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Hypothesis Testing
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A hypothesis refers to a claim or an opinion given about an item or issue .This claim has
to be tested statistically to establish whether it is correct or not. The two hypothesis to be tested
are the null hypothesis (H0 ) and the alternative hypothesis (H1 ). Both the null and alternative
hypotheses are mutually exclusive and exhaustive. They cannot share the same outcome in
common. Together they account for all possible outcomes.
The null hypothesis refers to the hypothesis being tested.eg. a bureau of standard may
walk into a sugar making company with the intention of confirming that 2kg bags of sugar
produced are actually 2kgs and no less. Hence the hypothesis test conducted has null hypothesis
being Ho = each bag weigh 2kgs. The test results will either confirm or disprove this.
The alternative hypothesis is a contradiction of the null hypothesis formulated on the fact
that the belief might be untrue. Therefore, we will reject it. When the null hypothesis is rejected,
we accept the alternative hypothesis.
The procedure to be followed in hypothesis testing include; formulating the hypothesis
H0 and H1; determining the analysis plan in form of the significance level; identifying the test
statistic of the sample data and finally formulating the decision. For small sample size where
(n<30) the arithmetic mean are not normally distributed. Hence, students t distribution must be
used to estimate the population mean.
In our case, we are going to carry out three research question and test their hypothesis
Question 1
A tyres manufacturing company by the name Super Trends Tyres Ltd introduced new brands of
tyres, which in their advertisement claim to be superior to their only competitor brand,
‘Roadsmaster.’ The Roadmaster brand manager disputed this for quite some time and termed it
as an advertisement scheme. To settle this, the brand managers for both companies agreed to run
a road test for each brand using ten Nissan vehicles of uniform weight and identical
specifications with each car fitted with both brands tyres; one brand in front and the other at the
rear. The cars were then driven to cover a distance of 5000 kilometers each and the trend wear
was recorded. Results were as follows:
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Super Trendy tyres
Roadmaster Tyres
Centimeters
Centimeters
1
1.08
1.12
Paired
Difference
(x)
-0.04
2
1.06
1.09
-0.03
3
1.24
1.16
0.08
4
1.20
1.24
-0.04
5
1.17
1.23
-0.06
6
1.21
1.25
-0.04
7
1.18
1.20
-0.02
8
1.10
1.15
-0.05
9
1.22
1.19
0.03
10
1.09
1.13
-0.04
Sum -0.21
The hypothesis to test is:
H0=Super trendy tyres are not superior to Roadmasters
H1=Super Trendy tyres are superior to Roadmasters
We compute the sum paired differences of the samples.
N =10
Sum =-0.21
Compute the Mean
-0.21/10
x =-0.021
Standard Deviation
4
S=
 X  X 
2
n 1
for n < 30
=0.0431
Calculating the test statistics:
Test statistic assuming the sampled data is normally distributed with mean of 0
t. calculated =
X 
SX
where
= - 0.21/ (0.431/√10)
=-1.541
From the t tables
t value=t9. 0.995
=3.25
Conclusion:
Accept H0 :Super Trendy tyres are not superior to Roadmaster tyres
From this we note that the Super trend advertising may be just a marketing gimmick that is
trying to outdo the Roadmaster tyre
Question 2
A research was carried out in a city hospital .A doctor claims that boys at the age if 17 year have
an average body temperature that is higher than the commonly accepted average human
temperature of 98.60 Fahrenheit. To test this, a sample of 25 boys, each of 17 years of age is
selected. The temperature of the 17 year olds boys is found to be an average of 98.9 0, and with a
standard deviation of 0.6 0.
Hypothesis to be tested
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The claim being investigated is that;
-The average body temperature of 17 year olds is > 98.60
That is x ≥ 98.6.
The negation of this is that:
-The population average is not greater than 98.6 degrees.
That this is x < 98.6.
Null hypothesis H0 : x = 98.6.
The alternative hypothesis, or H1: x >98.6.
let alpha be a 5% level of significance and alpha will be equal to 0.05.
Choice of Test Statistic and Distribution
We will use the standard normal distribution. A table of z-scores will hence be necessary.
Here n=25, which has square root of 5,
standard error is 0.6/5 = 0.12.
The test statistic is therefore z = (98.9-98.6)/.12 = 2.5
At a 5% significance level, the critical value for a one tailed test is found from the z table of zscores to be 1.645.
Since the test statistic does fall within the critical region, we reject the null hypothesis.
Using the P Value
The z-score of 2.5 has a p-value of 0.0062. Since this is less than the significance level of 0.05,
we reject the null hypothesis.
Conclusion
In conclusion, we see from the statistical evidence above that either a rare event has occurred, or
that the average temperature of 17 year olds is in fact greater than 98.60.
Question 3
6
I carried out a research on advertising. The research was on a company by the name Star beam
Advertising Company has prepared two different television Commercials advert for Sean’s
designer women’s jeans. One of the commercial is humorous, and the other is Serious. A test
screening involves eight consumers who are asked to view and rate the commercials by
using a standard scale of measurement with the higher scores indicating more favorable and a
lower score indicating unfavorable responses. The results are listed below. A 0.05 level of
significance is used.
The sample data used is as follows:
User
Commercial
with humor
Serious
Commercial
A
26.2
B
20.3
C
25.4
D
19.6
E
28.3
F
21.5
G
23.7
H
24.0
24.1
21.3
23.7
18.0
20.1
25.8
21.4
22.4
Hypothesis tested:
Ho = Mean difference is zero
H1 ≠mean difference of Zero
Assumption
The assumption made is that the differences between the ratings for each and every consumer are
from a population that has a normal distribution
To get the we compute the sum paired differences of the samples.
Compute the Mean
Mean of paired difference
x =9.6
Standard Deviation
Sd=18.98654
Using p value:
To compute the test statistics or p value
P(x_Bar < -9.6)
=P(Z< (-9.6-0) / (10/√8)
7
=3.822
δ
Critical T value = x̄ + Zα/2√𝑛
= x̄ ±2.365
P =0.0065
Conclusion
Reject Ho: Mean difference is zero
There is sufficient evidence to warrant rejection of the claim that the differences between the
ratings of commercial have a mean of zero. There does not appear to be a difference and because
humorous commercials have a generally higher rating, this implies the humorous commercials
are better.