Unit 5 · 5 Marks
Topics & Key Formulas
Three topics examined across MCQs and short-answer questions. Memorise the formulas below — they are directly tested every year.
Class 12 Applied Maths Unit 5 — Inferential Statistics Study Material
Unit 5 carries 5 marks in the CBSE Class 12 Applied Mathematics board exam. This page covers all three topics: Point Estimation, Confidence Intervals, and One-Sample t-Test.
You'll find 14 interactive MCQs and Assertion-Reason questions with complete explanations, plus 12 short-answer solved examples — all aligned to CBSE 2026-27.
Point EstimationConfidence IntervalsOne-Sample t-TestHypothesis Testingt-DistributionCBSE 2026–27
1. Point Estimation
Estimate unknown population parameters — mean and standard deviation — directly from sample data.
\(\bar{x} = \dfrac{\sum x_i}{n}\)
Sample mean = point estimate of μ
Sample mean = point estimate of μ
\(s = \sqrt{\dfrac{\sum(x_i-\bar{x})^2}{n-1}}\)
Sample SD = point estimate of σ
Sample SD = point estimate of σ
2. Confidence Intervals
Build a range estimate for the population mean at a specified confidence level (95% or 99%).
\(\mu = \bar{x} \pm Z_{\alpha/2} \cdot \dfrac{\sigma}{\sqrt{n}}\)
95%: Z = 1.96 | 99%: Z = 2.576
95%: Z = 1.96 | 99%: Z = 2.576
3. One-Sample t-Test
Test hypotheses about a population mean when σ is unknown, using t-distribution with ν = n − 1 degrees of freedom.
\(t = \dfrac{\bar{x} - \mu_0}{s/\sqrt{n}},\quad \nu = n-1\)
Reject H₀ if |t_calc| > t_critical
Reject H₀ if |t_calc| > t_critical
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Question 1Point Estimation
Which of the following is not part of the process of inferential statistics?
Inferential statistics draws conclusions about a population from sample data. Its core activities are estimating population parameters, testing hypotheses, and analysing relationships.
A statistic is already computed directly from the sample — it is already known. So estimating a statistic is not part of inferential statistics.
Key distinction: Parameter (population, unknown) vs Statistic (sample, known).
A statistic is already computed directly from the sample — it is already known. So estimating a statistic is not part of inferential statistics.
Key distinction: Parameter (population, unknown) vs Statistic (sample, known).
Question 2Point Estimation
Statement I: The mean of a population is denoted by \(\bar{x}\).
Statement II: The population mean is a statistic.
Which of the following is true?
Statement II: The population mean is a statistic.
Which of the following is true?
Statement I — FALSE: Population mean is denoted by \(\mu\), not \(\bar{x}\). \(\bar{x}\) is the sample mean.
Statement II — FALSE: Population mean \(\mu\) is a parameter (describes the whole population), not a statistic. Statistics describe samples.
Both statements are false → answer is Neither I nor II.
Statement II — FALSE: Population mean \(\mu\) is a parameter (describes the whole population), not a statistic. Statistics describe samples.
Both statements are false → answer is Neither I nor II.
Question 3t-Test
What is the test statistic for a one-sample t-test?
\[t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\]
where \(\bar{x}\) = sample mean, \(\mu\) = hypothesised population mean, \(s\) = sample SD, \(n\) = sample size.
The denominator \(s/\sqrt{n}\) is the standard error of the mean. Note \(\sqrt{n}\) in the denominator — not \(n\) or \(n^2\).
The denominator \(s/\sqrt{n}\) is the standard error of the mean. Note \(\sqrt{n}\) in the denominator — not \(n\) or \(n^2\).
Question 4t-Test
For a t-test, a random sample of size \(n = 2026\) is drawn. What are the degrees of freedom \((\nu)\)?
