What is the difference between a parameter and a statistic?

A parameter is a numerical value that describes a characteristic of a population (e.g., population mean μ, population standard deviation σ). A statistic is a numerical value that describes a characteristic of a sample (e.g., sample mean x̄, sample standard deviation s). Parameters are usually unknown and are estimated using statistics.

What is a confidence interval in inferential statistics?

A confidence interval is a range of values, derived from sample data, that is likely to contain the true population parameter with a specified probability (confidence level). For example, a 95% confidence interval means that if we repeated the sampling process many times, about 95% of the resulting intervals would contain the true population mean.

When do we use a t-test instead of a z-test?

We use a t-test when: (1) the population standard deviation is unknown, and (2) the sample size is small (typically n < 30). In this case, we use the sample standard deviation s as an estimate for σ, and the test statistic follows a t-distribution with n−1 degrees of freedom.

What is the formula for the one-sample t-test statistic?

The one-sample t-test statistic is: t = (x̄ − μ₀) / (s / √n), where x̄ is the sample mean, μ₀ is the hypothesised population mean, s is the sample standard deviation, and n is the sample size. The degrees of freedom are ν = n − 1.

What are Type I and Type II errors in hypothesis testing?

A Type I error (α error) occurs when we reject a true null hypothesis — a false positive. A Type II error (β error) occurs when we fail to reject a false null hypothesis — a false negative. The significance level α is the probability of making a Type I error.

Class 12 Applied Maths Unit 5: Inferential Statistics - Free Resources

Unit 5: Inferential Statistics – Class 12 Applied Maths | Free MCQs & Solutions | Boundless Maths

Unit 5: Inferential Statistics carries 10 marks in the Class 12 Applied Mathematics board examination, making it one of the most important units to prepare thoroughly.

This unit covers three interconnected topics: Point Estimation, where you learn to estimate unknown population parameters such as the mean and standard deviation using sample data; Interval Estimation and Confidence Intervals, which extends point estimates by building a range that is likely to contain the true population value at a specified confidence level (95% or 99%); and Hypothesis Testing using the t-Distribution, which involves setting up null and alternative hypotheses, computing the one-sample t-test statistic, determining degrees of freedom, comparing with critical values, and drawing conclusions about population means.

This page includes MCQs with full solutions (including Assertion–Reason questions), and short-answer questions with complete step-by-step workings across all sub-topics.

📋 All questions and solutions are aligned to the latest CBSE Applied Maths syllabus 2026-27.

MCQs & AR Questions

Short Answer Questions

Board Exam Marks

Key Topics

📊

Topics Covered

📈 Point Estimation

Estimate population parameters from sample data — sample mean and sample standard deviation as point estimates.

Point estimate of \(\mu\): \(\bar{x} = \dfrac{\sum x_i}{n}\)

Point estimate of \(\sigma\): \(s = \sqrt{\dfrac{\sum(x_i - \bar{x})^2}{n-1}}\)

📉 Confidence Intervals

Build interval estimates for population mean using margin of error at 95% and 99% confidence levels.

\(\mu = \bar{x} \pm Z_{\alpha/2} \cdot \dfrac{\sigma}{\sqrt{n}}\)

95%: \(Z = 1.96\) | 99%: \(Z = 2.576\)

🔢 One-Sample t-Test

Test hypotheses about population means when \(\sigma\) is unknown using the t-distribution.

\(t = \dfrac{\bar{x} - \mu_0}{s/\sqrt{n}}\), \(\quad \nu = n - 1\)

Reject \(H_0\) if \(|t_{\text{calc}}| > t_{\text{critical}}\)

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📝 MCQs & Assertion-Reason (14) 💡 Short Answer Questions (12)

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Multiple Choice Questions

Question 1

Inferential statistics is a process that involves all of the following except:

aEstimating a parameter
bEstimating a statistic
cTest a hypothesis
dAnalyze relationships

✓ Correct Answer: (b) Estimating a statistic

Explanation: Inferential statistics involves drawing conclusions about a population from sample data. Its core activities are:
• Estimating population parameters (not statistics — statistics are already known from the sample)
• Testing hypotheses about population parameters
• Analysing relationships between variables
A statistic is computed directly from the sample and is already known, so estimating it is not part of inferential statistics.

