CAIIB ABM UNIT 7 MCQs – Estimation

Correct Answer: B. Inferential Statistics. Inferential statistics involves using data from a sample to make estimations or test hypotheses about the larger population from which the sample was drawn. For example, measuring the average height of 100 students (sample) to estimate the average height of all students in a school (population) is an application of inferential statistics.

Correct Answer: C. Point Estimate. A point estimate is a single numerical value used to estimate an unknown population parameter. Here, ₹5,400 is a single value estimating the average spending for all households. For instance, stating the average mileage of a car model is 18 km/litre based on a sample test is a point estimate.

Correct Answer: D. It indicates the potential error and reliability of the estimate. An interval estimate provides a range of values and is usually accompanied by a confidence level, giving a sense of how likely the true population parameter falls within that range, thus indicating potential error and reliability. For example, stating that average spending is between ₹5,200 and ₹5,600 with 95% confidence is more informative about reliability than just stating ₹5,400.

Correct Answer: C. Estimator. An estimator is a rule or formula (a statistic) based on sample data used to guess the value of a population parameter. The sample mean (calculated as Sum of observations / Number of observations) is a common estimator for the population mean. For example, the formula for sample variance, s², is an estimator for the population variance, σ².

Correct Answer: B. Estimate. An estimate is the specific numerical value obtained by applying an estimator to a particular sample. In this case, 1,200 hours is the calculated value of the sample mean (the estimator) from the specific sample of 50 bulbs. It is the specific guess for the average lifespan of all such bulbs.

Correct Answer: B. The mean of its sampling distribution is equal to the population parameter being estimated. An unbiased estimator does not systematically overestimate or underestimate the population parameter. If we were to take many samples and calculate the estimate each time, the average of all these estimates would be equal to the true parameter value. For example, the sample mean is an unbiased estimator of the population mean.

Correct Answer: D. Estimator A is more efficient. Efficiency refers to the size of the standard error of an estimator; a smaller standard error indicates less variability and thus higher efficiency. Since Estimator A has a smaller standard error (2.5 < 3.1), it is considered more efficient than Estimator B. For example, if we are estimating the population mean, the sample mean usually has a smaller standard error than the sample median (for symmetric distributions), making the sample mean more efficient.

Correct Answer: C. Its value approaches the population parameter as the sample size increases. A consistent estimator becomes more accurate and reliable as more data is collected (i.e., as the sample size grows larger). For example, the sample mean becomes a more reliable estimate of the population mean as the number of observations in the sample increases.

Correct Answer: D. Sufficiency. A sufficient estimator captures all the information present in the sample relevant to estimating the parameter; no other estimator calculated from the same sample can provide additional information about the population parameter. For instance, for a normally distributed population, the sample mean is a sufficient estimator for the population mean.

Correct Answer: D. A point estimate to an interval estimate. The initial prediction of ₹45 Lakhs is a single value, representing a point estimate. The revised prediction of ₹42 Lakhs to ₹48 Lakhs provides a range, which is characteristic of an interval estimate, also conveying a level of certainty about the range containing the true value.

Correct Answer: C. Point Estimator. A point estimator is a single numerical value calculated from sample data that serves as the best guess for an unknown population parameter. For example, using the average weight of 50 sampled apples (say, 150 grams) to estimate the average weight of all apples in an orchard is using a point estimator.

Correct Answer: C. Sample Mean (x̄). The sample mean (x̄) is typically the preferred estimator for the population mean (μ) because it possesses desirable properties such as being unbiased, consistent, and the most efficient among common estimato₹ For instance, if we measure the heights of a sample of students, the average height of this sample is the best single value to estimate the average height of all students in the school.

Correct Answer: C. 102 syringes. The point estimate for the population mean (μ) using the sample mean (x̄) is calculated as the sum of observations (Σx) divided by the sample size (n). Here, x̄ = Σx ⁄ n = 3570 ⁄ 35 = 102. This single value, 102, is the best estimate for the average number of syringes per carton in the entire population based on this sample.

Correct Answer: D. s² = Σ(x − x̄)² ⁄ (n − 1). To obtain an unbiased estimate of the population variance from a sample, the sum of squared deviations from the sample mean should be divided by n − 1 (sample size minus one), not just n. Using n − 1 corrects the tendency of the sample variance calculated with n to slightly underestimate the true population variance.

