ISS 2016 Statistics Paper-2 Solution: Question 34 (Chi-Square Goodness-of-Fit Test)

Continuing our question-by-question walk through the ISS Statistics Paper-2 previous year papers, we now arrive at Question 34 of the 2016 paper. This one looks deceptively short, but it hides a very common point of confusion for beginners: knowing which statistical test to reach for when you're told a variable "follows" a particular distribution.

Quick Summary

• Topic: Choosing the correct hypothesis test — testing distributional assumptions

• Question reference: ISS 2016, Statistics Paper-2, Question 34

• Correct answer: (b) Chi-square test as a test of goodness of fit

Here is the question exactly as it appears in the paper:

Question 34.The Manager of a cyber café says that the number of customers visiting on the weekdays followed a Binomial distribution. Which one of the following techniques can be used to test the hypothesis at a given level of significance?

• (a) Test of significance of mean

• (b) Chi-square test as a test of goodness of fit

• (c) Test of significance of difference between two means

• (d) Correlation analysis

Step 1: Understand exactly what is being claimed

Read the question slowly. The manager is not making a claim about a single number, like "the average number of customers is 40." He is making a claim about the entire shape of the distribution: he says the whole pattern of customer counts across weekdays behaves like a Binomial random variable.

That distinction — a claim about one parameter (like a mean) versus a claim about an entire distributional form — is exactly what determines which test applies. So before even looking at the four options, we should set up the hypothesis in words:

• H0: The number of customers visiting on weekdays follows a Binomial distribution.

• H1: The number of customers visiting on weekdays does not follow a Binomial distribution.

We need a test that compares an observed pattern of data against an entire hypothesized probability model — not a test that compares just one summary statistic.

Step 2: Set up how such a test would actually work

Suppose the manager recorded, over many weekdays, how many customers walked in (say, out of some fixed number of "trials" or opportunities, consistent with a Binomial setup with parameters n and p). From this data we can build a frequency table: for each possible customer count value i = 0, 1, 2, ..., n, count how many weekdays actually had exactly i customers. Call this the observed frequency, Oi.

Next, under the assumption that H0 is true (the counts really are Binomial(n, p)), we can calculate what frequency we would expect to see in each category. If N is the total number of weekdays observed, the expected frequency for category i is:

Ei = N × P(X = i), where P(X = i) = nCi pi (1−p)n−i

(If p is not given, we estimate it from the sample data itself, usually as p̂ = sample mean / n.)

Step 3: Compare observed versus expected — this is the goodness-of-fit idea

Now we have two columns of numbers side by side: what we actually observed (Oi) and what the Binomial model predicts we should observe (Ei). If the Binomial assumption is correct, these two columns should be close to each other, differing only by chance sampling fluctuation. If the assumption is wrong, the columns should differ by more than chance alone can explain.

The standard way to measure "how far apart are these two columns, overall" is the chi-square statistic:

χ2 = Σ (Oi − Ei)2 / Ei

This statistic, under H0, follows (approximately, for large samples) a chi-square distribution with (k − 1 − m) degrees of freedom, where k is the number of categories used and m is the number of parameters that had to be estimated from the sample data itself (m = 1 here if p was estimated, m = 0 if p was already fully specified by the manager).

We then compare our computed χ2 value to the critical value from chi-square tables at the chosen level of significance α. If our computed value exceeds the critical value, we reject H0 and conclude the data does not really follow a Binomial distribution; otherwise we do not reject the Binomial claim.

This entire procedure — comparing an observed frequency distribution to a hypothesized theoretical distribution using the χ2 statistic — is precisely what is called the chi-square test of goodness of fit.

Step 4: Eliminate the other three options

Now let's go back and check why each wrong option fails to do the job:

• (a) Test of significance of mean:This test (a z-test or t-test on the sample mean) only checks whether the population mean equals some hypothesized value. It says nothing about whether the full shape of the distribution — its skewness, its variance structure, the relative frequency of each possible count — actually matches a Binomial curve. Two very different distributions can share the same mean.

• (c) Test of significance of difference between two means:This compares the means of two separate samples or populations (for example, weekday customers versus weekend customers). Here we only have one claim, about one set of weekday data, being checked against one theoretical model — there is no second group to compare against.

• (d) Correlation analysis:Correlation measures the strength and direction of a linear association between two different variables (like customer count and, say, temperature). There is only one variable being discussed here, and no second variable to correlate it with.

Only option (b) actually tests the thing the manager claimed: that the underlying distribution itself is Binomial.

Pro Tip: The trickiest part of this question isn't spotting that chi-square is involved — most students get that far. It's remembering that the degrees of freedom must be reduced by one for

Final Answer

The correct answer is (b) Chi-square test as a test of goodness of fit. This matches the official answer key for Series A of the 2016 Statistics Paper-2 booklet, where Question 34 is keyed as (b).

Why This Question Matters

This question is a favourite because it tests something more fundamental than a formula: it tests whether you can correctly translate a plain-English claim ("this variable follows distribution X") into the right statistical framework. In real applied statistics — quality control, survey sampling, genetics, insurance claims modelling — you very often need to check whether an assumed distribution (Poisson arrivals, Binomial defect counts, Normal errors) actually matches observed data before you can trust any further analysis built on top of that assumption. The chi-square goodness-of-fit test is the standard first tool reached for in exactly these situations, which is why ISS examiners return to this concept across multiple years in slightly different disguises.

What is the chi-square goodness-of-fit test used for?

It is used to check whether an observed set of data is consistent with a specific hypothesized probability distribution, such as Binomial, Poisson, or Normal. It compares observed frequencies against the frequencies you would expect if the hypothesized distribution were true.

How do you decide the degrees of freedom for this test?

Start with (number of categories − 1), then subtract one further degree of freedom for every parameter of the hypothesized distribution that had to be estimated from the sample data itself, rather than being specified in advance.

Why can't a test on the mean alone answer this kind of question?

A test of significance of the mean only checks one summary number. Two completely different-shaped distributions can share an identical mean, so matching the mean does not prove the whole distribution matches. The goodness-of-fit test checks the entire shape, category by category.

What should you do if an expected frequency in some category is very small?

A common rule of thumb is to pool or merge adjacent categories so that each expected frequency is at least 5. Very small expected frequencies make the chi-square approximation to the test statistic's distribution unreliable.

Is the chi-square goodness-of-fit test the same as the chi-square test of independence?

No, though both use the same underlying chi-square statistic formula. Goodness-of-fit compares one sample's distribution to a single hypothesized theoretical distribution. The test of independence instead uses a two-way contingency table to check whether two categorical variables are associated with each other.

Can this same technique be used if the claimed distribution were Poisson or Normal instead of Binomial?

Yes. The chi-square goodness-of-fit procedure works the same way regardless of which theoretical distribution is being tested — only the formula used to calculate the expected frequencies Ei changes to match the claimed distribution.

If any step above felt unclear, drop your doubt in the comments below — working through exactly where the confusion starts is usually the fastest way to fix it. And if you know a fellow ISS aspirant who's grinding through these same past papers, feel free to share this solution with them too.