Question 2

Which of the following statements are true?
Statement I: The mean of a population is denoted by \(\bar{x}\)
Statement II: The population mean is a statistic

aI only
bII only
cBoth I and II
dNone

✓ Correct Answer: (d) None

Explanation:
• Population mean is denoted by \(\mu\) (mu), not \(\bar{x}\) — \(\bar{x}\) is the sample mean → Statement I is FALSE
• Population mean is a parameter (it describes the whole population), not a statistic → Statement II is FALSE
Therefore both statements are false, so the answer is (d).

Question 3

The test statistic for a one-sample t-test, denoted by \(t\), is defined as:

a\(t = \dfrac{\bar{x} - \mu}{s/\sqrt{n}}\)
b\(t = \dfrac{\bar{x} - \mu}{s/n}\)
c\(t = \dfrac{\bar{x} - \mu}{s^2/n}\)
d\(t = \dfrac{\bar{x} - \mu}{s/n^2}\)

✓ Correct Answer: (a)

Explanation: The standard t-test statistic formula is \(t = \dfrac{\bar{x} - \mu}{s/\sqrt{n}}\), where:
• \(\bar{x}\) = sample mean
• \(\mu\) = hypothesised population mean
• \(s\) = sample standard deviation
• \(n\) = sample size
The denominator \(s/\sqrt{n}\) is called the standard error of the mean.

Question 4

For the purpose of a t-test of significance, a random sample of size \(n = 2025\) is drawn from a normal population. The degree of freedom \((\nu)\) is:

a\(2025^{2025}\)
b\(2024^{2025}\)
c\(2025\)
d\(2024\)

✓ Correct Answer: (d) 2024

Explanation: For a one-sample t-test, degrees of freedom \(\nu = n - 1\).
Given \(n = 2025\), therefore \(\nu = 2025 - 1 = \mathbf{2024}\).
We lose one degree of freedom because we use \(\bar{x}\) (estimated from the data) to calculate \(s\).

Question 5

What is the purpose of a null hypothesis?

aTo provide an alternative explanation for the data
bTo state that there is no significant difference or relationship
cTo predict the outcome of the study
dTo provide a summary of the data

✓ Correct Answer: (b) To state that there is no significant difference or relationship

Explanation: The null hypothesis \(H_0\) is a statement of no effect or no difference. It is the default assumption that we attempt to disprove using sample evidence. We test against the alternative hypothesis \(H_1\) (or \(H_a\)) and either reject or fail to reject \(H_0\) based on the test statistic.

Question 6

A grain wholesaler visits the granary market. While going around to make a good purchase, he takes a handful of rice from different sacks of rice, in order to inspect the quality of farmers' produce. The handful of rice taken from a sack for quality inspection is a:

aStatistic
bPopulation
cParameter
dSample

✓ Correct Answer: (d) Sample

Explanation: A sample is a subset of the population selected for analysis. Here:
• The population = all rice in the sack
• The sample = the handful taken for inspection
The wholesaler uses the sample to make inferences (decisions) about the quality of the entire sack.

Question 7

A population consists of four observations 1, 3, 5, 7. What is the variance?

✓ Correct Answer: (c) 5

Explanation:
• Population mean: \(\mu = \dfrac{1+3+5+7}{4} = \dfrac{16}{4} = 4\)
• Squared deviations: \((1-4)^2 = 9,\; (3-4)^2 = 1,\; (5-4)^2 = 1,\; (7-4)^2 = 9\)
• Population variance: \(\sigma^2 = \dfrac{9+1+1+9}{4} = \dfrac{20}{4} = \mathbf{5}\)
Note: Since this is a population, we divide by \(N\) (not \(N-1\)).

Question 8

Which of the following is the true relation between sample mean \((\bar{x})\) and population mean \((\mu)\)?

a\(|\bar{x} - \mu|\) increases when the size of sample increases
b\(\bar{x} = \mu\), for all sample sizes
c\(|\bar{x} - \mu|\) decreases when the size of sample decreases
d\(|\bar{x} - \mu|\) decreases when the size of sample increases

✓ Correct Answer: (d) |x̄ − μ| decreases when the size of sample increases

Explanation: As sample size \(n\) increases, the sample mean \(\bar{x}\) becomes a more accurate estimate of the population mean \(\mu\), so \(|\bar{x} - \mu|\) decreases. This is the essence of the Law of Large Numbers. The standard error \(\sigma/\sqrt{n}\) also shows that as \(n\) increases, the spread of \(\bar{x}\) around \(\mu\) shrinks.

