Unit 5: Inferential Statistics – Class 12 Applied Maths | Free MCQs & Solutions | Boundless Maths
📊 Unit 5: Inferential Statistics
Class 12 Applied Mathematics — Point Estimation, Confidence Intervals & One-Sample t-Test
📌 Weightage: 5 Marks in Board Exam
Class 12 Applied Maths Unit 5 — Inferential Statistics Study Material
This page covers all topics in Unit 5 of CBSE Class 12 Applied Mathematics — carrying 5 marks in the board exam.
You'll find 14 MCQs and Assertion-Reason questions with detailed explanations and 12 step-by-step solved short-answer questions
on Point Estimation, Confidence Intervals (95% and 99%), and One-Sample t-Test.
All content is aligned to the Class 12 Applied Maths CBSE Syllabus for 2026–27.
Point EstimationConfidence IntervalsHypothesis TestingOne-Sample t-Testt-DistributionAssertion-ReasonCBSE 2026-27
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Board Exam Marks
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Key Topics
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MCQs & AR Questions
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Short Answer Questions
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Topics Covered in Unit 5: Inferential Statistics
Master these 3 key topics and their formulas for the board exam
📈 Point Estimation
Estimate unknown population parameters — mean and standard deviation — directly from sample data.
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 an interval estimate for the population mean at a specified confidence level (95% or 99%).
Practice MCQs with Answers — Unit 5 Inferential Statistics
Attempt each question, then click Show Answer to reveal the explanation
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
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, and analysing relationships.
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
Population mean is denoted by \(\mu\), not \(\bar{x}\) — \(\bar{x}\) is the sample mean → Statement I is FALSE.
Population mean is a parameter (describes the whole population), not a statistic → Statement II is FALSE.
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)
The one-sample t-test statistic 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 the standard error of the mean.
Question 4
For a t-test, a random sample of size \(n = 2025\) is drawn from a normal population. The degrees of freedom \((\nu)\) are:
a\(2025^{2025}\)
b\(2024^{2025}\)
c\(2025\)
d\(2024\)
✓ Correct Answer: (d) 2024
Degrees of freedom for a one-sample t-test: \(\nu = n - 1\).
Given \(n = 2025\): \(\nu = 2025 - 1 = \mathbf{2024}\).
We lose one degree of freedom because we estimate \(\mu\) using \(\bar{x}\) when computing \(s\).
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)
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 either reject or fail to reject \(H_0\) based on the test statistic compared to the critical value.
Question 6
A grain wholesaler takes a handful of rice from different sacks to inspect quality. The handful taken from a sack is a:
aStatistic
bPopulation
cParameter
dSample
✓ Correct Answer: (d) Sample
A sample is a subset of the population selected for analysis.
Here the population = all rice in the sack, and the sample = the handful taken for inspection.
The wholesaler uses the sample to make inferences about the quality of the entire sack.
Question 7
A population consists of four observations 1, 3, 5, 7. What is the variance?
a2
b4
c5
d6
✓ Correct Answer: (c) 5
Population mean: \(\mu = \dfrac{1+3+5+7}{4} = 4\)
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: This is a population, so 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)
As sample size \(n\) increases, \(\bar{x}\) becomes a more accurate estimate of \(\mu\), so \(|\bar{x} - \mu|\) decreases — this is the Law of Large Numbers. The standard error \(\sigma/\sqrt{n}\) also shrinks as \(n\) increases.
Question 9
A simple random sample consists of five observations 2, 4, 6, 7, 6. The point estimate of population standard deviation is:
\(H_0: \mu \geq 20\); \(H_1: \mu < 20\). A sample of 64 gives \(\bar{x} = 19.5\), population \(\sigma = 2\). The test statistic is:
a\(-2.5\)
b\(-2\)
c\(2\)
d\(-1.5\)
✓ Correct Answer: (b) −2
\[t = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} = \frac{19.5 - 20}{2/\sqrt{64}} = \frac{-0.5}{2/8} = \frac{-0.5}{0.25} = \mathbf{-2}\]
Since \(H_1: \mu < 20\), this is a left-tailed test. The negative t-value confirms the sample mean falls below the hypothesised mean.
Question 11
A machine makes car wheels; in a sample of 26 wheels the test statistic is 3.07. At 5% significance level, the quality of wheels is? [Use \(t_{25}(0.05) = 2.06\)]
aSuperior quality
bInferior quality
cSame quality
dCannot say
✓ Correct Answer: (a) Superior quality
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 \(H_0\) (assumed equal/standard quality).
The positive t-value indicates the sample mean exceeds the standard → superior quality.
