Samacheer Kalvi Class 12 Business Maths Solutions Chapter 9 Applied Statistics More Ques

 

Question 1. Using three yearly moving averages, Determine the trend values from the following data.

Answer:

The trend values are calculated using three-yearly moving averages. This method smooths out short-term fluctuations to show the overall trend over time.

In simple words: To find the trend, we add up the profit for three years and then divide by three to get an average. We do this for each year, moving one year at a time. The first and last years cannot have a three-year average because there isn't enough data.

🎯 Exam Tip: Remember to center the moving total/average against the middle year of the period being averaged. For odd-period moving averages, this is straightforward; for even-period, it requires a centering step.

 

Question 2. From the following data, calculate the trend values using fourly moving averages.

Answer:

We will calculate the trend values using four-yearly moving averages. This method involves two steps: first, calculating the four-yearly moving totals and averages, and then centering these averages to align them with specific time points.

In simple words: When finding a four-year moving average, we first add up four years of sales and divide by four. Because this average sits between two years, we then take two of these four-year averages and average them again to get a "centered" average that lines up with a specific year.

🎯 Exam Tip: For even-period moving averages (like 4-yearly), always remember the second step of centering the averages. This ensures the trend values align with actual time points.

 

Question 3. Fit a straight line trend by the method of least squares to the following data.

Answer:

To fit a straight-line trend using the method of least squares, we first set up a table to calculate the necessary sums for the trend equation \( y = a + bx \). We choose the mid-point of the years as our origin to simplify calculations for \( x \). The average of the years 1980-1987 is \( \frac{1980+1987}{2} = 1983.5 \). We then divide by 0.5 because the years are incrementing by 1, and we need our \( x \) values to be integers (or centered). Here, we take \( x = \frac{\text{Year} - 1983.5}{0.5} \). This makes our calculations easier by ensuring that \( \Sigma x = 0 \).

Now, we calculate the constants \( a \) and \( b \) for the trend line equation:
\( a = \frac{\Sigma y}{n} = \frac{447.9}{8} = 55.9875 \)
\( b = \frac{\Sigma xy}{\Sigma x^2} = \frac{139.5}{168} = 0.830 \)
The equation for the straight line trend is:
\( y = a + bx \)
\( y = 55.9875 + 0.830 x \)
This equation helps us predict sales for future years.
Now we find the trend values (\( y_t \)) for each year:
When \( x = 1980 \):
\( y = 55.9875 + 0.83 (\frac{1980 - 1983.5}{0.5}) \)
\( = 55.9875 + 0.83 (-7) \)
\( = 55.9875 - 5.81 \)
\( = 50.1775 \)
When \( x = 1981 \):
\( y = 55.9875 + 0.83 (\frac{1981 - 1983.5}{0.5}) \)
\( = 55.9875 + 0.83 (-5) \)
\( = 55.9875 - 4.15 \)
\( = 51.8375 \)
When \( x = 1982 \):
\( y = 55.9875 + 0.83 (-3) \)
\( = 55.9875 - 2.49 \)
\( = 53.4975 \)
When \( x = 1983 \):
\( y = 55.9875 + 0.83 (\frac{1983 - 1983.5}{0.5}) \)
\( = 55.9875 + 0.83 (-1) \)
\( = 55.9875 - 0.83 \)
\( = 55.1575 \)
When \( x = 1984 \):
\( y = 55.9875 + 0.83 (\frac{1984 - 1983.5}{0.5}) \)
\( = 55.9875 + 0.83 (1) \)
\( = 56.8175 \)
When \( x = 1985 \):
\( y = 55.9875 + 0.83 (\frac{1985 - 1983.5}{0.5}) \)
\( = 55.9875 + 0.83 (3) \)
\( = 55.9875 + 2.49 \)
\( = 58.4775 \)
When \( x = 1986 \):
\( y = 55.9875 + 0.83 (\frac{1986 - 1983.5}{0.5}) \)
\( = 55.9875 + 0.83 (5) \)
\( = 55.9875 + 4.15 \)
\( = 60.1375 \)
When \( x = 1987 \):
\( y = 55.9875 + 0.83 (\frac{1987 - 1983.5}{0.5}) \)
\( = 55.9875 + 0.83 (7) \)
\( = 55.9875 + 5.81 \)
\( = 61.7975 \)In simple words: The least squares method finds the "best fit" straight line through the data points. This line helps us see the general direction or trend of the sales over the years. We calculate two special numbers, 'a' and 'b', which tell us where the line starts and how steeply it goes up or down.

🎯 Exam Tip: When using the least squares method for time series data, choose the origin and unit of \( x \) carefully. Centering \( x \) by taking the mean of the time period as the origin often simplifies calculations, especially when \( \Sigma x = 0 \).

