IGNOU MMPC-005 Solved Question Paper PDF Download

IGNOU MMPC-005 Previous Year Solved Question Paper in Hindi

IGNOU MMPC-005 Previous Year Solved Question Paper in English

Q1. What do you understand by time series ? What are the four major components in a time series analysis ? Is the relationship between these four components multiplicative or additive or could be any of them ? Give the equations of the relationship.

Ans. A time series is a sequence of data points collected at successive, equally spaced points in time. It is a statistical technique that deals with the analysis of data over time to forecast future values. For example, a company’s quarterly sales, monthly rainfall data, or daily stock prices are all examples of time series. There are four major components in a time series analysis:

• Secular Trend (T): This represents the long-term direction of the time series, which could be upward (growth), downward (decline), or stable. For instance, a long-term increase in sales due to population growth or technological improvements.
• Seasonal Variation (S): These are regular and periodic fluctuations that occur within a year. They are often caused by weather, holidays, or social customs. For example, increased sales of warm clothing in winter and air conditioners in summer.
• Cyclical Variation (C): These are fluctuations that occur over a period of more than one year, often associated with the business cycle (prosperity, recession, depression, recovery). They are not as regular as seasonal variations.
• Irregular Variation (I): These are unpredictable, sporadic, and random events that affect the time series. They can be caused by events like strikes, floods, earthquakes, or wars. They cannot be forecasted.

The relationship between these four components can be

multiplicative

or

additive

.

1. Additive Model:

In this model, the four components are assumed to be independent of each other. The value of the time series is the sum of these four components. It is used when the magnitude of seasonal and cyclical fluctuations does not depend on the level of the trend.

Equation:

Y = T + S + C + I

2. Multiplicative Model:

This model is appropriate when the magnitude of the fluctuations is proportional to the level of the trend. This is the most common model for business and economic data, as seasonal effects are often expressed as a percentage of the sales or production level.

Equation:

Y = T × S × C × I

In conclusion, the relationship could be any of them, but the multiplicative model is more frequently used in practice, especially for business and economic applications, as it often provides a more realistic description of the data.

Q2. What is a frequency distribution ? What are the general guidelines of forming a frequency distribution with particular reference to the choice of class intervals and number of classes ?

Ans. A frequency distribution is a tabular presentation that summarizes and organizes a large amount of data. It divides the data into various classes or groups and shows the number of observations (frequency) falling into each class. Its primary purpose is to reveal the underlying pattern of the data, such as where the data is concentrated, its spread, and its skewness. The general guidelines for forming a frequency distribution, with particular reference to the choice of class intervals and the number of classes, are as follows: 1. Determine the Number of Classes (k):

• The number of classes should not be too large or too small. Typically, a number between 5 and 20 is considered ideal.
• If classes are too few, important characteristics of the data may be hidden. If classes are too many, the distribution will appear too irregular and the pattern will not be clear.
• Sturges’ Rule provides a common guideline for estimating the number of classes (k): k = 1 + 3.322 * log10(N) , where N is the total number of observations. The result is rounded to the nearest integer.

2. Calculate the Range of the Data:

• The range is calculated as the difference between the highest value (H) and the lowest value (L) in the dataset. Range = H – L . This helps to understand the total spread that needs to be covered by the classes.

3. Choose the Class Interval or Width (i):

• The class interval or width (i) should preferably be the same for all classes for easier comparison.
• It can be approximated by dividing the range by the number of classes: i ≈ Range / k .
• It is common practice to round up the calculated value to a convenient number, such as 5, 10, 20, or 100, to make the table easy to read and interpret.
• The class intervals should cover all the values in the data.

4. Set the Class Limits:

• Classes should be mutually exclusive , meaning no data value can fall into more than one class.
• Classes should be collectively exhaustive , meaning every data value must fall into some class.
• The lower limit of the lowest class should be slightly less than or equal to the smallest data value.
• For continuous data, the ‘exclusive method’ (e.g., 10-20, 20-30) is preferred, while for discrete data, the ‘inclusive method’ (e.g., 10-19, 20-29) can be used.

Q3. Seven methods of imparting business education were ranked by MBA students of two universities as follows : Method of Teaching | Rank by Students of Univ. A | Rank by Students of Univ. B I | 2 | 1 II | 1 | 3 III | 5 | 2 IV | 3 | 4 V | 4 | 7 VI | 7 | 5 VII | 6 | 6 Calculate Rank Correlation Coefficient and interpret the results.

