍 =
𝑥−𝜇
σ
𝑛

14. 2.33 –0– 
4. Critical Region
In case of lower tail test i.e. H1 𝑖𝑠 ˂.
Reject H0, if 𝑍 𝑐𝑎𝑙 ≤ −𝑍𝑡𝑎𝑏
Where 𝑍𝑡𝑎𝑏 = 𝑍 𝛼 = 𝑍0.01 = −2.33
𝑍 𝑐𝑎𝑙 ≤ −2.33
(Using inverse area of normal table)

15. 5. Computation
𝑍 =
𝑥−𝜇
σ
𝑛
𝑍 =
13000−14500
2100
25
= −3.57
6. Conclusion:
We should reject H0. The average life is
significantly less than the hypothesized
value.

16. Example-3:
The average commission charged by full-service
brokerage firms on a sale of common stock is $144, and
the standard deviation is $52.Joel Freelander has taken
a random sample of 121 trades by his clients and
determined that they paid an average commission of
$151. At a 0.10 significance level, can Joel conclude
that his clients’ commissions are higher than the
industry average?

17. Solution
n=121, σ =52, 𝑥=151, 𝛼=0.10, 𝜇=144
1. Hypothesis
H0: 𝜇 = 144
H1: 𝜇 > 144
2. Level of significance 𝛼 = 0.10
3. Test statistic 𝑍 =
𝑥−𝜇
σ
𝑛

18. 4. Critical Region
In case of upper tail test i.e. H1 𝑖𝑠 ˃.
Reject H0, if 𝑍 𝑐𝑎𝑙 ≥ 𝑍𝑡𝑎𝑏
Where 𝑍𝑡𝑎𝑏 = 𝑍 𝛼 = 𝑍0.10 = 1.28
𝑍 𝑐𝑎𝑙 ≥ 1.28(Using inverse area of normal table)
5. Computation
𝑍 =
𝑥−𝜇
σ
𝑛
𝑍 =
151−144
52
121
= 1.48
6. Conclusion:
Reject H0.Their commission are significantly
higher.
 –0– 1.28

19. Example-4:
A random sample of 8 cigarettes of a certain
brand has an average nicotine content of 4.2
milligrams and a standard deviation of 1.4
milligrams. Is this in line with the
manufacturer’s claim that the average nicotine
content does not exceed 3.5 milligrams?
Use α =0.01

20. Solution:
n=8, 𝑠=1.4, 𝑥=4.2, 𝛼=0.01, 𝜇=3.5
1. Hypothesis
H0: 𝜇 ≤ 3.5
H1: 𝜇 > 3.5
2. Level of significance 𝛼 = 0.01
3. Test statistic 𝑡 =
𝑥−𝜇
𝑠
𝑛

21. 4. Critical Region
In case of upper tail test i.e. H1 𝑖𝑠 ˃.
Reject H0, if 𝑡 𝑐𝑎𝑙 ≥ 𝑡𝑡𝑎𝑏
Where 𝑡𝑡𝑎𝑏 = 𝑡 𝛼 , 𝑛−1 = 𝑡0.01, 8−1 = 𝑡0.01,7=2.998
𝑡 𝑐𝑎𝑙 ≥ 2.998
5. Computation 𝑡 =
𝑥−𝜇
𝑠
𝑛
𝑡 =
4.2−3.5
1.4
8
= 1.41
6. Conclusion: Accept H0.(we accept that the average
nicotine content at most 3.5 milligrams).
 –0– 2.998