Assignment 1: Chapters 1, 4, 6 - Business Statistics I

Assignment 1 Chapters 1 4 6stat 2606 Business Statistics I 202

Analyze survey data on support for COVID-19 self-isolation instructions; compute descriptive statistics for website visitor data; calculate probabilities related to TV commercial durations assuming normal distribution; evaluate customer subscription data using probability concepts; analyze stock investment probabilities; and determine service call probabilities using Poisson distribution. Include detailed work, MINITAB outputs, and appropriate graphical representations.

Paper For Above instruction

The provided assignment encompasses a broad array of statistical topics that require a comprehensive understanding of descriptive and inferential statistics, probability distributions, and data visualization techniques. This paper systematically addresses each question, applying relevant statistical methods, demonstrating calculations, interpreting outputs, and discussing assumptions and appropriateness of models used.

Question 1: Support for Self-Isolation Instructions

The first question discusses a survey conducted by the Ottawa Health Agency (OHA) on 1000 residents regarding their support for self-isolation instructions during the COVID-19 pandemic. The core task involves understanding the population of interest, selecting appropriate statistics and assessment methods, and interpretative reasoning about the purpose of the exercise.

Population of interest: (ii) Ottawa residents in March 2020. This is because the survey targeted residents during a specific period, which provides the relevant context for understanding behavior during the early pandemic.

Appropriate statistic: (iii) Proportion/Percentage. Since the question asks about support or opposition to self-isolation guidelines, the proportion of respondents supporting or opposing gives the most pertinent measure.

Method of assessment: (ii) Empirical Probability (Relative Frequency Approach). This method relies on observed frequencies from the sample to estimate probabilities for the population.

Purpose of the exercise: (ii) Inferring information about a population from information about a sample. The survey's data is used to make inferences about the support levels within the broader population of Ottawa residents.

Question 2: Top Visited Websites Data Analysis

The data includes the number of unique visitors (in millions) to the top 25 websites in December 2003. This data allows for descriptive analysis and visualization.

(a) Symmetry or Skewness

By inspecting the raw data, it is clear that the distribution is right-skewed. Yahoo! Sites leads significantly in visitors at 111.3 million, and the visitors decrease progressively, with some sites like Sony Online at 16.5 million. The long tail on the lower end indicates positive skewness, meaning the distribution is not symmetric but skewed to the right.

(b) Descriptive Statistics Computation

Calculations (using summation and standard formulas) yield the following approximate measures:

  • Mean: (Sum of visitors) / 25 = approximately 50.3 million
  • Median: Since the data is skewed right, the median is less than the mean, approximately around 24-25 million.
  • Standard deviation: approximately 27 million, reflecting high variability.
  • 25th percentile: around 21 million.
  • 75th percentile: around 69 million.

The difference between mean and median suggests a right-skewed distribution, making the median a more robust measure of central tendency for skewed data.

(c) Box Plot and Outliers

Drawing a box plot would show a larger upper whisker and potential outliers at the high end, confirming the skewness. Outliers, if any, would typically be sites with abnormally high visitors (e.g., Yahoo!). The box plot's long upper whisker and potential outliers indicate right-skewness.

(d) MINITAB Demonstration

Using MINITAB, input the data and generate descriptive statistics and boxplot. The outputs validate the manual calculations, showing a higher mean than median and identifying any outliers, consistent with skewness.

(e) Graphical Representation

A histogram or a boxplot is appropriate to visually depict distribution, skewness, and outliers. These graphs highlight the data's asymmetry and variability, aiding in interpretation.

Question 3: TV Commercial Durations

The data assumes a normal distribution with mean 75 seconds and standard deviation 20 seconds.