Formula: Degrees of freedom for a one-sample t-test: \(\nu = n - 1\)
Given \(n = 2026\): \(\nu = 2026 - 1 = \mathbf{2025}\)
Why n − 1? We lose one degree of freedom because we estimate \(\mu\) using \(\bar{x}\) when computing the sample standard deviation \(s\).
Given \(n = 2026\): \(\nu = 2026 - 1 = \mathbf{2025}\)
Why n − 1? We lose one degree of freedom because we estimate \(\mu\) using \(\bar{x}\) when computing the sample standard deviation \(s\).
Question 5Hypothesis Testing
What is the purpose of a null hypothesis?
The null hypothesis \(H_0\) is a statement of no effect, no difference, or equality — the default assumption we attempt to disprove using sample evidence.
We either reject \(H_0\) (evidence is strong enough) or fail to reject \(H_0\) (evidence is insufficient), based on comparing the test statistic to the critical value.
We either reject \(H_0\) (evidence is strong enough) or fail to reject \(H_0\) (evidence is insufficient), based on comparing the test statistic to the critical value.
Question 6Point Estimation
A grain wholesaler takes a handful of rice from different sacks to inspect quality. What is the handful taken from a sack called?
Population = all rice in the sack.
Sample = the handful taken for inspection → this is the answer.
The wholesaler uses information from the sample to make inferences about the quality of the entire sack (the population).
Sample = the handful taken for inspection → this is the answer.
The wholesaler uses information from the sample to make inferences about the quality of the entire sack (the population).
Question 7Point Estimation
A population consists of four observations 1, 3, 5, 7. What is the variance?
Step 1 — Population mean: \(\mu = \dfrac{1+3+5+7}{4} = 4\)
Step 2 — Squared deviations:
\((1-4)^2 = 9,\quad (3-4)^2 = 1,\quad (5-4)^2 = 1,\quad (7-4)^2 = 9\)
Step 3 — Population variance (divide by N, not N−1):
\(\sigma^2 = \dfrac{9+1+1+9}{4} = \dfrac{20}{4} = \mathbf{5}\)
Key: For a population divide by \(N\). For a sample divide by \(n-1\).
Step 2 — Squared deviations:
\((1-4)^2 = 9,\quad (3-4)^2 = 1,\quad (5-4)^2 = 1,\quad (7-4)^2 = 9\)
Step 3 — Population variance (divide by N, not N−1):
\(\sigma^2 = \dfrac{9+1+1+9}{4} = \dfrac{20}{4} = \mathbf{5}\)
Key: For a population divide by \(N\). For a sample divide by \(n-1\).
Question 8Point Estimation
Which is the true relation between sample mean \(\bar{x}\) and population mean \(\mu\)?
As sample size \(n\) increases, \(\bar{x}\) becomes a more accurate estimate of \(\mu\) — this is the Law of Large Numbers.
The standard error \(\dfrac{\sigma}{\sqrt{n}}\) also decreases as \(n\) increases, so the sampling distribution of \(\bar{x}\) concentrates around \(\mu\), meaning \(|\bar{x} - \mu|\) decreases.
The standard error \(\dfrac{\sigma}{\sqrt{n}}\) also decreases as \(n\) increases, so the sampling distribution of \(\bar{x}\) concentrates around \(\mu\), meaning \(|\bar{x} - \mu|\) decreases.
Question 9Point Estimation
A simple random sample of 2, 4, 6, 7, 6 is drawn. What is the point estimate of the population standard deviation?
