Chapter Exam

Chapter 9: Applied Statistics — Online MCQ Test

Q: A sudden decline in sales of a company due to an unprecedented flood is an example of which component of a time series?

Answer: D. Irregular variation — Irregular or random variations are caused by unpredictable, sudden, non-recurring events like floods, earthquakes, strikes, or fires.

Q: Given the following annual data of production: N = 5, ∑Y = 150, ∑X² = 10, and ∑XY = 20, where X represents deviations from the mid-year (2020). What is the projected trend value for the year 2023?

Answer: B. 36 — Here, mid-year 2020 is origin (X=0). Thus, 2021 is X=1, 2022 is X=2, and 2023 is X=3. Since a = ∑Y/N = 150/5 = 30 and b = ∑XY/∑X² = 20/10 = 2, the trend equation is Y = 30 + 2X. For 2023, Y = 30 + 2(3) = 36.

Q: In a quality control inspection, 20 samples of size 100 each were taken from a production line. The total number of defectives observed across all samples was 160. Determine the Upper Control Limit (UCL) for the np-chart (number of defectives chart).

Answer: C. 16.14 — Mean fraction defective p-bar = 160 / (20 * 100) = 0.08. Central line np-bar = 100 * 0.08 = 8. S.E.(np) = √(np-bar * (1 - p-bar)) = √(8 * 0.92) = 2.713. UCL = np-bar + 3 * S.E.(np) = 8 + 3(2.713) = 16.14.

BUSINESS MATHS AND STATISTICS · CLASS 12th · Tamil Nadu State Board

Practice Chapter 9: Applied Statistics with a free chapter-wise online MCQ test for Tamil Nadu State Board CLASS 12th BUSINESS MATHS AND STATISTICS. This chapter covers: Focusing on economic data analytics this chapter details time series analysis index numbers and statistical quality control. Students compute trend values seasonal indices price qu.... AI-generated questions from basic to board-exam level, with instant results and explanations.

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Chapter 9: Applied Statistics — Important Questions & Answers (FAQ)

Frequently asked questions from Tamil Nadu State Board CLASS 12th BUSINESS MATHS AND STATISTICS — Chapter 9: Applied Statistics, with answers and explanations. These are sample questions; the exam has its own separate question set.

A sudden decline in sales of a company due to an unprecedented flood is an example of which component of a time series?

A. Secular trend
B. Seasonal variation
C. Cyclical variation
D. Irregular variation ✓

Answer: D. Irregular variation
Irregular or random variations are caused by unpredictable, sudden, non-recurring events like floods, earthquakes, strikes, or fires.

Fisher's Price Index Number is called the 'Ideal Index' because it satisfies which of the following tests?

A. Only the Time Reversal Test
B. Only the Factor Reversal Test
C. Both the Time Reversal and Factor Reversal Tests ✓
D. Neither the Time Reversal nor the Factor Reversal Test

Answer: C. Both the Time Reversal and Factor Reversal Tests
Fisher's index is theoretically sound as it satisfies both the Time Reversal Test (TRT) and the Factor Reversal Test (FRT).

In a 4-yearly moving average method of trend determination, why is 'centering' performed?

A. To make the trend values perfectly linear
B. To synchronize and align the trend values with the original time periods ✓
C. To eliminate irregular variations completely
D. To ensure the sum of residuals equals zero

Answer: B. To synchronize and align the trend values with the original time periods
For an even-period moving average (like 4-yearly), the averages lie between two periods. Centering aligns the averages with actual years.

Given the following annual data of production: N = 5, ∑Y = 150, ∑X² = 10, and ∑XY = 20, where X represents deviations from the mid-year (2020). What is the projected trend value for the year 2023?

A. 34
B. 36 ✓
C. 38
D. 40

Answer: B. 36
Here, mid-year 2020 is origin (X=0). Thus, 2021 is X=1, 2022 is X=2, and 2023 is X=3. Since a = ∑Y/N = 150/5 = 30 and b = ∑XY/∑X² = 20/10 = 2, the trend equation is Y = 30 + 2X. For 2023, Y = 30 + 2(3) = 36.

In a quality control inspection, 20 samples of size 100 each were taken from a production line. The total number of defectives observed across all samples was 160. Determine the Upper Control Limit (UCL) for the np-chart (number of defectives chart).

A. 12.16
B. 14.85
C. 16.14 ✓
D. 8.00

Answer: C. 16.14
Mean fraction defective p-bar = 160 / (20 * 100) = 0.08. Central line np-bar = 100 * 0.08 = 8. S.E.(np) = √(np-bar * (1 - p-bar)) = √(8 * 0.92) = 2.713. UCL = np-bar + 3 * S.E.(np) = 8 + 3(2.713) = 16.14.

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