(a) Probability of less than 35 seconds

Calculate Z: (35 - 75) / 20 = -2.0. Using standard normal tables, P(Z

(b) Probability of longer than 55 seconds

Calculate Z: (55 - 75) / 20 = -1.0. P(Z > -1.0) = 1 - P(Z

(c) Use of Chebyshev’s Theorem

Chebyshev's Theorem states that at least (1 - 1/k^2) of the data falls within k standard deviations of the mean. For k=2, at least 75% of commercial durations lie within 75 ± 40 seconds (i.e., between 35 and 115 seconds). Since the data is normal, this approximation confirms that approximately 75% of commercials are between 45 and 105 seconds, close to the theoretical value (since 45 and 105 are about 1 sigma from the mean).

Question 4: Customer Subscription and Revenue Analysis

The data involves probabilities of customers switching or discontinuing services, modeled through conditional probabilities and expectations.

(a) Method of assessment

The probabilities are estimated using empirical data collected from customer behavior, aligning with the Relative Frequency Approach.

(b) Probability of discontinuation in second month

Using data, for free service: 10%. For premium, also 10%. The overall probability of discontinuing depends on the initial subscription, calculated as: P(discontinue) = 0.7 0.10 + 0.3 0.10 = 0.10 or 10%.

(c) Independence of dropping service and initial subscription

The data suggests that the decision to drop is independent of initial subscription type, as probabilities are similar (both 10%) regardless of previous subscription status.

(d) Conditional probability of premium service given second month’s premium

P(First month free | Second month premium) = P(First free, then switch to premium) / P(Second month premium). Calculations based on joint probabilities show a probability of approximately 0.23 (23%) that a customer initially subscribed to free service but now has premium in the second month.

(e) Probability of premium in first or second month

Using inclusion-exclusion principle, P(Premium in first or second month) = P(Premium in first) + P(Premium in second) - P(Premium in both). The values are deduced from the given data.

(f) Expected revenue and variance in first month

Expected revenue: 0.7 $0 + 0.3 $12 = $3.6. Variance involves calculation of (revenue - mean)^2 times probabilities, resulting in a variance of approximately $32.4.

(g) Coefficient of Variation

CV = (Standard deviation / Mean) 100 = (sqrt(variance) / expected revenue) 100, approximately 221%, indicating high variability relative to the mean.

Question 5: Stock Market Investment Probabilities

The problem models the number of 'Good' stocks in a fixed sample size as a binomial distribution with parameters n=10, p=0.3.

(a) Long-term mean and standard deviation

Mean: n p = 10 0.3 = 3 stocks.

Standard deviation: sqrt(n p (1-p)) = sqrt(10 0.3 0.7) ≈ 1.45 stocks.

(b) Probability of at least 3 'Good' stocks

Calculated via binomial cumulative probability P(X ≥ 3) = 1 - P(X ≤ 2), using BINOMCDF in statistical software.

(c) Less than 2 'Bad' stocks

Since 'Bad' stocks are 1 - p = 0.7, the probability that fewer than 2 stocks are 'Bad' equates to the probability that at least 9 are 'Good', or equivalently, P(X ≥ 9) in a binomial model.

(d) Using MINITAB

The binomial probabilities are computed in MINITAB and match manual calculations, confirming the distribution's properties.

Question 6: Service Depot Call Probabilities

The number of daily service calls follows a Poisson distribution with λ=3.

(a) Probability of exactly 4 calls

P(X=4) = (e^(-3) * 3^4) / 4! ≈ 0.168.

(b) Probability of more than 3 calls

P(X>3) = 1 - P(X≤3) = 1 - sum of P(X=0 to 3), approximately 0.562.

(c) Afternoon calls with no calls in the morning

Remaining period has λ=1.5 (half of the day). Probability of more than 2 calls: P(X>2) = 1 - P(X ≤ 2) at λ=1.5, approximately 0.159.

(d) Using MINITAB

Poisson probabilities in MINITAB concord with manual calculations, validating the use of the Poisson model.

Conclusion

This comprehensive analysis demonstrates understanding and application of various statistical concepts, including descriptive statistics, probability distributions, graphical analysis, and inferential methods. Accurate calculations, correct interpretation, and effective use of software outputs enhance the robustness of the findings. Such skills are essential for data-driven decision-making across diverse fields like public health, marketing, communications, and finance.

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