Unit 5: Inferential Statistics
Master Point and Interval Estimation, One sample t-Tests
10 Marks in Board ExamTopics Covered
📈 Point Estimation
Find Statistical Point estimates of the Population based on the sample
📉 Confidence Intervals
Margin of error, 95% and 99% confidence levels
🔢 Hypothesis Testing, t-Distribution
Null and alternative hypotheses, degrees of freedom, One-sample t-test
Multiple Choice Questions
Question 1
Inferential statistics is a process that involves all of the following except:
✓ Correct Answer: (b) Estimating a statistic
Explanation: Inferential statistics involves estimating population parameters (not statistics), testing hypotheses, and analyzing relationships. Statistics are already known from the sample data.
Question 2
Which of the following statements are true?
Statement I: The mean of a population is denoted by x̄
Statement II: The population mean is a statistic
Statement I: The mean of a population is denoted by x̄
Statement II: The population mean is a statistic
✓ Correct Answer: (d) None
Explanation:
• Population mean is denoted by μ (mu), not x̄ → Statement I is FALSE
• Population mean is a parameter, not a statistic → Statement II is FALSE
Therefore, both statements are false.
• Population mean is denoted by μ (mu), not x̄ → Statement I is FALSE
• Population mean is a parameter, not a statistic → Statement II is FALSE
Therefore, both statements are false.
Question 3
The test statistic for a one-sample t-test, denoted by t, is defined as:
✓ Correct Answer: (a) t = (x̄ - μ) / (s/√n)
Explanation: This is the standard formula for the t-test statistic, where x̄ is the sample mean, μ is the population mean, s is the sample standard deviation, and n is the sample size.
Question 4
For the purpose of t-test of significance, a random sample of size (n) 2025 is drawn from a normal population, then the degree of freedom (v) is:
✓ Correct Answer: (d) 2024
Explanation: For a one-sample t-test, degree of freedom is: v = n - 1
Given: n = 2025
Therefore: v = 2025 - 1 = 2024
Given: n = 2025
Therefore: v = 2025 - 1 = 2024
Question 5
What is the purpose of a null hypothesis?
✓ Correct Answer: (b) To state that there is no significant difference or relationship
Explanation: The null hypothesis (H₀) is a statement that assumes no significant difference or relationship exists. It serves as the baseline for statistical testing.
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 rice taken from a sack for quality inspection is a:
✓ Correct Answer: (d) Sample
Explanation: A sample is a subset of the population selected for inspection or analysis. The handful of rice taken from the sack represents a sample used to make inferences about the entire sack (population).
Question 7
A population consists of four observations 1, 3, 5, 7. What is the variance?
✓ Correct Answer: (c) 5
Explanation:
Mean μ = (1 + 3 + 5 + 7)/4 = 4
Variance σ² = Σ(x - μ)²/N
Squared deviations: (1-4)² = 9, (3-4)² = 1, (5-4)² = 1, (7-4)² = 9
Variance = (9 + 1 + 1 + 9)/4 = 20/4 = 5
Mean μ = (1 + 3 + 5 + 7)/4 = 4
Variance σ² = Σ(x - μ)²/N
Squared deviations: (1-4)² = 9, (3-4)² = 1, (5-4)² = 1, (7-4)² = 9
Variance = (9 + 1 + 1 + 9)/4 = 20/4 = 5
Question 8
CUET 2022
Which of the following is true relation between sample mean (x̄) and population mean (μ)?
✓ Correct Answer: (d) |x̄ - μ| decreases when increases the size of sample
Explanation: As sample size increases, the sample mean becomes closer to the population mean. Hence, |x̄ - μ| decreases with increase in sample size. This is a fundamental principle of the Law of Large Numbers.
Question 9
CUET 2023
A simple random sample consists of five observations 2, 4, 6, 7, 6. The point estimate of population standard deviation is:
✓ Correct Answer: (d) 2
Explanation: Point estimate of population standard deviation = sample standard deviation. Calculating the sample standard deviation for the given data (2, 4, 6, 7, 6) gives a value ≈ 2.
Question 10
CUET 2023
Consider the following hypothesis test: H₀ : μ ≥ 20 ; H₁ : μ < 20
A sample of 64 provided a sample mean of 19.5. The population standard deviation is 2.
The value of test statistic is:
A sample of 64 provided a sample mean of 19.5. The population standard deviation is 2.
The value of test statistic is:
✓ Correct Answer: (b) -2
Explanation:
t = (x̄ - μ) / (σ/√n)
t = (19.5 - 20) / (2/√64)
t = -0.5 / 0.25 = -2
t = (x̄ - μ) / (σ/√n)
t = (19.5 - 20) / (2/√64)
t = -0.5 / 0.25 = -2
Question 11
CBSE Sample Paper 2023
A machine makes car wheels and in a random sample of 26 wheels, the test statistic is found to be 3.07. As per t-distribution test (of 5% level of significance), what can you say about the quality of wheels produced by the machine? (Use t₂₅(0.05) = 2.06)
✓ Correct Answer: (a) Superior quality
Explanation: Calculated |t| = 3.07 > tabulated value 2.06. Since the calculated value exceeds the critical value, the null hypothesis is rejected, indicating significant improvement in quality.
Question 12
The Central Limit Theorem states that the sampling distribution of sample mean approaches a normal distribution if:
✓ Correct Answer: (b) Sample size is large
Explanation: According to the Central Limit Theorem, when the sample size is sufficiently large (typically n ≥ 30), the sampling distribution of the sample mean becomes approximately normal, irrespective of the population distribution.