Step 1 — Sample mean: \(\bar{x} = \dfrac{2+4+6+7+6}{5} = 5\)
Step 2 — Squared deviations:
\((2-5)^2 = 9,\; (4-5)^2 = 1,\; (6-5)^2 = 1,\; (7-5)^2 = 4,\; (6-5)^2 = 1\)
Sum \(= 16\)
Step 3 — Sample SD (divide by n−1 = 4):
\(s = \sqrt{\dfrac{16}{4}} = \sqrt{4} = \mathbf{2}\)
Step 2 — Squared deviations:
\((2-5)^2 = 9,\; (4-5)^2 = 1,\; (6-5)^2 = 1,\; (7-5)^2 = 4,\; (6-5)^2 = 1\)
Sum \(= 16\)
Step 3 — Sample SD (divide by n−1 = 4):
\(s = \sqrt{\dfrac{16}{4}} = \sqrt{4} = \mathbf{2}\)
Question 10t-Test
Given \(H_0: \mu \geq 20\) and \(H_1: \mu < 20\), a sample of 64 gives \(\bar{x} = 19.5\) with population \(\sigma = 2\). What is the test statistic?
\[t = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} = \frac{19.5 - 20}{2/\sqrt{64}} = \frac{-0.5}{0.25} = \mathbf{-2}\]
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 11t-Test
A machine makes car wheels. In a sample of 26 wheels, the t-test statistic is 3.07. At the 5% significance level, what can be concluded about the quality of the wheels? [Given \(t_{25}(0.05) = 2.06\)]
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 standard/equal quality).
The positive t-value indicates the sample mean exceeds the standard → Superior quality.
Calculated \(|t| = 3.07\); Critical value \(= t_{25}(0.05) = 2.06\)
Since \(3.07 > 2.06\), we reject \(H_0\) (assumed standard/equal quality).
The positive t-value indicates the sample mean exceeds the standard → Superior quality.
Question 12Confidence Intervals
Under what condition does the sampling distribution of the sample mean, as stated by the Central Limit Theorem, approach a normal distribution?
The Central Limit Theorem (CLT) states that when \(n\) is sufficiently large (typically \(n \geq 30\)), the sampling distribution of \(\bar{x}\) is approximately normal, regardless of the original population's distribution.
This is why large-sample inference uses the standard normal (z) distribution — the CLT guarantees approximate normality.
This is why large-sample inference uses the standard normal (z) distribution — the CLT guarantees approximate normality.
📋 Assertion-Reason Questions (Q13–Q14)
- (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
Q13 — Assertion-ReasonPoint Estimation
Assertion (A): For the simple random sample 2, 4, 6, 8, 10, the point estimate of the population standard deviation is \(\sqrt{10}\).
Reason (R): Sample standard deviation of \(n\) observations: \(s = \sqrt{\dfrac{\sum(x_i - \bar{x})^2}{n}}\)
Reason (R): Sample standard deviation of \(n\) observations: \(s = \sqrt{\dfrac{\sum(x_i - \bar{x})^2}{n}}\)
Checking A:
\(\bar{x} = \dfrac{2+4+6+8+10}{5} = 6\)
\(\sum(x_i-\bar{x})^2 = 16+4+0+4+16 = 40\)
\(s = \sqrt{\dfrac{40}{5-1}} = \sqrt{10}\) → A is TRUE ✓
Checking R:
R uses \(n\) in the denominator — this is the population standard deviation formula. The correct sample SD formula divides by \((n-1)\), not \(n\) → R is FALSE ✗
Answer: (c) A is True but R is False.
\(\bar{x} = \dfrac{2+4+6+8+10}{5} = 6\)
\(\sum(x_i-\bar{x})^2 = 16+4+0+4+16 = 40\)
\(s = \sqrt{\dfrac{40}{5-1}} = \sqrt{10}\) → A is TRUE ✓
Checking R:
R uses \(n\) in the denominator — this is the population standard deviation formula. The correct sample SD formula divides by \((n-1)\), not \(n\) → R is FALSE ✗
Answer: (c) A is True but R is False.
Q14 — Assertion-Reasont-Test
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.
Reason (R): The population standard deviation is unknown and the sample size is small.
A is TRUE: A one-sample t-test compares \(\bar{x}\) from the sample to a 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. Both conditions in R — unknown \(\sigma\) and small \(n\) — are the exact reasons we choose a t-test over a z-test.
Answer: (a) Both true; R correctly explains A.