Ans. To find the relationship between the rankings given by the two universities, we will calculate Spearman’s Rank Correlation Coefficient (R) . The formula is: R = 1 – [6 Σ(d²)] / [n (n² – 1)] Where,

• d = Difference between the two ranks (R1 – R2)
• n = Number of pairs of observations

Steps for Calculation:

1. First, we calculate the difference (d) between the ranks and the square of the difference (d²) for each method of teaching.

2. Now, we substitute the values into the formula:

• n = 7 (seven methods of teaching)
• Σd² = 28

R = 1 – [6

28] / [7

(7² – 1)]

R = 1 – [168] / [7 * (49 – 1)]

R = 1 – [168] / [7 * 48]

R = 1 – 168 / 336

R = 1 – 0.5

R = 0.5

Interpretation of the Results:

The rank correlation coefficient is

+0.5

. The value ranges from +1 to -1.

• The positive sign (+): This indicates a positive correlation between the rankings of the two universities. This means, in general, that if one university ranks a method highly, the other university tends to rank it highly as well.
• The magnitude (0.5): This value represents a moderate degree of positive correlation. It is not a strong relationship (close to 1), but it is better than no relationship (close to 0).

Conclusion:

There is a moderate level of agreement between the MBA students of the two universities regarding their preferences for the teaching methods. Their opinions are somewhat consistent but not perfectly aligned.

Q4. Find the median and 3rd quartile for the following distribution : Class Interval | Frequency 50—60 | 2 60—70 | 4 70—80 | 5 80—90 | 3 90—100 | 5 100—110 | 6 110—120 | 4 120—130 | 6 130—140 | 4 140—150 | 4 150—160 | 2

Ans. To calculate the median and the 3rd quartile (Q3), we first need to create a cumulative frequency (cf) table.

1. Calculation of Median:

Median item = N/2 = 45/2 = 22.5

The 22.5th item falls in the cumulative frequency of 25, so the

Median Class is 100-110

.

Formula for Median:

Median = L + [ (N/2 – cf) / f ] * i

Where:

• L = Lower limit of the median class = 100
• N = Total frequency = 45
• cf = Cumulative frequency of the class preceding the median class = 19
• f = Frequency of the median class = 6
• i = Class width = 10

Median = 100 + [ (22.5 – 19) / 6 ] * 10

Median = 100 + [ 3.5 / 6 ] * 10

Median = 100 + 0.5833 * 10

Median = 100 + 5.833

Median = 105.83

2. Calculation of 3rd Quartile (Q3):

Q3 item = 3N/4 = (3 * 45) / 4 = 135 / 4 = 33.75

The 33.75th item falls in the cumulative frequency of 35, so the

Q3 Class is 120-130

.

Formula for Q3:

Q3 = L + [ (3N/4 – cf) / f ] * i

Where:

• L = Lower limit of the Q3 class = 120
• N = 45
• cf = Cumulative frequency of the class preceding the Q3 class = 29
• f = Frequency of the Q3 class = 6
• i = Class width = 10

Q3 = 120 + [ (33.75 – 29) / 6 ] * 10

Q3 = 120 + [ 4.75 / 6 ] * 10

Q3 = 120 + 0.7917 * 10

Q3 = 120 + 7.917

Q3 = 127.92

Thus, for the given distribution, the

Median is 105.83

and the

3rd Quartile is 127.92

.

Q5. The customer accounts of a certain departmental store have an average balance of ₹ 120 and a standard deviation of ₹ 40. Assuming that the account balances are normally distributed, what proportion of the accounts is over ₹ 150? (Given the area between mean and required value for calculated statistic is 0.2734.)

Ans. To solve this problem, we will use the properties of the Normal Distribution and the Z-score. Given:

• Average balance (Mean, μ) = ₹ 120
• Standard deviation (σ) = ₹ 40
• We need to find the proportion of accounts over ₹ 150, i.e., P(X > 150).

Step 1: Calculate the Z-score.

The Z-score formula is:

Z = (X – μ) / σ

Where X is the value we are examining (₹ 150).

Z = (150 – 120) / 40

Z = 30 / 40

Z = 0.75

Step 2: Find the area under the standard normal curve.