Question 9

A simple random sample consists of five observations 2, 4, 6, 7, 6. The point estimate of population standard deviation is:

a4
b2.5
c5
d2

✓ Correct Answer: (d) 2

Explanation: The point estimate of population standard deviation = sample standard deviation \(s\).
• \(\bar{x} = \dfrac{2+4+6+7+6}{5} = \dfrac{25}{5} = 5\)
• \(\sum(x_i - \bar{x})^2 = (2-5)^2+(4-5)^2+(6-5)^2+(7-5)^2+(6-5)^2 = 9+1+1+4+1 = 16\)
• \(s = \sqrt{\dfrac{16}{5-1}} = \sqrt{\dfrac{16}{4}} = \sqrt{4} = \mathbf{2}\)

Question 10

Consider the following hypothesis test: \(H_0 : \mu \geq 20\); \(H_1 : \mu < 20\)
A sample of 64 provided a sample mean of 19.5. The population standard deviation is 2.
The value of the test statistic is:

a−2.5
b−2
c2
d−1.5

✓ Correct Answer: (b) −2

Explanation:
\[t = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} = \frac{19.5 - 20}{2/\sqrt{64}} = \frac{-0.5}{2/8} = \frac{-0.5}{0.25} = \mathbf{-2}\] Since \(H_1: \mu < 20\), this is a left-tailed test. The negative sign confirms the sample mean falls below the hypothesised mean.

Question 11

A machine makes car wheels and in a random sample of 26 wheels, the test statistic is found to be 3.07. As per the t-distribution test (at 5% level of significance), what can you say about the quality of wheels produced by the machine? [Use \(t_{25}(0.05) = 2.06\)]

aSuperior quality
bInferior quality
cSame quality
dCannot say

✓ Correct Answer: (a) Superior quality

Explanation:
• Degrees of freedom \(\nu = 26 - 1 = 25\)
• Calculated \(|t| = 3.07\); Critical value \(t_{25}(0.05) = 2.06\)
• Since \(3.07 > 2.06\), we reject the null hypothesis (which assumed equal/standard quality).
• The positive t-value indicates the sample mean exceeds the standard, so the machine is producing superior quality wheels.

Question 12

The Central Limit Theorem states that the sampling distribution of sample mean approaches a normal distribution if:

aAll possible samples are selected
bSample size is large
cSample size is small
dNone of the above

✓ Correct Answer: (b) Sample size is large

Explanation: The Central Limit Theorem (CLT) states that when the sample size is sufficiently large (typically \(n \geq 30\)), the sampling distribution of \(\bar{x}\) becomes approximately normal, regardless of the shape of the original population distribution. This is why large-sample inference uses the normal (z) distribution.

📋 Assertion-Reason Questions

Statement I is the Assertion (A) and Statement II is the Reason (R). Choose the correct option:

(a) Both A and R are True and R is the correct explanation of A
(b) Both A and R are True but R is not the correct explanation of A
(c) A is True but R is False
(d) A is False but R is True

Assertion-Reason 1

Assertion (A): A simple random sample consists of five observations 2, 4, 6, 8, 10. The point estimate of population standard deviation is \(\sqrt{10}\).

Reason (R): Sample standard deviation of \(n\) observations, \(s = \sqrt{\dfrac{\sum(x_i - \bar{x})^2}{n}}\)

aBoth A and R are True and R is the correct explanation of A
bBoth A and R are True but R is not the correct explanation of A
cA is True but R is False
dA is False but R is True

✓ Correct Answer: (c) A is True but R is False

Checking Assertion (A):
• \(\bar{x} = \dfrac{2+4+6+8+10}{5} = 6\)
• \(\sum(x_i - \bar{x})^2 = (2-6)^2+(4-6)^2+(6-6)^2+(8-6)^2+(10-6)^2 = 16+4+0+4+16 = 40\)
• \(s = \sqrt{\dfrac{40}{5-1}} = \sqrt{10}\) → Assertion is TRUE

Checking Reason (R):
The formula given in R uses \(n\) in the denominator: \(s = \sqrt{\dfrac{\sum(x_i-\bar{x})^2}{n}}\). This is the population standard deviation formula, not the sample standard deviation formula. The correct sample standard deviation formula divides by \((n-1)\). → Reason is FALSE.

Assertion-Reason 2

Assertion (A): A one-sample t-test is used to compare the mean of a sample to a known population mean.

