Expected count

An expected count $E$ is what a cell would hold on average if the null hypothesis were exactly true. In a two-way table the null says one variable tells you nothing about the other, so a cell's share of the table is its row share times its column share, and multiplying by the grand total gives $E = \frac{(\text{row total}) \times (\text{column total})}{\text{grand total}}$. The expected counts are built from the margins of your own table, so the observed and expected tables always carry identical row totals, column totals, and grand total.

Two schools are surveyed about biking to school. School A has 45 bikers out of 100, School B 45 out of 140, so the column totals are 90 bikers and 150 non-bikers out of 240. School A bikers have $E = \frac{100 \times 90}{240} = 37.5$ and School A non-bikers $E = \frac{100 \times 150}{240} = 62.5$, which add back to that row's 100. The second row gives 52.5 and 87.5.

The objection to answer: "expected counts have to be whole numbers, because there is no such thing as 37.5 students." Round 37.5 up to 38 and the chi-square statistic changes. $E$ is an average across repeated samples under the null, not a forecast of any single table, and it goes into $\frac{(O - E)^2}{E}$ with its decimals intact. If the expected counts fail to add back to the margins, the arithmetic is wrong.

The condition for trusting the chi-square approximation is that every expected count is at least 5, and it is checked on the expected counts, never the observed ones. A cell holding 2 observed is fine as long as its $E$ clears 5. When some fall short, categories get combined or more data collected rather than reading the p-value anyway.

Building these counts is AP Statistics topic 3.14, Setting Up a Chi-Square Test for Homogeneity or Independence, and topic 3.15 carries the test out. A goodness-of-fit test instead uses $E = np_0$ from claimed proportions, which is standard college content but no longer on the AP exam.