Hypothesis Testing in Statistics

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Hypothesis Testing in Statistics
Course Name: Introduction to Statistics
Course Code: STATS 101/101G/108
Assignment: 3
Semester: Second
Year 2018

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Answer
a) Sample Size = 18 batteries
Independent or treatment variable was two possible choice of batteries, Energizer
batteries or Ultra-cell batteries. The response or dependent variable was the battery life in
the electronic game in hours
Independent t-test was used to test the difference in battery hours of the two types of
batteries.
Let the average battery life for an Energizer battery is μ1 and average life for Ultra-cell
battery is μ2 . The Energizer batteries (M = 8.28 hours, SD = 0.22 hours, n = 9) were
observed to be to some extent better than the Ultra-cell batteries (M = 8.24 hours, SD =
0.16 hours, n = 9).

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Independent t-test to test the difference in battery life:
1. The difference in average battery life of Energizer and Ultra-cell batteries = μd =μ1−μ2
2. Null Hypothesis: H0: μd =μ1−μ2=0
3. Two tail Alternate Hypothesis: HA: μd =μ1−μ2≠0
4. From the provided samples, difference in battery life was x1
−
−x2
−
= 8.28 - 8.24 = 0.036
5. Standard error of the mean of the difference in battery life = 0.0905 (t‐
procedures tool on Canvas)
The test statistic =
t= estimate−hypothesised value
s tan dard error = 0 .0356−0
0. 0905 =0. 39 with degrees of
freedom = min (9 – 1, 9 – 1) = 8.
6. The P-value for two tail = P (|t|>0 .39 ) =0 . 7044 (t‐procedures tool)
7. P-value explanation: The calculated p-value was 0.7044 and was greater than 0.05.
Hence, enough evidence for rejecting the null hypothesis was not present. The observed
difference of 0.036 hours of life times for the two types of batteries was not statistically
significant at 5% level of significance. Consequently, the null hypothesis failed to get
rejected, and it was concluded that difference in life time between the two types of
batteries was almost equal to zero.
8. At 5% level of significance, estimated confidence interval for μd was
= ( μ1−μ2 ) ±t −multiplier∗S tan dard Error=0 . 0356±2. 306∗0 .0905= [ −0 .173 , 0 . 244 ]
(The t-multiplier at 5% level of significance was calculated from t‐procedure tool).
9. Confidence interval explanation: With 95% probability, it was inferred that the average
battery hours of Energizer batteries would be roughly 0.17 hours less than that of Ultra-
cell batteries. Similarly, battery hours of Energizer batteries would be roughly 0.24 hours
more than average battery hours of Ultra-cell batteries. The result was practically
significant from also.
10. Conclusion: The sample of the battery hours for playing an electronic game did not have
enough evidence to establish any significant difference in battery life of Energizer and
Ultra-cell batteries.

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b) The confidence interval at 95% level of significance estimated that the value of the
population mean of difference in life hours of the two types of batteries was not within
the boundaries. The hypothesized value of the difference in average battery life of the
two types of batteries was assumed to be zero, and the non inclusion of population
parameter in the confidence interval indicated that the conclusion from p-value was true.
Answer
a) The sampling situation for analysing the difference in proportion was the case of one
sample with four different response categories.

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b) The scenario was tested by t-test for difference between two proportions as follows.
1. It was assumed that p1 , and p2 denotes the proportions of responses for supporting the
legislation, and refuting it.
Hence, p1 - p2 = Difference of the proportions
2. Null hypothesis: H0: p
^¿d
¿= p1 - p2 = 0
3. Two tailed Alternate hypothesis: HA: p1 - p2 ¿ 0
4. Estimated difference: p1
^¿
¿- p2
^¿
¿=
384
500 −79
500 =0 . 61
5. The t-test statistic was,
t= estimated difference−hypothesized difference
s tan dard error
Estimated difference = 0.61, hypothesized difference = 0
Standard error of mean = 0.0333 at 5% level of significance (t‐procedures tool).
The test statistic, t= 0 .61−0
0. 0333 =18 .318 with degrees of freedom = ∞
6. Two tail P-value = P (|t|>18 .318 ) =0 . 0000 (from t‐procedures)
7. Explanation of P-value:
The p-value was less than 0.05, at 5% level of significance. Hence, there was very strong
enough evidence to accept the alternate hypothesis opposed to the null hypothesis. Hence,
the null hypothesis was rejected based on the evidence from the difference in proportions
of adult New Zealanders in support and against the legislation.
8. The confidence interval was calculated as
CI =¿ ¿
The t-estimate = 1.96 was calculated from t‐procedures tool.
9. Interpretation of Confidence Interval:
With 95% confidence, the estimated value of difference in proportions of support and
against views of the New Zealanders was anywhere between 0.545 and 0.675. The
support was 54.47% to 67.52% higher than against views, which was also practically
significant.

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10. Conclusion:
The null hypothesis was rejected at 5% level of significance. The response
of New Zealanders in favor of the legislation was significantly greater than that of the
views against.