ISS 2016 Statistics Paper-1 Solution: Question 33 (Chi-Square Test for 2x2 Contingency Table)

Continuing our question-by-question walkthrough of the ISS Statistics Paper-1 previous year papers, we now move to Question 33 of the 2016 exam. This one takes us into categorical data analysis — testing whether two attributes are independent using a 2 × 2 contingency table.

Quick Summary

• Topic: Chi-square test of independence for a 2 × 2 contingency table

• Question Reference: ISS 2016, Statistics Paper-1, Question 33

• Correct Answer: Option (a) — 0.79

The Question

Consider the 2 × 2 contingency table on two attributes A and B, with columns A1 and A2, and rows B1 and B2: row B1 has the values 10 (under A1) and 20 (under A2); row B2 has the values 30 (under A1) and 40 (under A2).

What is the value of χ2 for testing the independence of the attributes A and B?

• (a) 0.79

• (b) 0.81

• (c) 0.83

• (d) 0.85

Step 1: Write Down the Observed Frequencies Clearly

Before touching any formula, lay the data out exactly as given, because a sign error or a swapped cell here ruins everything downstream:

• Cell (B1, A1) = 10

• Cell (B1, A2) = 20

• Cell (B2, A1) = 30

• Cell (B2, A2) = 40

Row totals: Row B1 = 10 + 20 = 30. Row B2 = 30 + 40 = 70.

Column totals: Column A1 = 10 + 30 = 40. Column A2 = 20 + 40 = 60.

Grand total: N = 30 + 70 = 40 + 60 = 100. Both ways of adding up must agree — if they don't, you've made an arithmetic slip and need to redo the totals before continuing.

Step 2: Understand What "Independence" Means Here

The null hypothesis is that attribute A and attribute B are statistically independent. If two attributes are truly independent, the frequency in each cell should be proportional to the product of its row total and column total, divided by the grand total. This gives us what statisticians call the expected frequency under the null hypothesis — the count we would expect to see in each cell if there were absolutely no association between A and B.

The general formula for the expected frequency of any cell is:

E(row, column) = (Row Total × Column Total) / Grand Total

Step 3: Calculate Each Expected Frequency

Apply the formula above to all four cells, one at a time, showing full working so nothing is skipped:

E(B1, A1) = (30 × 40) / 100 = 1200 / 100 = 12

E(B1, A2) = (30 × 60) / 100 = 1800 / 100 = 18

E(B2, A1) = (70 × 40) / 100 = 2800 / 100 = 28

E(B2, A2) = (70 × 60) / 100 = 4200 / 100 = 42

As a quick sanity check, the expected frequencies must also sum row-wise and column-wise to the same totals as the observed table: 12 + 18 = 30, 28 + 42 = 70, 12 + 28 = 40, and 18 + 42 = 60. Everything matches, so we can proceed with confidence.

Step 4: Apply the Chi-Square Formula

The chi-square statistic for a contingency table is:

χ2 = Σ [ (O − E)2 / E ]

where O is the observed frequency and E is the expected frequency in each cell. We compute this term by term:

Cell (B1, A1): (10 − 12)2 / 12 = (−2)2 / 12 = 4 / 12 = 0.3333

Cell (B1, A2): (20 − 18)2 / 18 = (2)2 / 18 = 4 / 18 = 0.2222

Cell (B2, A1): (30 − 28)2 / 28 = (2)2 / 28 = 4 / 28 = 0.1429

Cell (B2, A2): (40 − 42)2 / 42 = (−2)2 / 42 = 4 / 42 = 0.0952

Adding these four contributions together:

χ2 = 0.3333 + 0.2222 + 0.1429 + 0.0952 = 0.7936

Rounded to two decimal places, χ2 ≈ 0.79.

Pro Tip: Notice that every single (O − E) difference in this table came out to exactly ±2. That is not a coincidence — in any 2 × 2 contingency table, all four cells share the same absolute deviation from their expected value, because the row and column totals are fixed. Once you spot that pattern, you don't even need to compute all four (O − E) values separately; you can find just one and reuse its square across every term. This is exactly the kind of pattern-recognition habit that's usually drilled into students early, since it turns a four-step calculation into a much faster check during a timed exam.

Step 5: The Shortcut Formula (Cross-Check)

For any 2 × 2 table with cells labelled a, b (top row) and c, d (bottom row), there's a direct formula that avoids computing expected frequencies altogether:

χ2 = N(ad − bc)2 / [(a + b)(c + d)(a + c)(b + d)]

Here a = 10, b = 20, c = 30, d = 40, and N = 100.

ad − bc = (10 × 40) − (20 × 30) = 400 − 600 = −200

(ad − bc)2 = (−200)2 = 40,000

Denominator = (a + b)(c + d)(a + c)(b + d) = (30)(70)(40)(60)

30 × 70 = 2100. 2100 × 40 = 84,000. 84,000 × 60 = 5,040,000.

χ2 = (100 × 40,000) / 5,040,000 = 4,000,000 / 5,040,000 = 0.7936

This confirms our earlier answer exactly: χ2 ≈ 0.79.

Final Answer and Confirmation

Both methods agree: χ2 ≈0.79, which is option (a). This matches the official ISS 2016 Statistics Paper-1 answer key, which also lists option (a) as correct for Question 33.

Why This Question Matters

The chi-square test of independence is one of the most frequently tested topics across ISS, IIT JAM, and other statistics entrance exams, because it sits at the intersection of categorical data analysis and hypothesis testing. Beyond just plugging numbers into a formula, examiners use small 2 × 2 tables like this one to check whether a candidate genuinely understands what "expected frequency under independence" means, rather than just memorizing a formula. Being comfortable with both the long method (via expected frequencies) and the shortcut formula also builds the kind of calculation speed and cross-checking habit that pays off heavily under exam time pressure.

What is the chi-square test of independence used for?

It is used to test whether two categorical attributes, such as A and B in this question, are statistically independent of each other, based on observed frequency data arranged in a contingency table. A significantly large chi-square value suggests the attributes are associated rather than independent.

How many degrees of freedom does a 2 x 2 contingency table have?

A 2 × 2 contingency table has (r − 1)(c − 1) = (2 − 1)(2 − 1) = 1 degree of freedom, where r is the number of rows and c is the number of columns.

Why do we divide by the expected frequency in each term of the chi-square formula?

Dividing by E standardizes each squared deviation relative to the scale of that cell's expected count. Without this division, cells with naturally larger expected frequencies would dominate the statistic simply because their raw counts are bigger, not because they show more association.

Is there a shortcut formula for 2 x 2 tables specifically?

Yes. For a table with cells a, b, c, d, the formula χ2 = N(ad − bc)2 / [(a+b)(c+d)(a+c)(b+d)] gives the same result directly from the observed values, without needing to compute expected frequencies as an intermediate step.

What does a chi-square value of 0.79 with 1 degree of freedom tell us?

At 1 degree of freedom, the critical value for rejecting independence at the 5% significance level is 3.841. Since 0.79 is far smaller than this, we would fail to reject the null hypothesis, meaning there is no significant evidence of association between attributes A and B in this data.

What is the most common mistake students make on this type of question?

The most frequent error is mismatching rows and columns while computing expected frequencies, or forgetting to divide by the grand total N when calculating E. Double-checking that the expected frequencies sum back to the same row and column totals as the observed table, as shown in Step 3 above, catches this mistake immediately.

If anything in this derivation felt unclear, drop a comment below and we'll walk through it further — and if you found this useful, consider sharing it with a fellow ISS aspirant who's working through the same paper.