Reason (R): The population standard deviation is unknown, and the sample size is small.

aBoth A and R are True and R is the correct explanation of A
bBoth A and R are True but R is not the correct explanation of A
cA is True but R is False
dA is False but R is True

✓ Correct Answer: (a) Both A and R are True and R is the correct explanation of A

Explanation:
• Assertion is TRUE — a one-sample t-test compares \(\bar{x}\) (from the sample) to a known or hypothesised value of \(\mu\).
• Reason is TRUE — we use the t-distribution (instead of the z-distribution) precisely because \(\sigma\) is unknown and we estimate it using \(s\), especially when \(n\) is small. This is the correct explanation for why the t-test is used.

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Short Answer Questions (2/3 Marks)

Question 1

The following data are from a simple random sample: 5, 8, 10, 7, 10, 14.
(i) What is the point estimate of the population mean?
(ii) What is the point estimate of the population standard deviation?

Solution:

Given: sample data = 5, 8, 10, 7, 10, 14 and \(n = 6\)

(i) Point estimate of population mean

The point estimate of the population mean is the sample mean.

\(\bar{x} = \dfrac{\sum x_i}{n} = \dfrac{5+8+10+7+10+14}{6} = \dfrac{54}{6} = 9\)

(ii) Point estimate of population standard deviation

The point estimate of population standard deviation is the sample standard deviation \(s\).

\(\sum(x_i - \bar{x})^2 = (5-9)^2+(8-9)^2+(10-9)^2+(7-9)^2+(10-9)^2+(14-9)^2\)

\(= 16+1+1+4+1+25 = 48\)

\(s = \sqrt{\dfrac{48}{6-1}} = \sqrt{\dfrac{48}{5}} = \sqrt{9.6} \approx 3.1\)

Answers:
(i) Point estimate of population mean \(= \mathbf{9}\)
(ii) Point estimate of population standard deviation \(\approx \mathbf{3.1}\)

Question 2

A student measuring the boiling temperature of a certain liquid observes the readings (in °C): 102.5, 101.7, 103.1, 100.9, 100.5 and 102.2 on 6 different samples of the liquid. If the standard deviation for this procedure is 1.2°C, find the interval estimation for the population mean at a 95% confidence level.

Solution:

Given: \(n = 6,\; \sigma = 1.2,\;\) confidence level = 95%

Observations: 102.5, 101.7, 103.1, 100.9, 100.5, 102.2

Sample mean: \(\bar{x} = \dfrac{102.5+101.7+103.1+100.9+100.5+102.2}{6} = \dfrac{610.9}{6} \approx 101.82\)

Confidence level = 95% \(\Rightarrow (1-\alpha) = 0.95 \Rightarrow \alpha = 0.05 \Rightarrow \alpha/2 = 0.025\)

From standard normal table: \(Z_{0.025} = 1.96\)

Margin of error \(= Z_{0.025} \times \dfrac{\sigma}{\sqrt{n}} = 1.96 \times \dfrac{1.2}{\sqrt{6}} = 1.96 \times 0.49 \approx 0.96\)

Confidence interval: \(\mu = \bar{x} \pm 0.96 = 101.82 \pm 0.96\)

95% Confidence Interval: (100.86, 102.78)°C

Question 3

A random sample of the heights of 10 students from a large number of students in a school has a mean height of 5.6 ft with a standard deviation of 0.75 ft. Find:
(i) 95% confidence limits (ii) 99% confidence limits for the average height of all students.
[Use \(t_9(0.05) = 2.262\); \(t_9(0.01) = 3.250\)]

Solution:

Given: \(\bar{x} = 5.6,\; s = 0.75,\; n = 10\), degrees of freedom \(\nu = 10-1 = 9\)

Standard error: \(\dfrac{s}{\sqrt{n}} = \dfrac{0.75}{\sqrt{10}} = \dfrac{0.75}{3.162} \approx 0.237\)

(i) 95% Confidence Limits

Margin of error: \(E = t_{0.05} \times \dfrac{s}{\sqrt{n}} = 2.262 \times 0.237 \approx 0.536\)

Confidence interval: \(\mu = 5.6 \pm 0.536\)

(ii) 99% Confidence Limits

Margin of error: \(E = t_{0.01} \times \dfrac{s}{\sqrt{n}} = 3.250 \times 0.237 \approx 0.770\)

Confidence interval: \(\mu = 5.6 \pm 0.770\)

Answers:
(i) 95% confidence limits: (5.064, 6.136) ft
(ii) 99% confidence limits: (4.83, 6.37) ft

Question 4

Consider the following hypothesis test:
\(H_0 : \mu \leq 12\); \(H_a : \mu > 12\)
A sample of 25 provided \(\bar{x} = 14\) and \(s = 4.32\).
(i) Compute the value of the test statistic.
(ii) What is the rejection rule? What is your conclusion? [Given \(t_{0.05} = 1.711\)]

Solution:

Given: \(\bar{x} = 14,\; \mu_0 = 12,\; s = 4.32,\; n = 25\)

This is a right-tailed t-test (\(H_a: \mu > 12\)). Degrees of freedom \(\nu = 25-1 = 24\).