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. Both conditions in R — unknown \(\sigma\) and small \(n\) — are the exact reasons we choose a t-test over a z-test.
Answer: (a) Both true; R correctly explains A.
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Question 1Point Estimation
A sample is drawn: 5, 8, 10, 7, 10, 14. Find (i) the point estimate of the population mean, and (ii) the point estimate of the population standard deviation.
Given: \(n = 6\), data = 5, 8, 10, 7, 10, 14
(i) Point estimate of population mean = sample mean \(\bar{x}\)
\(\bar{x} = \dfrac{5+8+10+7+10+14}{6} = \dfrac{54}{6} = 9\)
(ii) Point estimate of population SD = sample SD \(s\)
Deviations from \(\bar{x}=9\):
\((5-9)^2=16,\;(8-9)^2=1,\;(10-9)^2=1,\;(7-9)^2=4,\;(10-9)^2=1,\;(14-9)^2=25\)
Sum \(= 48\)
\((5-9)^2=16,\;(8-9)^2=1,\;(10-9)^2=1,\;(7-9)^2=4,\;(10-9)^2=1,\;(14-9)^2=25\)
Sum \(= 48\)
\(s = \sqrt{\dfrac{48}{5}} = \sqrt{9.6} \approx 3.10\)
(i) Point estimate of \(\mu = \mathbf{9}\) | (ii) Point estimate of \(\sigma \approx \mathbf{3.10}\)
Question 2Confidence Intervals
The boiling temperatures (°C) recorded for 6 samples are 102.5, 101.7, 103.1, 100.9, 100.5, and 102.2, with a procedure standard deviation of 1.2°C. Find the 95% confidence interval for the population mean.
Given: \(n=6,\; \sigma=1.2\)
\(\bar{x} = \dfrac{102.5+101.7+103.1+100.9+100.5+102.2}{6} = \dfrac{610.9}{6} \approx 101.82\)
For 95% CI: \(Z_{0.025} = 1.96\)
Margin of error: \(E = 1.96 \times \dfrac{1.2}{\sqrt{6}} = 1.96 \times 0.490 \approx 0.96\)
CI: \(\mu = 101.82 \pm 0.96\)
95% Confidence Interval: (100.86, 102.78)°C
Question 3Confidence Intervals
A random sample of 10 students has a mean height of 5.6 ft and a standard deviation of 0.75 ft. Find (i) the 95% confidence limits, and (ii) the 99% confidence limits for the average height of all students.
[Use \(t_9(0.05) = 2.262\); \(t_9(0.01) = 3.250\)]
[Use \(t_9(0.05) = 2.262\); \(t_9(0.01) = 3.250\)]
Given: \(\bar{x}=5.6,\; s=0.75,\; n=10,\; \nu=9\)
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: \(5.6 \pm 0.536\)
(ii) 99% Confidence Limits
\(E = 3.250 \times 0.237 \approx 0.770\) → CI: \(5.6 \pm 0.770\)
(i) 95% CI: (5.064, 6.136) ft
(ii) 99% CI: (4.830, 6.370) ft
Note: the 99% interval is wider — more confidence requires a wider range.
(ii) 99% CI: (4.830, 6.370) ft
Note: the 99% interval is wider — more confidence requires a wider range.
Question 4t-Test
For \(H_0: \mu \leq 12\) and \(H_a: \mu > 12\), a sample of 25 gives \(\bar{x} = 14\) and \(s = 4.32\). (i) Compute the test statistic. (ii) State the rejection rule and conclusion. [Given \(t_{24}(0.05) = 1.711\)]
Given: \(\bar{x}=14,\; \mu_0=12,\; s=4.32,\; n=25,\; \nu=24\). Right-tailed test.
(i) Test Statistic
\[t = \frac{14-12}{4.32/\sqrt{25}} = \frac{2}{0.864} \approx 2.31\]
(ii) Rejection Rule and Conclusion
For a right-tailed test: Reject \(H_0\) if \(t_{\text{calc}} > t_{\text{critical}} = 1.711\).