📋 Assertion-Reason Questions
Statement I is called Assertion (A) and Statement II is called 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 √10
Reason (R): Sample standard deviation of n observations, s = √[Σ(xᵢ - x̄)² / n]
Reason (R): Sample standard deviation of n observations, s = √[Σ(xᵢ - x̄)² / n]
✓ Correct Answer: (c) A is True but R is False
Explanation:
Mean, x̄ = (2 + 4 + 6 + 8 + 10)/5 = 30/5 = 6
Σ(xᵢ - x̄)² = (2-6)² + (4-6)² + (6-6)² + (8-6)² + (10-6)² = 16 + 4 + 0 + 4 + 16 = 40
Sample standard deviation: s = √[40/(5-1)] = √10 → Assertion is TRUE
However, the formula in Reason is incorrect. The correct formula is: s = √[Σ(xᵢ - x̄)²/(n-1)]
Therefore, Reason is FALSE.
Mean, x̄ = (2 + 4 + 6 + 8 + 10)/5 = 30/5 = 6
Σ(xᵢ - x̄)² = (2-6)² + (4-6)² + (6-6)² + (8-6)² + (10-6)² = 16 + 4 + 0 + 4 + 16 = 40
Sample standard deviation: s = √[40/(5-1)] = √10 → Assertion is TRUE
However, the formula in Reason is incorrect. The correct formula is: s = √[Σ(xᵢ - x̄)²/(n-1)]
Therefore, 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.
Reason (R): The population standard deviation is unknown, and the sample size is small.
✓ Correct Answer: (a) Both A and R are True and R is the correct explanation of A
Explanation: The assertion is true - a one-sample t-test compares the sample mean to a known population mean. The reason is also true and correctly explains why we use the t-test: when the population standard deviation is unknown and sample size is small, we use the t-distribution instead of the z-distribution.
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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?
(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 = sample mean
x̄ = Σxᵢ/n = (5 + 8 + 10 + 7 + 10 + 14)/6
x̄ = 54/6 = 9
(ii) Point estimate of population standard deviation = sample standard deviation
s = √[Σ(xᵢ - x̄)²/(n-1)]
Σ(xᵢ - x̄)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 = √(48/5) = √9.6 = 3.1
Answers:
(i) Point estimate of population mean = 9
(ii) Point estimate of population standard deviation = 3.1
(i) Point estimate of population mean = 9
(ii) Point estimate of population standard deviation = 3.1
Question 2
Suppose a student measuring the boiling temperature of a certain liquid observes the reading (in degree Celsius) 102.5, 101.7, 103.1, 100.9, 100.5 and 102.2 on 6 different samples of the liquid. If he knows that the standard deviation for this procedure is 1.2°C, what is the interval estimation for the population mean at a 95% confidence level?
Solution:
Given: n = 6, σ = 1.2, confidence level = 95%
Observations: 102.5, 101.7, 103.1, 100.9, 100.5, 102.2
Sample mean: x̄ = (102.5 + 101.7 + 103.1 + 100.9 + 100.5 + 102.2)/6 = 610.9/6 = 101.82
Confidence level = 95% ⇒ (1 - α) = 0.95 ⇒ α = 0.05 ⇒ α/2 = 0.025
Z₀.₀₂₅ = 1.96 (from standard normal table)
Margin of error = Z₀.₀₂₅ × (σ/√n) = 1.96 × (1.2/√6) = 1.96 × 0.49 = 0.96
μ = x̄ ± margin of error = 101.82 ± 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 have 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 t₉(0.05) = 2.262 ; t₉(0.01) = 3.250]
(i) 95% confidence limits
(ii) 99% confidence limits for the average height of all students
[Use t₉(0.05) = 2.262 ; t₉(0.01) = 3.250]
Solution:
Given: x̄ = 5.6, s = 0.75, n = 10
Standard error: s/√n = 0.75/√10 = 0.75/3.162 = 0.237
(i) 95% Confidence limits
Margin of error: E = t₀.₀₅ × (s/√n) = 2.262 × 0.237 = 0.536
μ = x̄ ± E = 5.6 ± 0.536 = (5.064, 6.136)
(ii) 99% Confidence limits
Margin of error: E = t₀.₀₁ × (s/√n) = 3.250 × 0.237 = 0.770
μ = x̄ ± E = 5.6 ± 0.770 = (4.83, 6.37)
Answers:
(i) 95% confidence limits: (5.064, 6.136) ft
(ii) 99% confidence limits: (4.83, 6.37) ft
(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:
H₀ : μ ≤ 12 ; Hₐ : μ > 12
A sample of 25 provided a sample mean x̄ = 14 and a sample standard deviation s = 4.32
(i) Compute the value of the test statistic
(ii) What is the rejection rule using the critical value? What is your conclusion? [Given t₀.₀₅ = 1.711]
H₀ : μ ≤ 12 ; Hₐ : μ > 12
A sample of 25 provided a sample mean x̄ = 14 and a sample standard deviation s = 4.32
(i) Compute the value of the test statistic
(ii) What is the rejection rule using the critical value? What is your conclusion? [Given t₀.₀₅ = 1.711]
Solution:
Given: x̄ = 14, μ = 12, s = 4.32, n = 25
This is a right-tailed t-test
(i) Test statistic
t = (x̄ - μ)/(s/√n)
t = (14 - 12)/(4.32/√25) = 2/(4.32/5) = 2/0.864 = 2.31
(ii) Rejection Rule and Conclusion
Given: t₀.₀₅ = 1.711
Rejection Rule: As it is right-tailed test, if t_calc > t_critical ⇒ Reject H₀
Calculated value: t = 2.31
Since 2.31 > 1.711
Conclusion: Reject the null hypothesis H₀. There is sufficient evidence that μ > 12.
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