This document provides examples and explanations of common hypothesis testing techniques including: - Z tests for large samples with known population variance to test claims about population means - T tests for small samples with unknown population variance - Tests comparing two population means using Z tests for large samples and T tests for small samples - One-tailed and two-tailed tests at various significance levels (e.g. 5%, 10%) Step-by-step solutions and calculations are shown for multiple examples testing claims about means, differences in means, and whether sample data is consistent with hypothesized population parameters.

1. HYPOTHESIS TESTING EXAMPLES

2. TEST OF HYPOTHESIS CONCERNING NORMAL POPULATION,
INFINITE, LARGE SAMPLES, WITH KNOWN POPULATION
VARIANCE.
( Z TEST)
If the size of sample exceeds 30 it should be
regarded as a large sample.
a) Testing Hypothesis about population mean μ.
Z= x - μ
S.E. of x
Z= x– μ
σp/√n x = sample mean
σp = population s.d.
n= sample size
Population size is infinite
.

3. SOME HYPOTHESIS TESTING EXAMPLES
ILLUSTRATION1 (TWO-TAILED TEST)
The mean lifetime of a sample of 100 light tubes produced by a
company is found to be 1,580 hours Test the hypothesis at 5% level
of significance that the mean lifetime of the tubes produced by the
company is 1,600 hours with standard deviation of 90 hours.
Solution:
z= -2.22.
Critical value of Z = ± 1.96

4. ILLUSTRATION2 ONE TAILED (UPPER TAILED)
An insurance company is reviewing its current policy rates.
When originally setting the rates they believed that the average
claim amount will be maximum Rs180000. They are concerned
that the true mean is actually higher than this, because they
could potentially lose a lot of money. They randomly select 40
claims, and calculate a sample mean of Rs195000. Assuming
that the standard deviation of claims is Rs50000 and set
α= .05, test to see if the insurance company should be
concerned or not.

5. SOLUTION
Step 1: Set the null and alternative hypotheses
H0 : μ≤ 180000
H1 : μ > 180000
Step 2: Calculate the test statistic
z= = x – μ
σ/√n
= 1.897
Step 3: Set Rejection Region

6. 1.65

7. Step 4: Conclude
We can see that 1.897 > 1.65, thus our test statistic
is in the rejection region. Therefore we fail to
accept the null hypothesis. The insurance company
should be concerned about their current
policies.

8. ILLUSTRATION3 ONE TAILED (LOWER TAILED)
Trying to encourage people to stop driving to
campus, the university claims that on average it
takes at least 30 minutes to find a parking space on
campus. I don’t think it takes so long to find a spot.
In fact I have a sample of the last five times I drove
to campus, and I calculated x = 20. Assuming that
the time it takes to find a parking spot is normal,
and that σ = 6 minutes, then perform a hypothesis
test with level α= 0.10 to see if my claim is correct.

9. SOLUTION
Step 1: Set the null and alternative hypotheses
H0 : μ ≥ 30
H1 : μ < 30
Step 2: Calculate the test statistic
Z= x– μ
σ/√n
= -3.727

10. STEP 3: SET REJECTION REGION

11. STEP 4: CONCLUDE
We can see that -3.727 <-1.28 ( or absolute value is
higher than the critical value) , thus our test
statistic is in the rejection region. Therefore we
Reject the null hypothesis in favor of the alternative.
We can conclude that the mean is significantly less
than 30, thus I have proven that the mean time to
find a parking space is less than 30.

12. EXERCISE

13. REJECTION REGION AND CONCLUSION

14. TEST OF HYPOTHESIS CONCERNING NORMAL
POPULATION, FINITE, LARGE SAMPLES, WITH
KNOWN POPULATION VARIANCE.
( Z TEST)
Z= x– μ
( σ/√n) √(N-n)/(N-1)
N= population size

15. ILLUSTRATION 4
Suppose we are interested in a population of 20
industrial units of the same size, all of which are
experiencing excessive labour turnover problems.
The past record show that the mean annual turnover
is 320 employees, with a standard deviation of 75
employees. A sample of 5 of these industrial units is
taken at random which gives a mean of annual
turnover as 300 employees. Is the sample mean
consistent with the population mean? Test at 5% level.

16. TESTING HYPOTHESIS ABOUT DIFFERENCE
BETWEEN TWO POPULATION MEANS
We assume that the populations are normally
distributed.
The null hypothesis is H0 : μ1 = μ2
i.e. H0 : μ1 -μ2 = 0
Z = x1 - x2
√σ1
2/ n1 + σ2
2 / n2
In case σ1
2 and σ2
2 are not known then s1
2 and s2
2 can be
used.

17. ILLUSTRATION 1
A test given to two groups of boys and girls gave
the following information:
Gender Mean score S.D. Sample Size
Girls 75 10 50
Boys 70 12 100
Is the difference in the mean scores of boys and girls statistically significant?
Test at 1% level.
Z=2.695, table value Z =2.58.

18. ILLUSTRATION 2
Suppose you are working as a purchase manager
for a company. The following information has been
supplied to you by two manufacturers of electric
bulbs:
Company A Company B
Mean life ( in hours) 1,300 1,288
Standard deviation( in hours) 82 93
Sample size 100 100
Which brand of bulbs are you going to purchase if you desire to take
a risk of 5%

19. SOLUTION HINT
Take the null hypothesis that there is no significant difference
In the quality of two brands of bulbs i. e. H0 : μ1 = μ2
Z= 0.968

20. POPULATION NORMAL, POPULATION INFINITE,
SAMPLE SIZE SMALL (30 OR LESS) AND VARIANCE
OF POPULATION UNKNOWN
t statistic.
t = x – μ
( σs/√n) with degree of freedom =(n-1)
σs = ∑(X-X)2
(n-1)

21. ILLUSTRATION
A tea stall near the New Delhi Railway Station is
making a sales of 500 tea cups per day. Because of
the development of bus stand nearby , it expects to
increase its sales. During the first 12 days after the
start of the bus stand, the daily sales were recorded
which are as under:
550, 570,490, 615,505,580,570,460,600,580,530,526.
On the basis of this sample information, can one
conclude that the tea stall’s sales have increased?
Use 5% level of significance.
t= 3.558

22. TESTING HYPOTHESIS ABOUT DIFFERENCE
BETWEEN TWO POPULATION MEANS
We assume that the populations are normally
distributed.
Variance of population is unknown.
The null hypothesis is H0 : μ1 = μ2
Small Sample t test
t = x 1 – x 2
s√ (1/n1 + 1/n2)
s= ∑(x1 - x1 ) 2 + ∑(x2 – x2 ) 2
(n1 + n2 -2 )
with degree of freedom = (n1 + n2 -2 )