Hypothesis Testing

HYPOTHESIS TESTING
ONE TAIL & TWO TAIL TESTS

HYPOTHESIS
 Hypothesis: “Proposition or Supposition made from known facts
as basis for reasoning or investigation” – The Oxford Dictionary.
 Hypothesis Testing begins with an assumption, called a
hypothesis, that we make about a population parameter.
 Hypothesis Testing involves making two quantitative
statements about some population parameter:
 One based on a claim or belief or suspicion – Null Hypothesis
 Other statement is opposite of the first – Alternative Hypothesis
 Determining whether the null hypothesis is true or false is
based on the sample information, but not by only intuition.
 Null & Alternative Hypothesis covers all possible values of
parameter.
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BirinderSingh,AssistantProfessor,PCTE
Ludhiana

HYPOTHESIS
 Null Hypothesis can be inferred from alternative hypothesis
and vice a versa.
 Fuel Consumption of Maruti Car is atleast 18 km/ltr
 Null Hypothesis : FC ≥ 18 km/l or H0 : µ ≥ 18
 Alternative Hypothesis : FC < 18 km/l or H1 : µ < 18
 ‘18’ is called hypothesized value and is denoted by µH0
 Plant fills exactly 500 ml of milk in every bottle
 H0 : µ = 500
 H1 : µ < 500, µ > 500, µ ≠ 500,
 µH0 = 500
 My Secretary does not make more than 5 errors in a letter
 H0 : µ ≤ 5
 H1 : µ > 5
 µH0 = 5 3
Ludhiana

LEVEL OF SIGNIFICANCE
 LOS is denoted by α = 1 – CL
 Probability levels are used to decide whether to accept or
reject null hypothesis
 LOS is probability of rejecting a true null hypothesis
 LOS = 5% means that experimenter runs the risk of
rejecting a correct hypothesis of 5 out of 100 times.
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Ludhiana

REGION OF ACCEPTANCE OR REJECTION
 LOS is mainly employed in decision making problems
because it distinguishes between true regions known as
region of acceptance or rejection.
 A sample of values such that if sample statistic falls in
range, the null hypothesis is accepted & is called Range of
Acceptance and where rejected is called Range of Rejection.
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Ludhiana

COMPUTATIONAL PROCEDURE IN
HYPOTHESIS TESTING
 Define Null & Alternative Hypothesis
 Select estimator & determine its distribution
 Select Significance Level
 Define decision rule (One Tail or Two Tail Test)
 Compute Critical Value/s
 In case of Two Tail Test
 CVL =µH0 − 𝒛α/ 𝟐 𝝈 𝒙
 CVR =µH0 + 𝒛α/ 𝟐 𝝈 𝒙
 In case of One Tail Test
 CV =µH0 − 𝒛 𝝈 𝒙 (Left Tail Test)
 CV =µH0 + 𝒛 𝝈 𝒙 (Right Tail Test)
 Make the Statistical Decision by checking whether the
sample mean lies in region of acceptance or rejection
 Make the managerial decision 6
Ludhiana

HYPOTHESIS TESTING OF MEANS
PRACTICE PROBLEMS – TWO TAIL
 A bottling machine is supposed to fill 500 ml of milk in a
bottle. However, the contents are not always exact. The
quantity is known to be normally distributed with a
standard deviation of 10 ml. A sample check of 36 bottles
revealed mean of 515 ml. The manager of bottling plant
wants to know if machine is working satisfactorily? Test at
5% LOS. (496.73≤ µ≤ 503.27)
(Rejection Region)
(Machine not OK)
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Ludhiana

PRACTICE PROBLEMS – ONE TAIL
 A manufacturer claims that the mean life of a tube is 4000 hrs.
(at least). A retailer wants to find out if manufacturer claim is
correct or not. Random sample of 65 tubes checked by him
revealed a mean life of 3982 hrs with a standard deviation of
80 hrs. Should the tubes be accepted at 5% level of confidence?
(CV = 3983.67)
(Rejection Region)
(Retailer rejects)
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Ludhiana

 There are 1000 customers who are authorized credit purchase
from a grocery store. The store has targeted that, on an
average, credit sales in a month should not exceed Rs. 10000. It
is known from past experience that SD of credit sales is Rs.
500. A random sample of 81 such customers revealed a mean of
Rs. 10250 towards credit sales. Would you conclude at 5%
significance level that store is achieving its targets?
(CV = 10087.65)
(Rejection Region)
(Not achieving)
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Ludhiana

 A TV documentary telecast on ‘Nikat Darshan’ channel claimed
that adult Punjabi males are 10 kg over weight on an average.
18 randomly selected individuals showed an excess weight of
11.6 kg with SD of 2.7 kg. Check at LOS of 0.01, is there any
reason to doubt validity of claim. Assume normal distribution
of excess weight.
(8.16≤ µ≤ 11.84)
(Acceptance Region)
(Claim is correct)
 Here t α but not tα/2 as area mentioned is under both tails
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Ludhiana

 A manufacturer has claimed that life of a refrigerator produced
by his company is atleast 14500 hrs. From a sample of 25
refrigerators, sample mean was found to be 13600 hrs with SD
of 2100 hrs. Would you accept the manufacturer’s claim at 0.01
LOS?
(CVL = 13453.36)
(Acceptance Region)
(Claim is correct)
 Here t 2α but not tα as area mentioned is under one tail
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Ludhiana

HYPOTHESIS TESTING OF PROPORTIONS
 A B-School has advertised that at least 80% of the students are
placed in good jobs during interviews. In a sample survey of
100 ex-students, it emerged that 74 were placed in good jobs
during recruitment. Validate claim of B School at 0.05 LOS.
(CVL = 0.734)
(Acceptance Region)
(Claim is correct)
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Ludhiana

HYPOTHESIS TESTING OF PROPORTIONS
 HR Sol. Pvt. Ltd. feels that in NCR region, 15% of MBA
students opts for HR. In a sample survey of 120 students, it
was found that 22 of them had opted for HR in 2nd year. At 0.02
LOS, is there any evidence to suggest that firm’s perception is
correct?
(0.074≤ pH0 ≤ 0.226)
(Acceptance Region)
(Perception is correct)
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Ludhiana

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Ludhiana

DECISION FLOW DIAGRAM -
ESTIMATION
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Ludhiana
Start
Is
n≥30
Is pop.
Known to
be
normally
distributed
Use ‘Z’ table Stop
Use a
Statistician
Is SD
known
?
Use ‘Z’
table
Stop
Use ‘t’
table
Stop

Hypothesis Testing

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