Question 12
The Central Limit Theorem states that the sampling distribution of sample mean approaches normal 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
The Central Limit Theorem (CLT) states that when \(n\) is sufficiently large (typically \(n \geq 30\)), the sampling distribution of \(\bar{x}\) becomes approximately normal regardless of the original population distribution — which 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
Checking R:
R uses \(n\) in the denominator — this is the population standard deviation formula. The correct sample standard deviation formula divides by \((n-1)\). → R 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
A is TRUE — a one-sample t-test compares \(\bar{x}\) from the sample to a known/hypothesised \(\mu\). R is TRUE and explains A — we use the t-distribution (not z) precisely because \(\sigma\) is unknown and we estimate it using \(s\), especially when \(n\) is small.
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Short Answer Questions with Step-by-Step Solutions
Attempt each question, then click Show Solution to reveal full working
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?
(i) Point estimate of population mean \(= \mathbf{9}\)
(ii) Point estimate of population standard deviation \(\approx \mathbf{3.1}\)
Question 2
A student observes boiling temperatures (°C): 102.5, 101.7, 103.1, 100.9, 100.5, 102.2 on 6 samples. If the procedure standard deviation is 1.2°C, find the 95% confidence interval for the population mean.
A random sample of 10 students has mean height 5.6 ft with SD 0.75 ft. Find:
(i) 95% confidence limits (ii) 99% confidence limits for average height of all students.
[Use \(t_9(0.05) = 2.262\); \(t_9(0.01) = 3.250\)]
Solution:
Given: \(\bar{x} = 5.6,\; s = 0.75,\; n = 10,\; \nu = 10-1 = 9\)
Standard error: \(\dfrac{s}{\sqrt{n}} = \dfrac{0.75}{\sqrt{10}} \approx 0.237\)
(i) 95% Confidence Limits
\(E = 2.262 \times 0.237 \approx 0.536\)
CI: \(\mu = 5.6 \pm 0.536\)
(ii) 99% Confidence Limits
\(E = 3.250 \times 0.237 \approx 0.770\)
CI: \(\mu = 5.6 \pm 0.770\)
(i) 95% confidence limits: (5.064, 6.136) ft
(ii) 99% confidence limits: (4.830, 6.370) ft
Question 4
\(H_0: \mu \leq 12\); \(H_a: \mu > 12\). A sample of 25 gives \(\bar{x} = 14\), \(s = 4.32\).
(i) Compute the test statistic. (ii) State the rejection rule and your conclusion. [Given \(t_{24}(0.05) = 1.711\)]
Solution:
Given: \(\bar{x} = 14,\; \mu_0 = 12,\; s = 4.32,\; n = 25\). Right-tailed test. \(\nu = 24\).
For a right-tailed test: Reject \(H_0\) if \(t_{\text{calc}} > t_{\text{critical}}\).
Since \(2.31 > 1.711\), we reject \(H_0\).
Conclusion: There is sufficient evidence at 5% significance level to conclude that \(\mu > 12\).
📚 More Detailed Solutions Available in the Question Bank
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✅ All 8 Units covered
✅ MCQs, Assertion-Reason, Unit-wise CBSE Sample Paper & Previous Year Questions with Solutions
Common mistakes examiners flag every year in CBSE Class 12 Applied Maths
✅ Tip 1
Always use \(n-1\) in the denominator when computing sample standard deviation. The most common board exam mistake is dividing by \(n\) instead of \((n-1)\) when computing \(s\). We lose one degree of freedom because we estimate \(\mu\) using \(\bar{x}\). This applies for both point estimation questions and the t-test statistic calculation — always divide by \(n-1\) for samples.
🔒 More exam tips for Unit 5 — including how to identify tails, how to write conclusions for full method marks, and the most common t-table lookup errors — are in the Question Bank.
Answers to questions students commonly ask about Class 12 Applied Maths Unit 5
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 point estimates of the unknown population parameters.
A confidence interval is a range \((\bar{x} - E,\; \bar{x} + E)\) likely to contain the true population parameter. A 95% CI means: if we repeated the experiment many times, about 95% of resulting intervals would contain the true \(\mu\). The interval is built as: \(\mu = \bar{x} \pm Z_{\alpha/2} \cdot \dfrac{\sigma}{\sqrt{n}}\), where \(Z = 1.96\) for 95% and \(Z = 2.576\) for 99%.
Use a t-test when: (1) the population standard deviation \(\sigma\) is unknown, and (2) the sample size is small (typically \(n < 30\)). We substitute \(s\) for \(\sigma\), 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 that the CLT applies.
\[t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}\]
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 (\(\alpha\)) occurs when we reject a true null hypothesis — a false positive. The probability of a Type I error equals the significance level \(\alpha\) (e.g., 5%).
A Type II error (\(\beta\)) 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 is mostly a conceptual question.
Unit 5: Inferential Statistics carries 10 marks in the Class 12 Applied Mathematics CBSE board examination. Questions typically appear as MCQs (1 mark each), short-answer questions (2–3 marks), and a longer problem on confidence intervals or the t-test.
A two-tailed test is used when \(H_1: \mu \neq \mu_0\) — 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 identify the direction of \(H_1\) first, then choose the correct critical value from the t-table.
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