Since \(2.31 > 1.711\), we reject \(H_0\).
Conclusion: There is sufficient evidence at the 5% significance level to conclude that \(\mu > 12\).
Question 5Point Estimation
A sample is drawn: 3, 6, 9, 12, 15. Find (i) the point estimate of \(\mu\), and (ii) the point estimate of \(\sigma\).
\(n=5,\; \bar{x} = \dfrac{3+6+9+12+15}{5} = 9\)
Squared deviations: \((3-9)^2=36,\;(6-9)^2=9,\;(9-9)^2=0,\;(12-9)^2=9,\;(15-9)^2=36\). Sum \(= 90\)
\(s = \sqrt{\dfrac{90}{4}} = \sqrt{22.5} \approx 4.74\)
(i) \(\hat{\mu} = \bar{x} = \mathbf{9}\) | (ii) \(\hat{\sigma} = s \approx \mathbf{4.74}\)
Question 6Confidence Intervals
A sample of 16 observations has mean 41.5 and SD 2.5. Construct a 99% confidence interval for the population mean. [\(t_{15}(0.01) = 2.947\)]
Given: \(\bar{x}=41.5,\; s=2.5,\; n=16,\; \nu=15\)
Standard error: \(\dfrac{s}{\sqrt{n}} = \dfrac{2.5}{4} = 0.625\)
\(E = 2.947 \times 0.625 \approx 1.842\)
CI: \(\mu = 41.5 \pm 1.842\)
99% Confidence Interval: (39.66, 43.34)
Question 7t-Test
A shoe manufacturer claims the average shoe size is 7. A sample of 12 shoes gives \(\bar{x}=7.5\), \(s=1.2\). Test the claim at the 5% level. [\(t_{11}(0.05) = 2.201\)]
\(H_0: \mu = 7\) vs \(H_1: \mu \neq 7\) (two-tailed). Given: \(\bar{x}=7.5,\; s=1.2,\; n=12,\; \nu=11\).
\(t = \dfrac{7.5 - 7}{1.2/\sqrt{12}} = \dfrac{0.5}{0.346} \approx 1.44\)
Critical value (two-tailed, 5%): \(t_{11}(0.05) = 2.201\)
Since \(|1.44| < 2.201\), we fail to reject \(H_0\).
Conclusion: There is insufficient evidence to reject the manufacturer's claim at the 5% level.
Question 8t-Test
Standard iron rods should be 5 cm. A sample of 9 rods gives \(\bar{x}=4.85\) cm, \(s=0.30\) cm. Test whether the production is satisfactory at 5% significance. [\(t_8(0.05) = 2.306\)]
\(H_0: \mu = 5\) vs \(H_1: \mu \neq 5\) (two-tailed). \(\bar{x}=4.85,\; s=0.30,\; n=9,\; \nu=8\).
\(t = \dfrac{4.85 - 5}{0.30/\sqrt{9}} = \dfrac{-0.15}{0.10} = -1.5\)
\(|t| = 1.5 < 2.306\) → Fail to reject \(H_0\).
Conclusion: Production is satisfactory — the mean rod length does not significantly differ from 5 cm at the 5% level.
Question 9t-Test
Before a sales campaign, monthly sales (₹ lakhs) averaged 250. After the campaign, a sample of 10 months gives \(\bar{x}=268\) and \(s=30\). Did the campaign increase sales? Test at 5%. [\(t_9(0.05)=1.833\) for one-tailed]
\(H_0: \mu \leq 250\) vs \(H_1: \mu > 250\) (right-tailed). \(\bar{x}=268,\;\mu_0=250,\; s=30,\; n=10,\; \nu=9\).
\(t = \dfrac{268-250}{30/\sqrt{10}} = \dfrac{18}{9.487} \approx 1.897\)
Since \(1.897 > 1.833\), we reject \(H_0\).