We need to find the value of P(Z > 0.75). This represents the area to the right of Z = 0.75 on the standard normal curve.

We know that the normal curve is symmetrical and the total area to the right of the mean (Z=0) is 0.5.

The question gives us the area between the mean (Z=0) and our calculated Z-score (Z=0.75) is 0.2734.

That is,

P(0 ≤ Z ≤ 0.75) = 0.2734

.

The area we want, P(Z > 0.75), can be calculated as:

P(Z > 0.75) = (Total area to the right of the mean) – (Area between the mean and Z=0.75)

P(Z > 0.75) = P(Z > 0) – P(0 ≤ Z ≤ 0.75)

P(Z > 0.75) = 0.5 – 0.2734

P(Z > 0.75) = 0.2266

Conclusion:

The proportion of accounts with a balance over ₹ 150 is 0.2266.

As a percentage, this is 0.2266 * 100 =

22.66%

.

Therefore, approximately

22.66%

of the customer accounts have a balance over ₹ 150.

Q6. Write short notes on any three of the following : (a) Precautions in the use of secondary data (b) Relative Skewness (c) Cluster Sampling (d) Mode (e) Two-tailed tests

Ans. (a) Precautions in the use of secondary data Secondary data is data that has already been collected by someone else. While it is cost-effective and time-saving, researchers must be cautious when using it. The main precautions are:

• Check for Reliability: Who collected the data? What was their reputation and purpose? Was the collecting agency unbiased? The source of the data is crucial in determining its reliability.
• Check for Suitability: Does the secondary data suit the objective of the current research? The objective for which the data was originally collected may be different from the current objective. For example, the definition of ‘income’ or ‘household’ might differ.
• Check for Adequacy: Is the scope and coverage of the data adequate for the current study? The data may be too old (obsolete) or the geographical area covered may be too limited, making it inadequate for the present investigation.
• Check Definitions and Units: The units of measurement (e.g., kgs, tonnes) and definitions of terms (e.g., ‘urban’, ‘literate’) used in the data must be carefully examined to ensure they align with the research needs.

(c) Cluster Sampling

Cluster sampling is a probability sampling technique where the entire population is divided into groups, or ‘clusters’. These clusters are often naturally occurring, such as cities, schools, or residential blocks. The process is as follows:

1. The total population is divided into mutually exclusive and collectively exhaustive clusters.
2. A random sample of these clusters is then selected.
3. Once clusters are selected, either all members within the chosen clusters are surveyed (one-stage cluster sampling), or a further random sample is drawn from within the selected clusters (two-stage cluster sampling).

This method is very useful and cost-effective when the population is geographically widespread. However, its disadvantage is that it can have a higher sampling error than simple random sampling if the clusters are internally homogeneous (elements within a cluster are too similar to each other).

(d) Mode

The mode is a measure of central tendency. It is the value that appears

most frequently

in a data set. It is one of the easiest measures to understand and calculate.

• A data set can have one mode (unimodal), two modes (bimodal), multiple modes (multimodal), or no mode if all values appear with the same frequency.
• Example: In the data set {2, 3, 4, 4, 5, 5, 5, 6}, the mode is 5 because it occurs three times, more than any other number.
• Formula for Grouped Data: In a frequency distribution, the mode is calculated using the formula: Mode = L + [ (f1 – f0) / (2 f1 – f0 – f2) ] i Where L is the lower limit of the modal class, f1 is the frequency of the modal class, f0 and f2 are the frequencies of the pre-modal and post-modal classes respectively, and i is the class width.

The mode is particularly useful for qualitative data (e.g., most popular car colour) and is not affected by extreme values (outliers).

Q7. Why is forecasting so important in business ? Identify applications of forecasting for the following : (a) Long-term decisions (b) Medium-term decisions (c) Short-term decisions

Ans. Forecasting in business is the process of estimating future events and trends. It is critically important because it enables management to reduce uncertainty, plan effectively, and make informed decisions. Without accurate forecasts, a business might face excess inventory, a shortage of resources, cash flow problems, or missed market opportunities. It impacts nearly every functional area of a business, including production, marketing, finance, and human resources. The applications of forecasting for different time horizons are as follows: (a) Long-term decisions (Strategic – 2+ years): Long-term forecasts provide the basis for strategic planning and major investment decisions.