(i) Test Statistic

\[t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{14 - 12}{4.32/\sqrt{25}} = \frac{2}{4.32/5} = \frac{2}{0.864} \approx 2.31\]

(ii) Rejection Rule and Conclusion

For a right-tailed test: Reject \(H_0\) if \(t_{\text{calc}} > t_{\text{critical}}\)

Critical value \(t_{0.05} = 1.711\)

Since \(2.31 > 1.711\), we reject \(H_0\).

Conclusion: There is sufficient evidence at 5% significance level to conclude that \(\mu > 12\).

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Questions covering shoe manufacturer trials, iron rod thickness, sales campaign analysis, copper wire tensile strength, and more hypothesis testing problems — with complete step-by-step solutions included in the Question Bank.

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Exam Tips for Unit 5 — Inferential Statistics

(Common mistakes examiners flag every year)

✅ Tip 1

Always use \(n-1\) (not \(n\)) in the denominator when calculating sample standard deviation. The most common mistake in board exams is dividing by \(n\) instead of \(n-1\) when computing \(s\). This is because we estimate the population mean (\(\mu\)) using \(\bar{x}\), which reduces the degrees of freedom by 1. This applies both for point estimation and for the t-test statistic calculation.

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❓ Frequently Asked Questions

A parameter is a numerical value describing a characteristic of a population — for example, population mean \(\mu\) and population standard deviation \(\sigma\). Parameters are usually unknown. A statistic is computed from sample data — for example, sample mean \(\bar{x}\) and sample standard deviation \(s\). We use statistics as estimates (point estimates) of the unknown population parameters.

A confidence interval is a range of values \((\bar{x} - E,\; \bar{x} + E)\) that is likely to contain the true population parameter. A 95% confidence interval means: if we repeated the experiment many times, about 95% of the resulting intervals would contain the true \(\mu\). It does not mean there is a 95% probability that \(\mu\) lies in this particular interval — the interval is random, \(\mu\) is fixed.

Use a t-test when: (1) the population standard deviation \(\sigma\) is unknown, and (2) the sample size is small (typically \(n < 30\)). In this case, we use \(s\) (sample SD) instead of \(\sigma\), and the test statistic follows a t-distribution with \(\nu = n-1\) degrees of freedom. Use a z-test when \(\sigma\) is known, or when \(n\) is large enough (by CLT, \(\bar{x}\) is approximately normal).

The one-sample t-test statistic is: \[t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}\] where \(\bar{x}\) = sample mean, \(\mu_0\) = hypothesised population mean, \(s\) = sample standard deviation, \(n\) = sample size. Degrees of freedom: \(\nu = n - 1\). Compare \(|t_{\text{calc}}|\) with \(t_{\text{critical}}\) from the t-table.

A Type I error (\(\alpha\) error) occurs when we reject a true null hypothesis — also called a false positive. The probability of a Type I error is the significance level \(\alpha\) (e.g., 0.05 or 5%). A Type II error (\(\beta\) error) occurs when we fail to reject a false null hypothesis — a false negative. In board problems, focus on Type I error and significance level; Type II error is conceptual.

Unit 5: Inferential Statistics carries 10 marks in the Class 12 Applied Mathematics CBSE board examination. Given this weightage, questions from this unit typically appear as 1–2 MCQs (1 mark each), 1 short-answer question (2–3 marks), and 1 long-answer question (4–5 marks).

A two-tailed test is used when the alternative hypothesis states \(H_1: \mu \neq \mu_0\) (the mean could be either higher or lower). Rejection region is split equally on both sides. A one-tailed test is used when \(H_1: \mu > \mu_0\) (right-tailed) or \(H_1: \mu < \mu_0\) (left-tailed). Always check the direction of \(H_1\) before choosing the critical value from the t-table.