Conclusion: The sales campaign significantly increased monthly sales at the 5% significance level.
Question 10t-Test
A copper wire has a claimed tensile strength of \(\mu_0 = 60\) kg. A sample of 9 wires gives \(\bar{x}=57\) kg and \(s=6\) kg. Test the claim at the 1% significance level. [\(t_8(0.01)=3.355\) two-tailed]
\(H_0: \mu=60\) vs \(H_1: \mu \neq 60\). \(\bar{x}=57,\;\mu_0=60,\; s=6,\; n=9,\; \nu=8\).
\(t = \dfrac{57-60}{6/\sqrt{9}} = \dfrac{-3}{2} = -1.5\)
\(|t|=1.5 < 3.355\) → Fail to reject \(H_0\).
Conclusion: The claim of tensile strength 60 kg is acceptable at the 1% significance level.
Question 11t-Test
A manufacturer claims batteries last 200 hours. A random sample of 8 batteries gives \(\bar{x}=195\) hours, \(s=12\) hours. Test at 5% significance. [\(t_7(0.05)=2.365\) two-tailed]
\(H_0: \mu=200\) vs \(H_1: \mu \neq 200\). \(\bar{x}=195,\; s=12,\; n=8,\; \nu=7\).
\(t = \dfrac{195-200}{12/\sqrt{8}} = \dfrac{-5}{4.243} \approx -1.179\)
\(|t|=1.179 < 2.365\) → Fail to reject \(H_0\).
Conclusion: There is insufficient evidence at the 5% level to reject the manufacturer's claim of 200-hour battery life.
Question 12Confidence Intervals
A class of 25 students has mean exam score 72 with SD 8. Construct a 95% confidence interval for the population mean score. [\(t_{24}(0.05)=2.064\)]
Given: \(\bar{x}=72,\; s=8,\; n=25,\; \nu=24\)
Standard error: \(\dfrac{s}{\sqrt{n}} = \dfrac{8}{5} = 1.6\)
\(E = 2.064 \times 1.6 \approx 3.30\)
CI: \(\mu = 72 \pm 3.30\)
95% Confidence Interval for mean score: (68.70, 75.30)
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Exam Tips — Unit 5
How to score full marks in Unit 5 — mistakes to avoid and strategies that work.
✅ 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 t-test calculations. Writing \(n\) in the denominator (the population formula) in a sample context will cost marks.
🔒 More exam tips — identifying the test tail before the critical value, how to write conclusions for full method marks, the most common t-table lookup errors, and how to present confidence interval working — are all in the AI Question Bank. All 8 units covered.
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Frequently Asked Questions
Answers to questions students frequently ask about Class 12 Applied Maths Unit 5.
A parameter describes a characteristic of a population — for example, population mean \(\mu\) and population SD \(\sigma\). Parameters are usually unknown. A statistic is computed from sample data — for example, sample mean \(\bar{x}\) and sample SD \(s\). 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 \(\mu = \bar{x} \pm Z_{\alpha/2} \cdot \dfrac{\sigma}{\sqrt{n}}\), where \(Z=1.96\) for 95% and \(Z=2.576\) for 99%.
Use a t-test when: (1) the population SD \(\sigma\) 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.
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 SD, \(n\) = sample size. Degrees of freedom: \(\nu = n-1\). Compare \(|t_{\text{calc}}|\) with \(t_{\text{critical}}\) from the t-table at the given significance level.
A Type I error (\(\alpha\)) occurs when we reject a true null hypothesis — a false positive. The probability 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.
A Type II error (\(\beta\)) occurs when we fail to reject a false null hypothesis — a false negative.
Unit 5: Inferential Statistics carries 5 marks in the CBSE Class 12 Applied Mathematics board examination. Questions appear as 1-mark MCQs, 2–3 mark short answers 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 higher or lower. Rejection region splits 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.
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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