• Capital Investment: Forecasting long-term demand is key to making decisions on building new plants, purchasing machinery, or expanding facilities.
• Product Diversification and Development: By forecasting market trends and consumer preferences, companies decide on launching new products or entering new markets.
• Research & Development (R&D): Resources are allocated to R&D activities based on anticipated technological changes and future market needs.
• Strategic Alliances: Decisions on mergers, acquisitions, or strategic partnerships are made by forecasting industry growth and the competitive landscape.

(b) Medium-term decisions (Tactical – 6 months to 2 years):

Medium-term forecasts help in tactical decisions that relate to the optimal utilization of resources.

• Budgeting: Forecasts are used to prepare annual and semi-annual budgets for sales, production, and finances.
• Production Planning: Determining production levels to meet demand over the next year, thereby avoiding stock-outs or excess inventory.
• Manpower Planning: Planning for hiring, training, and potential layoffs of staff based on projected business activity.
• Cash Flow Management: Forecasting revenue and expenses for the effective management of working capital.

(c) Short-term decisions (Operational – up to 6 months):

Short-term forecasts are vital for ensuring the smooth running of day-to-day operations.

• Inventory Control: Deciding on reorder points and order quantities for raw materials and finished goods based on short-term sales forecasts.
• Production Scheduling: Planning daily or weekly production runs to meet immediate demand.
• Workforce Scheduling: Assigning shifts, overtime, and vacations for workers.
• Sales Quotas: Setting realistic short-term targets for the sales team.

Q8. A marketing manager wants to know if display at point of purchase helps in increasing the sales of his product. Unless there is strong evidence to the contrary, he is likely to believe that such displays do not affect sales. He picks up 70 retail shops where there is no display and finds that the weekly sale in these shops has a mean of ₹ 6,000 and S.D. of ₹ 1,004. Similarly, he picks up a second sample of 36 retail shops with display at point of purchase and finds that the weekly sale in these shops has a mean of ₹ 6,500 and S.D. of ₹ 1,200. What should he conclude at a significance level of 5% ? (Given the tabulated value of test statistic is 1.645).

Ans. This is a hypothesis testing problem for comparing the means of two independent samples. We will use a Z-test to determine if the display has a significant effect on sales. Step 1: State the Hypotheses Let:

• μ1 = mean weekly sale in shops with no display
• μ2 = mean weekly sale in shops with display

The manager wants to know if the display helps in

increasing

sales, so this will be a one-tailed test.

• Null Hypothesis (H0): μ1 = μ2 (or μ2 – μ1 = 0). The display has no effect on sales.
• Alternative Hypothesis (H1): μ2 > μ1 (or μ2 – μ1 > 0). The display increases sales.

Step 2: Set the Significance Level

The significance level, α = 5% or 0.05.

Step 3: Collect Data and Choose Test Statistic

Sample 1 (No display):

• n1 = 70
• x̄1 = ₹ 6,000
• s1 = ₹ 1,004

Sample 2 (With display):

• n2 = 36
• x̄2 = ₹ 6,500
• s2 = ₹ 1,200

Since both sample sizes are large (n1 > 30 and n2 > 30), we can use the Z-test. The formula for the Z-test statistic is:

Z = (x̄2 – x̄1) / √[(s1²/n1) + (s2²/n2)]
Step 4: Calculate the Test Statistic

Z = (6500 – 6000) / √[(1004² / 70) + (1200² / 36)]

Z = 500 / √[(1008016 / 70) + (1440000 / 36)]

Z = 500 / √[14400.23 + 40000]

Z = 500 / √[54400.23]

Z = 500 / 233.24

Z ≈ 2.14

Step 5: Make a Decision

This is a one-tailed (right-tailed) test. For a 5% level of significance, the critical value of Z is

1.645

(as given in the question).

Decision Rule:

If the calculated Z (Z_calc) is greater than the critical Z (Z_crit), we reject the null hypothesis (H0).

Here,

Z_calc (2.14) > Z_crit (1.645)

.

Step 6: Conclusion

Since the calculated Z-value is greater than the critical Z-value, we reject the null hypothesis (H0).

This means there is sufficient statistical evidence at the 5% significance level to conclude that the

display at the point of purchase helps in increasing the sales of the product

. The marketing manager should conclude that the displays are an effective strategy.