Correct Answer: B. 0.00055. The unbiased point estimate for the population variance (σ²) is the sample variance (s²) calculated using the formula s² = Σ(x − x̄)² ⁄ (n − 1). Here, n = 5, so n − 1 = 4. Therefore, s² = 0.0022 ⁄ 4 = 0.00055.

Correct Answer: B. 0.08. The sample proportion (p̄) is used as the point estimator for the population proportion (p). It is calculated as the number of items with the characteristic (damaged cartons) divided by the total sample size (n). Here, p̄ = 4 ⁄ 50 = 0.08. This means our best single guess for the proportion of damaged cartons in the population is 0.08 or 8%.

Correct Answer: B. Sample standard deviation (s). The sample standard deviation (s), which is the square root of the unbiased sample variance (s² = Σ(x − x̄)² ⁄ (n − 1)), is the most frequently used point estimator for the population standard deviation (σ). For example, if we calculate s for the heights of a sample of students, this value is our estimate for the standard deviation of heights of all students.

Correct Answer: C. Sufficient. Sample mean (x̄) is unbiased, consistent, and the most efficient estimator for the population mean (μ). While sufficiency is a desirable property for estimators in general, it is not among the key properties for the sample mean.

Correct Answer: B. 4.35. The point estimate for the population mean (μ) is the sample mean (x̄), calculated as x̄ = Σx ⁄ n. Given Σx = 21.75 and n = 5, the sample mean is x̄ = 21.75 ⁄ 5 = 4.35. This value, 4.35, is the point estimate.

Correct Answer: C. Sampling is generally less expensive and faster. While estimation introduces an inherent error because not every item is measured, sampling is often used because counting or measuring everything in a large population can be prohibitively expensive and time-consuming. Sampling provides a practical way to learn about population characteristics.

Correct Answer: B. The probability that the interval estimate will include the population parameter. The confidence level expresses the degree of certainty that the calculated interval contains the true value of the parameter being estimated.

Correct Answer: C. Confidence limits. The upper boundary of the range is the upper confidence limit (UCL) and the lower boundary is the lower confidence limit (LCL).

Correct Answer: C. It increases the width of the confidence interval. A higher confidence level requires capturing the parameter with greater certainty, which necessitates a wider range of possible values.

Correct Answer: C. The confidence interval. The confidence interval is the specific range of values calculated from sample data within which the population parameter is estimated to lie.

Correct Answer: C. Normal distribution. The Central Limit Theorem allows the use of the normal distribution to approximate the sampling distribution of the mean for large samples, regardless of the population’s distribution.

Correct Answer: B. 5 hou₹ The standard error of the mean is calculated as σₓ̄ = σ ⁄ √n, so σₓ̄ = 50 ⁄ √100 = 50 ⁄ 10 = 5.

Correct Answer: B. 1.96. A 95 per cent confidence level corresponds to the central 95 per cent of the area under the standard normal curve, which leaves 2.5 per cent in each tail. The Z-value cutting off 2.5 per cent in the upper tail is approximately 1.96.

Correct Answer: B. 19.82 to 22.18 months. The interval is x̄ ± Z * σₓ̄. For 95 per cent confidence, Z = 1.96. So, 21 ± 1.96 * 0.6 = 21 ± 1.18, giving the interval (19.82, 22.18).

Correct Answer: C. σ̂. The ‘hat’ symbol typically indicates an estimated value. When the population standard deviation is unknown, the sample standard deviation (s) is used as its estimate, denoted σ̂.

Correct Answer: B. n – 1. Using n-1 in the denominator, s² = Σ(x – x̄)² ⁄ (n-1), provides an unbiased estimate of the population variance.

Correct Answer: C. When the sample size (n) is more than 5 per cent of the population size (N). The correction factor adjusts the standard error to account for the reduced variability when sampling a significant portion of a finite population.

Correct Answer: C. σ̂ₓ̄ = (σ̂ ⁄ √n) * (√((N-n) ⁄ (N-1))). This formula uses the estimated population standard deviation (σ̂) and incorporates the finite population correction factor.

Correct Answer: B. ₹ 11,587.50 to ₹ 12,012.50. The interval is x̄ ± Z * σ̂ₓ̄. So, 11800 ± 1.64 * 129.57 = 11800 ± 212.50, giving the interval (11587.50, 12012.50).

Correct Answer: C. The population proportion (p). The proportion of successes observed in a sample is used as a point estimate for the corresponding proportion in the entire population.

Correct Answer: B. Normal distribution. For sufficiently large sample sizes, the shape of the binomial distribution approaches that of the normal distribution, simplifying calculations for interval estimates.

