ISS 2016 Statistics Paper-1 Solution: Question 32 (Sample Size for Confidence Interval)
This post continues our ongoing question-by-question walkthrough of the ISS Statistics Paper-1 previous year papers. Today we pick up right after Question 31 from the 2016 paper and work through Question 32, which sits in the estimation theory portion of the syllabus and tests how comfortable you are converting a confidence-interval requirement into a minimum sample size.
Quick Summary
Topic: Sample size determination for estimating a population mean at a given confidence level
Question Reference: ISS 2016, Statistics Paper-1, Question 32
Correct Answer: Option (b) n ≥ 78
The Question, As Asked
In measuring reaction time, an experimenter estimates that the standard deviation is 0.09 s. What should be the value of sample size n so that the experimenter is 95% confident that the error of his estimate of the mean reaction time will not exceed 0.02 s?
(a) 0 < n < 50
(b) n ≥ 78
(c) 50 ≤ n ≤ 77
(d) n < 77
Step 1: Identify What Is Given and What Is Asked
Read the question slowly and pull out the numbers before touching any formula. We are told:
The population standard deviation is σ = 0.09 seconds. This is treated as a known constant in the problem, not something we estimate from a sample.
The maximum allowable error in estimating the population mean is E = 0.02 seconds. This is how far, at most, our sample mean is allowed to stray from the true mean.
The confidence level required is 95%.
We are asked to find the smallest sample size n that guarantees this level of precision.
Step 2: Recall Why a Confidence Interval Leads to a Sample Size Formula
When we estimate a population mean μ using a sample mean X̄drawn from a normal population (or a large enough sample, by the Central Limit Theorem), the sample mean itself is approximately normally distributed with mean μ and standard deviation σ/√n. This σ/√n is called the standard error of the mean.
A 95% confidence interval for μ is built as:
X̄± z0.025 × (σ/√n)
Here z0.025 is the value from the standard normal table such that 2.5% of the area lies beyond it on each tail, leaving 95% in the middle. This value is the well-known constant z = 1.96.
The term z × (σ/√n) is exactly the margin of error of our estimate. The question is telling us that this margin of error must not exceed 0.02. So we set up the inequality:
z × (σ/√n) ≤ E
Step 3: Substitute the Known Values
Plugging in z = 1.96, σ = 0.09, and E = 0.02:
1.96 × (0.09/√n) ≤ 0.02
Step 4: Solve the Inequality for n, One Algebra Step at a Time
We now isolate n using ordinary algebra. There is no shortcut here — every line follows directly from the one before it.
Step 4a.Multiply both sides by √n (which is always positive, so the inequality direction does not flip):
1.96 × 0.09 ≤ 0.02 × √n
Step 4b.Compute the left-hand side:
1.96 × 0.09 = 0.1764
So: 0.1764 ≤ 0.02 × √n
Step 4c.Divide both sides by 0.02:
0.1764 / 0.02 ≤ √n
8.82 ≤ √n
Step 4d.Square both sides (again valid since both sides are positive):
8.822 ≤ n
77.7924 ≤ n
Step 5: Round to a Whole Number, Correctly
A sample size has to be a whole number of observations — you cannot collect 77.79 people's reaction times. The value 77.7924 is the bare minimum that makes the inequality true; anything slightly less, like n = 77, would make the margin of error slightly larger than 0.02, which violates the requirement.
Pro Tip: This is the single most common slip in sample-size problems. Students instinctively round 77.79 to the nearest whole number, which gives 78 anyway here, and get lucky. But the correct rule is always to round
Rounding 77.7924 up to the next whole number gives n = 78. Since the requirement was n ≥ 77.7924, and 78 is the smallest integer satisfying it, the correct condition on n is:
n ≥ 78
This matches option (b), and it also matches the official ISS 2016 answer key, which lists B as the correct response for Question 32. Our derivation and the official key are in full agreement.
Why This Question Matters
Sample size determination questions like this one appear almost every year in ISS Paper-1 because they combine three ideas that examiners love testing together: the standard error of the mean, the standard normal z-value for a given confidence level, and careful handling of an inequality with a real-world rounding constraint. Once you have internalized the formula n = (zσ/E)2 and, more importantly, the rounding-up rule, you can solve almost any variant of this question within a minute, whether the paper changes the confidence level to 90% or 99%, or changes which quantity (σ, E, or n) is being solved for. This kind of drilling, done question by question until the pattern becomes automatic, is exactly how it tends to be taught to serious aspirants.
Frequently Asked Questions
What is the general formula used to find sample size for estimating a population mean?
The formula is n = (z × σ / E)2, where z is the standard normal value corresponding to the desired confidence level, σ is the population standard deviation, and E is the maximum allowable error (margin of error) in the estimate.
Why do we always round the computed sample size up rather than to the nearest whole number?
Because the computed value is the minimum n required to keep the error within the stated bound. Rounding down, or rounding to the nearest integer when that integer is smaller than the computed value, would make the margin of error exceed the allowed limit, so you must always round up to the next whole number to stay safely within the requirement.
Where does the value z = 1.96 come from for a 95% confidence level?
For a 95% confidence interval, 5% of the total probability is split equally into the two tails of the standard normal distribution, leaving 2.5% in each tail. The z-value that leaves exactly 2.5% in the upper tail is 1.96, which is why it is used as the multiplier for 95% confidence problems.
How would the required sample size change if the confidence level was raised to 99% instead of 95%?
A higher confidence level requires a larger z-value; for 99% confidence, z = 2.576 instead of 1.96. Since sample size is proportional to the square of z, increasing the confidence level from 95% to 99% would noticeably increase the minimum required sample size, all else staying the same.
Why is the population standard deviation treated as known in this problem instead of being estimated?
The question states that the experimenter estimates the standard deviation to be 0.09 s and asks us to use it directly in a z-based formula, which is the standard convention in this type of sample-size planning question. In practice, this value often comes from a pilot study or prior data, and it is plugged in as if it were the true population standard deviation for the purpose of planning the main study.
What would happen to the answer if the allowable error E was halved, to 0.01 instead of 0.02?
Since n is inversely proportional to E2, halving E would multiply the required sample size by four. Redoing the calculation with E = 0.01 gives n ≥ 311.17, so the minimum required sample size would jump to 312, showing how sensitive sample size is to tightening the precision requirement.
If any step above felt unclear, or if you solved it a different way and want a second opinion, drop a comment below — and if this walkthrough helped, consider sharing it with a fellow ISS aspirant who is also working through the Paper-1 previous year questions.
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