 

Question 4. Fit a straight line trend by the method of least squares to the following data.

Answer:

We need to calculate various products of prices (\( P \)) and quantities (\( Q \)) for both the base year (denoted by subscript 0) and the current year (denoted by subscript 1). This is essential for computing price index numbers like Laspeyres', Paasche's, and Fisher's.

Now we can calculate the various price index numbers:
**Laspeyres' price index number** \( P_{01}^L \):
\( P_{01}^L = \frac{\Sigma P_1 Q_0}{\Sigma P_0 Q_0} \times 100 \)
\( = \frac{239945.23}{480244.71} \times 100 \)
\( = 49.96 \)
**Paasche's price index number** \( P_{01}^P \):
\( P_{01}^P = \frac{\Sigma P_1 Q_1}{\Sigma P_0 Q_1} \times 100 \)
\( = \frac{325150.5}{645496.98} \times 100 \)
\( = 50.37 \)
**Fisher's price index number** \( P_{01}^F \):
\( P_{01}^F = \sqrt{\frac{\Sigma P_1 Q_0}{\Sigma P_0 Q_0} \times \frac{\Sigma P_1 Q_1}{\Sigma P_0 Q_1}} \times 100 \)
\( = \sqrt{\frac{239945.23 \times 325150.5}{480244.71 \times 645496.98}} \times 100 \)
\( = \sqrt{\frac{3572000}{1827840}} \times 100 \)
\( = \sqrt{1.9542} \times 100 \)
\( = 1.3979 \times 100 \)
\( = 139.79 \approx 139.8 \)
**Time reversal test**:
This test checks if an index number formula yields the same result when the base and current periods are swapped. It is satisfied if \( P_{01} \times P_{10} = 1 \).
\( P_{01} \times P_{10} = \sqrt{\frac{\Sigma P_1 Q_0}{\Sigma P_0 Q_0} \times \frac{\Sigma P_1 Q_1}{\Sigma P_0 Q_1}} \times \sqrt{\frac{\Sigma P_0 Q_1}{\Sigma P_1 Q_1} \times \frac{\Sigma P_0 Q_0}{\Sigma P_1 Q_0}} \)
\( = \sqrt{\frac{1900 \times 1880}{1360 \times 1344}} \times \sqrt{\frac{1344 \times 1360}{1880 \times 1900}} \)
\( = \sqrt{\frac{1900 \times 1880 \times 1344 \times 1360}{1360 \times 1344 \times 1880 \times 1900}} \)
\( = \sqrt{1} = 1 \)
Thus, Fisher's Ideal Index satisfies the Time Reversal Test.In simple words: We calculate different ways to measure how prices change between a starting year and a current year. Laspeyres' index uses quantities from the starting year, while Paasche's uses quantities from the current year. Fisher's index combines both of these. Then, we check if Fisher's index works correctly if we swap the starting and current years, which it does.

🎯 Exam Tip: When calculating index numbers, clearly label all sums (\( \Sigma P_0 Q_0 \), \( \Sigma P_1 Q_0 \), etc.) as they are crucial for accuracy. Remember that Fisher's index is the geometric mean of Laspeyres' and Paasche's indices and satisfies important tests like Time Reversal.

 

Question 5. Using the following data, construct Fisher's Ideal Index Number and Show that it satisfies Factor Reversal Test and Time Reversal Test?

Answer:

To construct Fisher's Ideal Index and test its properties, we first calculate the necessary products of prices (\( P \)) and quantities (\( Q \)) for both the base year (0) and the current year (1).