Correct Answer: B. n * p ≥ 5 and n * q ≥ 5. This rule ensures that the sample size is large enough relative to the proportions p and q for the normal approximation to be reasonably accurate.

Correct Answer: A. √((p * q) ⁄ n). This formula measures the standard deviation of the sampling distribution of the sample proportion.

Correct Answer: B. By substituting the sample proportion (p̄) and q̄ for p and q. The estimated standard error is calculated as σ̂ₚ̄ = √((p̄ * q̄) ⁄ n).

Correct Answer: D. 0.057. Here, q̄ = 1 – p̄ = 1 – 0.4 = 0.6. The estimated standard error is σ̂ₚ̄ = √((0.4 * 0.6) ⁄ 75) = √(0.24 ⁄ 75) = √0.0032 ≈ 0.057.

Correct Answer: D. 2.58. A 99 per cent confidence level corresponds to the central 99 per cent of the area under the standard normal curve, leaving 0.5 per cent in each tail. The Z-value cutting off 0.5 per cent in the upper tail is approximately 2.58.

Correct Answer: B. 0.547. The UCL is p̄ + Z * σ̂ₚ̄. For 99 per cent confidence, Z = 2.58. So, UCL = 0.4 + 2.58 * 0.057 = 0.4 + 0.147 = 0.547.

Correct Answer: C. 0.253. The LCL is p̄ – Z * σ̂ₚ̄. For 99 per cent confidence, Z = 2.58. So, LCL = 0.4 – 2.58 * 0.057 = 0.4 – 0.147 = 0.253.

Correct Answer: B. Confidence Level. The confidence level is explicitly defined as the probability associated with an interval estimate.

Correct Answer: C. 1.64. A 90 per cent confidence level leaves 5 per cent in each tail. The Z-value corresponding to an area of 0.45 (half of 0.90) from the mean to the limit is approximately 1.64.

Correct Answer: C. If many samples were taken, about 95 per cent of the calculated confidence intervals would contain the true population mean. The confidence level applies to the procedure of constructing intervals over many samples, not to a single calculated interval.

Correct Answer: C. Point estimate. A point estimate is a single numerical value calculated from sample data that serves as the best guess for an unknown population parameter. For example, calculating the average height of 5 students (say, 165 cm) to estimate the average height of all students in a school is a point estimate.

Correct Answer: D. Interval estimate. An interval estimate provides a range of values within which the population parameter is expected to lie, along with a certain level of confidence. For instance, stating that the average height of students is likely between 162 cm and 168 cm is an interval estimate.

Correct Answer: B. Estimator. An estimator is the rule or formula (a sample statistic) used to calculate an estimate of a population parameter. For example, the formula for the sample mean (x̄ = Σx ⁄ n) is an estimator for the population mean (µ).

Correct Answer: C. Estimate. An estimate is the actual numerical value computed from sample data using an estimator. If we use the sample mean formula (estimator) on a sample of heights {160, 165, 170} cm, the calculated value x̄ = (160+165+170)⁄3 = 165 cm is the estimate.

Correct Answer: D. Unbiasedness. An estimator is unbiased if the mean of its sampling distribution is exactly equal to the population parameter it aims to estimate. For example, the sample mean (x̄) is an unbiased estimator of the population mean (µ) because, over many samples, the average of sample means will equal the population mean.

Correct Answer: C. Efficiency. Efficiency compares two unbiased estimators; the one with the smaller variance (or standard error) for a given sample size is considered more efficient because it provides estimates closer to the population parameter more consistently. If estimator A has a standard error of 2 and estimator B has a standard error of 3 for the same sample size, estimator A is more efficient.

Correct Answer: D. Consistency. A consistent estimator’s value gets progressively closer to the true population parameter as the sample size (n) grows larger. For instance, the sample mean becomes a more precise estimate of the population mean as we include more observations in our sample.

Correct Answer: A. 4.35 cm. The point estimate for the population mean (µ) is the sample mean (x̄). x̄ = (4.33 + 4.37 + 4.36 + 4.32 + 4.37) ⁄ 5 = 21.75 ⁄ 5 = 4.35 cm.

Correct Answer: B. 0.00055 cm². First calculate deviations from mean (4.35): -0.02, 0.02, 0.01, -0.03, 0.02. Square these: 0.0004, 0.0004, 0.0001, 0.0009, 0.0004. Sum of squares Σ(x-x̄)² = 0.0004 + 0.0004 + 0.0001 + 0.0009 + 0.0004 = 0.0022. Sample variance s² = 0.0022 ⁄ (5-1) = 0.0022 ⁄ 4 = 0.00055 cm².

