Here are the detailed solutions for the assignment questions from MMPC-005: Quantitative Analysis for Managerial Applications.
Question 1: Describe briefly the questionnaire method of collecting primary data. State the essentials of a good questionnaire.
Solution: The questionnaire method involves gathering primary data by asking a set of questions to the target respondents. This method can be used in surveys, interviews, or direct mailing to collect valuable insights about the subject under study. Primary data refers to the data collected for the first time directly from the source.
Essentials of a good questionnaire:
- Clear Objectives: The purpose of the questionnaire should be clear, ensuring that every question serves to meet the overall goal of the research.
- Simple and Unambiguous: The questions should be easy to understand, without any complex or confusing language.
- Logical Sequence: Questions should flow logically, with the most general questions first, followed by more specific ones.
- Avoid Leading Questions: Questions should not be biased or suggest a particular answer.
- Mutually Exclusive and Exhaustive Options: In multiple-choice questions, options should not overlap and should cover all possible answers.
- Length: The questionnaire should be short enough to maintain the interest of respondents but long enough to gather sufficient information.
- Pre-testing: The questionnaire should be tested on a small group to identify potential issues before distributing it to the target group.
Question 2: Discuss the importance of measuring variability for managerial decision-making.
Solution: Variability refers to the spread or dispersion of data points in a dataset. In managerial decision-making, understanding variability is crucial because it helps managers assess the risks and uncertainties in various decisions.
Importance of measuring variability:
- Risk Assessment: By analyzing the variability in outcomes, managers can estimate the level of risk associated with different decisions.
- Consistency Measurement: It helps in understanding the consistency of performance, product quality, or customer satisfaction.
- Forecasting: Variability helps in more accurate forecasting and budgeting by providing a range of expected outcomes.
- Resource Allocation: In project management, understanding variability in task completion times allows better resource allocation and scheduling.
- Performance Evaluation: Variability can help identify outliers or deviations from the norm, helping in quality control and process improvements.
Question 3: An investment consultant predicts that the odds against the price of a certain stock will go up during the next week are 2:1, and the odds in favor of the price remaining the same are 1:3. What is the probability that the price of the stock will go down during the next week?
Solution: The odds against the stock price going up are given as 2:1, which means the probability of the stock price going up is:
P(\text{Up}) = \frac{1}{2 + 1} = \frac{1}{3}
The odds in favor of the price remaining the same are given as 1:3, which means the probability of the stock price remaining the same is:
P(\text{Same}) = \frac{1}{1 + 3} = \frac{1}{4}
Now, the total probability must equal 1, so the probability of the stock price going down is:
P(\text{Down}) = 1 - P(\text{Up}) - P(\text{Same}) = 1 - \frac{1}{3} - \frac{1}{4}
First, calculate the sum of the probabilities of the stock going up and staying the same:
P(\text{Up}) + P(\text{Same}) = \frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12}
So, the probability of the stock price going down is:
P(\text{Down}) = 1 - \frac{7}{12} = \frac{12}{12} - \frac{7}{12} = \frac{5}{12}
Thus, the probability that the stock price will go down is 5/12.
Question 4: In practice, we find situations where it is not possible to make any probability assessment. What criterion can be used in decision-making situations where the probabilities of outcomes are unknown?
Solution: In situations where probability assessments are not possible, decision-making relies on non-probabilistic criteria. The following are some criteria used in such cases:
- Maximin Criterion: This is a pessimistic approach where the decision-maker chooses the option with the best of the worst possible outcomes.
- Maximax Criterion: This is an optimistic approach where the decision-maker chooses the option with the maximum possible outcome.
- Minimax Regret Criterion: This approach involves choosing the option that minimizes the maximum regret. Regret is the difference between the payoff of the chosen option and the best possible payoff.
- Laplace Criterion: This criterion assumes that all outcomes are equally likely and selects the option with the highest average payoff.
- Hurwicz Criterion: This is a compromise between the maximin and maximax criteria, where the decision-maker assigns a weight (α) to the best outcome and (1-α) to the worst outcome, and selects the option with the highest weighted payoff.
Question 5: A purchase manager knows that the hardness of castings from any supplier is normally distributed with a mean of 20.25 and a standard deviation of 2.5. He picks up 100 samples of castings from a supplier who claims that his castings have higher hardness and finds the mean hardness to be 20.50. Test whether the claim of the supplier is tenable.
Solution: We will use a hypothesis test to determine if the supplier’s claim of higher hardness is supported by the data.
- Null Hypothesis (H₀): The mean hardness of the supplier’s castings is 20.25 (i.e., no increase).
- Alternative Hypothesis (H₁): The mean hardness of the supplier’s castings is greater than 20.25 (i.e., there is an increase).
This is a one-tailed test. Given:
- Population mean (μ₀) = 20.25
- Sample mean (x̄) = 20.50
- Population standard deviation (σ) = 2.5
- Sample size (n) = 100
The test statistic (Z) is calculated as:
Z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} = \frac{20.50 - 20.25}{\frac{2.5}{\sqrt{100}}} = \frac{0.25}{\frac{2.5}{10}} = \frac{0.25}{0.25} = 1
For a one-tailed test at a significance level of 5%, the critical Z-value is 1.645. Since the calculated Z-value (1) is less than 1.645, we fail to reject the null hypothesis.
Conclusion: The supplier’s claim of higher hardness is not tenable based on the sample data