**Fisher's Ideal Index Number** \( P_{01} \):
\( P_{01} = \sqrt{\frac{\Sigma P_1 Q_0}{\Sigma P_0 Q_0} \times \frac{\Sigma P_1 Q_1}{\Sigma P_0 Q_1}} \times 100 \)
\( = \sqrt{\frac{1900}{1360} \times \frac{1880}{1344}} \times 100 \)
\( = \sqrt{1.39705 \times 1.39880} \times 100 \)
\( = \sqrt{1.9543} \times 100 \)
\( = 1.39796 \times 100 = 139.80 \)
**Time Reversal Test**: This test is satisfied if \( P_{01} \times P_{10} = 1 \).
\( P_{01} \times P_{10} = \sqrt{\frac{\Sigma P_1 Q_0}{\Sigma P_0 Q_0} \times \frac{\Sigma P_1 Q_1}{\Sigma P_0 Q_1}} \times \sqrt{\frac{\Sigma P_0 Q_1}{\Sigma P_1 Q_1} \times \frac{\Sigma P_0 Q_0}{\Sigma P_1 Q_0}} \)
\( = \sqrt{\frac{1900 \times 1880}{1360 \times 1344}} \times \sqrt{\frac{1344 \times 1360}{1880 \times 1900}} \)
\( = \sqrt{\frac{1900 \times 1880 \times 1344 \times 1360}{1360 \times 1344 \times 1880 \times 1900}} = \sqrt{1} = 1 \)
Hence, Fisher's Ideal Index satisfies the Time Reversal Test.
**Factor Reversal Test**: This test is satisfied if \( P_{01} \times Q_{01} = \frac{\Sigma P_1 Q_1}{\Sigma P_0 Q_0} \).
Here, \( Q_{01} = \sqrt{\frac{\Sigma Q_1 P_0}{\Sigma Q_0 P_0} \times \frac{\Sigma Q_1 P_1}{\Sigma Q_0 P_1}} \) (Fisher's quantity index).
\( Q_{01} = \sqrt{\frac{1344}{1360} \times \frac{1880}{1900}} \)
\( = \sqrt{0.988235 \times 0.989473} = \sqrt{0.97787} = 0.98887 \)
\( P_{01} \times Q_{01} = 1.39796 \times 0.98887 = 1.3823 \)
And \( \frac{\Sigma P_1 Q_1}{\Sigma P_0 Q_0} = \frac{1880}{1360} = 1.38235 \)
Since \( P_{01} \times Q_{01} \approx \frac{\Sigma P_1 Q_1}{\Sigma P_0 Q_0} \), Fisher's Ideal Index satisfies the Factor Reversal Test.In simple words: We find Fisher's index to see how prices changed. Then, we check two special rules. The first rule, "Time Reversal," means if we swap the start and end years, the index should still make sense. The second rule, "Factor Reversal," means if we multiply the price index by a quantity index, we should get the total value change. Fisher's index passes both of these important checks.

🎯 Exam Tip: Always remember the formulas for Fisher's price and quantity indices precisely. When proving the Time and Factor Reversal Tests, show each step clearly, as these proofs are critical for demonstrating a full understanding.

 

Question 6. Compute the consumer price index for 2015 on the basis of 2014 from the following data.

Answer:

To compute the consumer price index (CPI) for 2015 based on 2014, we need to calculate the total expenditure for each year using the given quantities. Let \( P_0 \) be the prices in 2015 (base year) and \( P_1 \) be the prices in 2016 (current year). Let \( Q \) be the quantities.

The formula for the Consumer Price Index (CPI) using the Laspeyres method (which uses base period quantities) is:
\( \text{CPI} = \frac{\Sigma P_1 Q_0}{\Sigma P_0 Q_0} \times 100 \)
In this case, since quantities are given as fixed for all items, we treat \( Q \) as \( Q_0 \).
\( \text{CPI} = \frac{174.00}{146.50} \times 100 \)
\( = 1.18778 \times 100 \)
\( = 118.77 \)In simple words: To find out how much prices have changed for everyday goods, we add up how much everything costs in the new year and compare it to how much it cost in the old year, using the same amounts of goods. The answer tells us that prices have gone up by nearly 19% from 2014 to 2015.

🎯 Exam Tip: When calculating CPI, always ensure you correctly identify the base year and current year prices, and use the appropriate quantity (usually base year quantities for a standard CPI calculation, unless specified otherwise).

 

Question 7. An Enquiry was made into the budgets of the middle class families in a city gave the following information. What changes in the cost of living have taken place in the middle class families of a city?

Answer:

To find the change in the cost of living, we use a weighted aggregate method, also known as the Family Budget Method. We first calculate the price relative (\( P_1/P_0 \times 100 \)) for each item and then multiply it by its weight (\( w \)). Finally, we sum these products and divide by the total weight.

The Cost of Living Index number is calculated as:
\( \text{Index} = \frac{\Sigma Pw}{\Sigma w} \)
\( = \frac{12610}{100} \)
\( = 126.10 \)
Since the index number is 126.10, the cost of living has increased by \( (126.10 - 100) = 26.10\% \) in 2011 compared to 2010.In simple words: We calculate how much more expensive daily needs have become by looking at how much prices changed for each item and how important that item is to a family's budget. We find that the overall cost of living has gone up by 26.10%.

🎯 Exam Tip: For cost of living index problems, it's vital to correctly calculate the price relative (\( P \)) for each item and then apply the weights to find the weighted average. Ensure the final interpretation of the index number (percentage increase or decrease) is accurate.

 

Question 8. From the following data, calculate the control limits for the mean and range chart.

Answer:

To calculate the control limits for the mean (\( \bar{x} \)) and range (\( R \)) charts, we first need to determine the overall mean of the sample means (\( \bar{\bar{x}} \)) and the average range (\( \bar{R} \)). The number of observations per sample (\( n \)) is 5, and the number of samples is 10. For \( n=5 \), the control chart constants are \( A_2 = 0.577 \), \( D_3 = 0 \), and \( D_4 = 2.114 \).