Correct Answer: C. 0.08. The point estimate for the population proportion (p) is the sample proportion (p̄). p̄ = (Number of successes) ⁄ (Sample size) = 4 ⁄ 50 = 0.08.

Correct Answer: B. The probability that the interval estimate contains the population parameter. The confidence level expresses how sure we are that the method used to construct the interval results in an interval containing the true parameter value. For example, a 95% confidence level means that if we took many samples and constructed intervals, about 95% of those intervals would contain the true population parameter.

Correct Answer: C. The interval becomes wider. To be more confident that the interval captures the true population parameter, we need to allow for a wider range of possible values. Increasing confidence level requires a larger margin of error, thus widening the interval.

Correct Answer: B. Confidence limits. The lowest value in the confidence interval is the Lower Confidence Limit (LCL), and the highest value is the Upper Confidence Limit (UCL).

Correct Answer: A. 0.6 months. The standard error of the mean measures the variability of sample means. σₓ̄ = σ ⁄ √n = 6 ⁄ √100 = 6 ⁄ 10 = 0.6 months.

Correct Answer: B. 19.82 to 22.18 months. The confidence interval is calculated as x̄ ± Z * σₓ̄. Lower Limit (LCL) = 21 – 1.96 * 0.6 = 21 – 1.18 = 19.82. Upper Limit (UCL) = 21 + 1.96 * 0.6 = 21 + 1.18 = 22.18.

Correct Answer: D. The sample standard deviation (s). When σ is unknown, the sample standard deviation (s), calculated as s = √{Σ(x-x̄)² ⁄ (n-1)}, is used as its estimate (σ̂ = s).

Correct Answer: B. Rs. 129.57. First, calculate the standard error without correction: s ⁄ √n = 950 ⁄ √50 ≈ 950 ⁄ 7.071 ≈ 134.35. Then apply the finite population correction factor: √{(N-n)⁄(N-1)} = √{(700-50)⁄(700-1)} = √(650⁄699) ≈ √0.9299 ≈ 0.9643. Estimated standard error σ̂ₓ̄ ≈ 134.35 * 0.9643 ≈ 129.57.

Correct Answer: B. Rs. 11,587.50 to Rs. 12,012.50. The confidence interval is x̄ ± Z * σ̂ₓ̄. LCL = 11800 – 1.64 * 129.57 = 11800 – 212.50 = 11587.50. UCL = 11800 + 1.64 * 129.57 = 11800 + 212.50 = 12012.50.

Correct Answer: C. Normal distribution. Although the binomial distribution is theoretically correct for proportions, the normal distribution provides a good approximation when the sample size (n) is large enough (typically when both np ≥ 5 and nq ≥ 5).

Correct Answer: B. √(p*q ⁄ n). The standard error of the proportion measures the variability of sample proportions around the population proportion (p), where q = 1-p and n is the sample size.

Correct Answer: B. By using √(p̄q̄ ⁄ n). When p is unknown, the sample proportion (p̄) and sample complement (q̄ = 1-p̄) are used to estimate the standard error. The formula becomes σ̂ₚ̄ = √(p̄q̄ ⁄ n).

Correct Answer: A. 0.057. First find q̄ = 1 – p̄ = 1 – 0.4 = 0.6. Then use the formula σ̂ₚ̄ = √(p̄*q̄ ⁄ n) = √(0.4 * 0.6 ⁄ 75) = √(0.24 ⁄ 75) = √0.0032 ≈ 0.057.

Correct Answer: B. 0.253 to 0.547. The confidence interval is p̄ ± Z * σ̂ₚ̄. LCL = 0.4 – 2.58 * 0.057 = 0.4 – 0.147 = 0.253. UCL = 0.4 + 2.58 * 0.057 = 0.4 + 0.147 = 0.547.

Correct Answer: A. 14.28. The point estimate of the mean is the sample mean. Sum = 14.1 + 8.8 + 14.0 + 21.3 + 7.9 + 12.5 + 20.6 + 16.3 + 13.0 = 128.5. Sample mean x̄ = 128.5 ⁄ 9 ≈ 14.28 (thousands).

Correct Answer: A. 23.33. s² = Σ(x-x̄)² ⁄ (n-1) = 186.64 ⁄ (9-1) = 186.64 ⁄ 8 ≈ 23.33.