Hypothesis Testing: STATS 101 Assignment, Second Semester 2018

This assignment solution covers hypothesis testing concepts in a statistics course (STATS 101/101G/108). It includes independent t-tests to compare battery life, analyzing the difference in proportions regarding legislation support in New Zealand, and examining productivity scores under different scenarios. The solutions provide detailed steps, including null and alternative hypotheses, test statistics, p-values, confidence intervals, and interpretations. ANOVA is used to analyze cyclist completion times across different age groups. The document also addresses assumptions, practical significance, and conclusions based on statistical evidence. This resource will help students understand and apply hypothesis testing methods.

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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.

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Answer:
a)
(i) The hypothesized difference in productivity score was considered to be zero.
(ii) At 5% level of significance, the estimated difference in productivity scores in Case1was
greater than the hypothesized value. The standard error of the sampling distribution was
0.682, and the deviation from average difference in scores constructed the confidence
interval signifying that the hypothesized value was not close to the estimated value.
b) At 5% level of significance, all the cases excluding Case 3 demonstrated that the sample
mean difference was statistically significant.
c) Practically significant and non significant cases were as follows.
(i) Case 1: Practically significant
(ii) Case 4 and Case 5: Not practically significant
d) Case 2 and Case 3: Nothing could be concluded, as lower limits of the confidence interval
were less than 10, the desired minimum change of productivity scores.
e) The mean difference was noteworthy at 5% level of significance. With 95% certainty, it was
assessed that mean difference of the two payout frameworks would be somewhere close to
0.96 hours to 5.58 hours. The real distinction of the methods for the two pay-out frameworks
would be as low as 0.96 hours and high as 5.58 hours. A statistical significance (Diff = 3.27,
p < 0.05) was noted for the Case 6 scenario, and Bonus pay-out framework would be
favored. The results of case 6 was not practically significant, and considering the
administration's thought about the distinction in efficiency score the organization might want
to adhere to its past model for pay-out.

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Answer 4
a) (i) Units: 12 short stories from the section of mystery, ironic, and literary.
(ii) Treatment: Enclosure or omission of spoiler paragraph.
(iii) Response variable: The enjoyment rating of the readers
b) (i) The following graphs of the two treatments and their difference were constructed
using iNZight tool.

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Figure 1: Side-by-Side Box Plots for the Two Treatments
Figure 2: Box Plot for Difference in Treatment Scores

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(ii) The side-by-side box plots reflected that median of enjoyment scores was undoubtedly
greater with inclusion of the spoiler paragraph compared to enjoyment score from story
without such paragraph. The median of enjoyment score was almost 7.0 for stories with
spoiler, while the median of enjoyment score was almost 6.0 for the stories without spoiler.
Distribution of enjoyment scores was highly left skewed for stories with spoiler paragraph,
whereas, stories without spoiler paragraph had comparatively less skewed distribution of
enjoyment scores.
The distribution of difference in the enjoyment scores was also somewhat left skewed. The
median of difference in enjoyment scores was approximately 0.5.
c) The difference between average enjoyment scores for the two treatments was verified by t-
test as follows. Let the average enjoyment score with spoiler was μ1 and average enjoyment
score without spoiler was μ2 . The average enjoyment score with spoiler (M = 6.217, SD =
1.22, n = 12) were observed to be to some extent better than the average enjoyment score
without spoiler (M = 5.725, SD = 1.26, n = 12).
1. Parameter: Let μ1 and μ2 are the average population enjoyment scores with and without
spoiler paragraph in stories. Let μd=μ1 - μ2 is the difference in average enjoyment scores.
2. Null hypothesis: H0: ( μd=μ1 - μ2 = 0)
3. The two-tailed Alternate hypothesis: HA: ( μd=μ1 - μ2≠0 )
4. Estimate: Difference in two average enjoyment scores =
x
−
1−x
−
2=6 . 217−5 . 725=0. 4917≃0 . 492
5. Test statistic: t=
( x
−
1−x
−
2 )−0
SEM = 0 . 4917−0
0 . 1003 =4 . 900 where standard error of mean = SEM =
0.1003 (from SPSS) with degrees of freedom = min (12-1, 12-1) = 11.
6. Calculated two tail P-value = P (|t|>0 . 492 ) =0 .000 (from SPSS)

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7. Explanation of P-value:
There was a statistical significant difference between the average enjoyment scores for
stories with and without spoiler paragraph. The evidence was very strongly against the null
hypothesis at 5% level of significance. Hence, the average enjoyment in reading stories
with spoiler paragraph was significantly greater compared to the without spoiler story
enjoyment scores.
8. Confidence Interval:
At 95% significance level, approximate confidence interval for μd was
CI = ( μd ) ±t multiplier∗SE=[ 0 . 27 , 0. 71 ] (from SPSS output)
9. Confidence Interval elucidation: With 95% confidence it was inferred that the average
enjoyment score with spoiler would be roughly 0.27 less than and 0.71 more than average
enjoyment score without spoiler. The result was practically significant (Bluman, 2009).
10. Conclusion: Very strong evidence was there to establish difference in enjoyment scores.
The null hypothesis was rejected at 5% significance level, stating that average enjoyment
score with spoiler was greater than that of the stories without spoiler.
d) Dependent variable: Difference of enjoyment scores with and without spoiler paragraph.
i. The difference scores were continuous in nature.
ii. There were two categorical values of the treatment attribute.
iii. No outlier scores were present
iv. Normally distributed enjoyment scores was noted. Shapiro-Wilk test (SW = 0.955, p =
0.712) ascertained that the differences of enjoyment scores were normally distributed.
Therefore, the fourth condition for dependent t-test was also satisfied.

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