First, let's calculate \( \bar{\bar{x}} \) and \( \bar{R} \):
\( \bar{\bar{x}} = \frac{\Sigma \bar{x}}{\text{number of samples}} = \frac{510}{10} = 51 \)
\( \bar{R} = \frac{\Sigma R}{n} = \frac{65}{10} = 6.5 \)
**Control limits for the \( \bar{x} \) chart:**
Upper Control Limit (UCL) \( = \bar{\bar{x}} + A_2 \bar{R} \)
\( = 51 + 0.577(6.5) \)
\( = 51 + 3.7505 \)
\( = 54.7505 \approx 54.75 \)
Central Line (CL) \( = \bar{\bar{x}} = 51 \)
Lower Control Limit (LCL) \( = \bar{\bar{x}} - A_2 \bar{R} \)
\( = 51 - 0.577(6.5) \)
\( = 51 - 3.7505 \)
\( = 47.2495 \approx 47.25 \)
**Control limits for the Range (\( R \)) chart:**
UCL \( = D_4 \bar{R} \)
\( = 2.114(6.5) \)
\( = 13.741 \)
CL \( = \bar{R} = 6.5 \)
LCL \( = D_3 \bar{R} \)
\( = 0(6.5) \)
\( = 0 \)In simple words: We find the upper, middle, and lower lines that help us check if a process is working correctly. For the average value of each sample, the upper limit is 54.75, the middle line is 51, and the lower limit is 47.25. For how spread out the numbers are (the range), the upper limit is 13.741, the middle line is 6.5, and the lower limit is 0. These limits help us see if anything goes wrong in the process.

🎯 Exam Tip: Always state the constants \( A_2, D_3, D_4 \) used for the given sample size \( n \). Clearly calculate \( \bar{\bar{x}} \) and \( \bar{R} \) first, as any error here will propagate through all control limit calculations.

 

Question 10. The following are the sample means and I ranges for 10 samples, each of size 5. Calculate; the control limits for the mean chart and range chart and state whether the process is in control or not.
Answer:

Now, let's find the control limits for the mean chart (X-bar chart):
\( \bar{\bar{x}} = \frac { \sum \bar{x} }{ \text{number of samples} } = \frac { 49.82 }{ 10 } = 4.982 \)
\( \bar{R} = \frac { \sum R }{ n } = \frac { 3.6 }{ 10 } = 0.36 \)
For a sample size of 5, the control chart constant \( A_2 = 0.577 \).
Upper Control Limit (UCL) for \( \bar{x} \) chart: \( \text{UCL}_{\bar{x}} = \bar{\bar{x}} + A_2 \bar{R} \)
\( = 4.982 + 0.577(0.36) \)
\( = 4.982 + 0.20772 \)
\( = 5.18972 \approx 5.19 \)
Central Line (CL) for \( \bar{x} \) chart: \( \text{CL}_{\bar{x}} = \bar{\bar{x}} = 4.982 \)
Lower Control Limit (LCL) for \( \bar{x} \) chart: \( \text{LCL}_{\bar{x}} = \bar{\bar{x}} - A_2 \bar{R} \)
\( = 4.982 - 0.577(0.36) \)
\( = 4.982 - 0.20772 \)
\( = 4.77428 \approx 4.774 \)

Next, let's find the control limits for the Range chart (R chart):
For a sample size of 5, the control chart constants are \( D_4 = 2.115 \) and \( D_3 = 0 \).
Upper Control Limit (UCL) for R chart: \( \text{UCL}_R = D_4 \bar{R} \)
\( = 2.115(0.36) \)
\( = 7.614 \)
Central Line (CL) for R chart: \( \text{CL}_R = \bar{R} = 3.6 \)
Lower Control Limit (LCL) for R chart: \( \text{LCL}_R = D_3 \bar{R} \)
\( = 0(0.36) \)
\( = 0 \)

Conclusion:
We see that all sample means (from 4.92 to 5.10) fall between the LCL of 4.774 and the UCL of 5.19 for the mean chart. Also, all sample ranges (from 0.1 to 0.8) fall between the LCL of 0 and the UCL of 7.614 for the range chart. Since all observed points are within their calculated control limits, the process is considered to be in control. Control charts help visualize if a process is stable over time.
In simple words: First, we find the average of all the sample means and ranges. Then, we use special numbers (constants) to calculate the upper and lower limits for both the mean and range charts. We check if all our sample points stay inside these limits. If they do, it means the process is stable and "in control."

🎯 Exam Tip: Always clearly state your conclusion about whether the process is in control by comparing all data points to the calculated control limits for both charts. Make sure to use the correct control chart constants \( A_2, D_3, D_4 \) for the given sample size.