Statistics Assignment: Confidence Intervals and Hypothesis Testing

This statistics assignment provides solutions to problems involving binomial distributions, confidence intervals, and hypothesis testing using both z-tests and t-tests. The assignment explores the application of these statistical methods, including the calculation of confidence intervals, and the interpretation of p-values. It also covers the concepts of Type I and Type II errors in hypothesis testing, and the conditions under which each is preferable. Furthermore, the assignment includes an analysis of A/B testing within the context of a marketing campaign and the evaluation of a conversion funnel, emphasizing the use of direct customer data for accurate assessment. The solution includes a bibliography citing relevant resources.

Statistics and Maths
Students Name:
Subject:
Professor’s Name:
Date:

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Question 1:
a.)
The probability distribution that would accurately portray the scenario is binomial distribution.
Binomial distribution is the likelihood of either a success or a failure in the outcome of a study or
a test that is repeated multiple of times. The conditions that the distribution follow is that each
observation is independent, the probability of success is the same for each outcome, and that
each observation represents one of two outcomes that is either a success or a failure,.
b.)
𝑝(𝑥) = 𝑛!
(𝑛 − 𝑋)! 𝑋!
(𝑝)𝑋 .(𝑞)𝑛−𝑥
𝑝(𝑥) = 10!
(10 − 3)! 3!
(0.4)3 .(0.6)10−3
𝑝(𝑥) = 0.215
Question 2:
a.)

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The main methodology to be used will be with the normal distribution or z distribution to
determine the confidence interval. A normal distribution occurs naturally in many situations. On
the other hand, the confidence interval is a statistic that is computed from the observed that
which may contain the true value of a population parameter that is unknown. Hence, the
confidence level consists of a range which has the potential values of the population parameter
that is unknown. The property of a normal distribution is that the normal curve is symmetrical
about the mean. The other property of a normal distribution is that it is completely evaluated by
its mean and that the standard deviation and the mean are at the middle and divides the area into
halves. The confidence interval for the population mean will, therefore, be determined using the
formula:
z = x̄/ ± z* σ / (√n)
b.)
The confidence interval is computed using the normal distribution / z distribution. The formula
to be used is as follows:
z = x̄/ ± z* σ / (√n)
x̄/ = 207
σ = 65
n = 100
Confidence Interval = 95%

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Thus, z = 1.96
Hence, 207 ± 1.96 * 65/ (√100)
CI (207 + 1.96 * 65/ (√100), 207 - 1.96 * 65/ (√100))
The CI is (219.74, 194.26)
Thus, with a confidence level of 95%, the population mean lies between 194.26 and 219.74.
Question 3:
a)
μ = 200
x̄/ = 207
σ = 65
n = 100
One Tailed One Sample T Test
H0: x̄/ = μ
H1: x̄/ ≠ μ
t = (x̄/ – μ)/( σ /sqrt(n))

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μ = 200
x̄/ = 207
σ = 65
n = 100
t = (207-200)/(65/sqrt (100))
t = 1.0769
df = 100 – 1
df = 99
p = 0.142
As a result, we accept the null hypothesis. Thus, the result has no statistically significant
difference between the sample mean and the population means.
One sample Z test
H0: μ = 200
H1: μ ≠ 200
z = (x̄/ – μ)/( σ /√n)
z = (207-200)/(65/√ 100)

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z = 1.0769
p = 0.140762
As a result, we accept the null hypothesis. Thus, the result has no statistically significant
difference between the sample mean and the population means.
b)
Type 1 error happens when the null hypothesis is right but is instead rejected. Hence, they can be
said to be false positives which happen when the tester validates a difference that is statistically
significant even though there is not one. On the other hand, type 2 error occur when inaccurately
assuming that no winner had been declared between a variation and a control version even
though there is actually a winner.
The condition in which a conducted hypothesis which has α and β as 0.05 and 0.45 respectively
is preferable is when the care of evidence is crucial. The condition that needs to be satisfied is
that inflating type 1 error will lead to weak evidence which will not convince in supporting the
hypotheses.
A reversed scenario where when conducting the hypothesis test with both α and β are fixed at
0.15 is preferred is when the evidence is not crucial. This scenario involves inflating type 1 error
from 0.05 to 0.15 and will, therefore, lead to the weakening of the evidence. Hence, the strength
of the evidence is not crucial.

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Question 4:
A marketing campaign and the conversion funnel is truly evaluated is when data is obtained form
the customers directly. A/B testing compares two versions of an email, web page or any other
asset of marketing with just one element that is varying1. A/B testing is used since it lets one
know which phrases, videos, images, testimonials, and other metrics work best. A simple change
in the elements can impact the conversion rate greatly. Thus, it is vital to know which other
element on the page might have an impact on the traffic, conversion and other metrics due to
such minor changes.
A/B testing is can be adopted by creating two versions of the same page where only the headline
is altered. After preparations of the variations, every version is presented to half of the visitors.
The assessment will be helpful in proving which version was the most prevalent among the
audiences on the basis of specific metrics such as time on page and conversion rate.
1 Dixon, Eleri, Emily Enos, and Scott Brodmerkle. "A/b testing of a webpage." U.S. Patent
7,975,000, issued July 5, 2011.

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Bibliography
Dixon, Eleri, Emily Enos, and Scott Brodmerkle. "A/b testing of a webpage." U.S. Patent
7,975,000, issued July 5, 2011.