Question

ms. truman has always provided her statistics students with a small index card for each exam. the students are permitted to write whatever they want on the index card and use it on the exam. she decides to conduct an experiment to investigate if the distribution of grades will be different if she provides students with a large index card. she has 80 students in her statistics classes this year. she randomly assigns 40 students to receive the typical small index card and 40 students to receive a large index card to use on the upcoming exam. the results of the exam are displayed in the table.
the random condition is met and the 10% condition is not needed because no random sampling took place. is the large counts condition met?
yes, the smallest expected count is 5, so all expected counts are at least 5.
yes, the smallest expected count is 7.5, so all expected counts are at least 5.
no, the smallest expected count is 4, so the expected counts are not all at least 5.
no, the smallest expected count is 0.2, so the expected counts are not all at least 5.

Explanation:

Step1: Calculate row totals

Row \(A\) total: \(8 + 22=30\)
Row \(B\) total: \(10 + 5 = 15\)
Row \(C\) total: \(10+5 = 15\)
Row \(D\) total: \(6 + 4=10\)
Row \(E\) total: \(6+4 = 10\)

Step2: Calculate column totals

Column \(Small\) total: \(8+10 + 10+6+6=40\)
Column \(Large\) total: \(22+5+5+4+4 = 40\)

Step3: Calculate expected counts

The formula for the expected count \(E_{ij}=\frac{\text{Row }i\text{ total}\times\text{Column }j\text{ total}}{\text{Grand total}}\)
For all cells, the grand total \(N = 80\)
For row \(A\), column \(Small\): \(E=\frac{30\times40}{80}=15\)
For row \(A\), column \(Large\): \(E=\frac{30\times40}{80}=15\)
For row \(B\), column \(Small\): \(E=\frac{15\times40}{80}=7.5\)
For row \(B\), column \(Large\): \(E=\frac{15\times40}{80}=7.5\)
For row \(C\), column \(Small\): \(E=\frac{15\times40}{80}=7.5\)
For row \(C\), column \(Large\): \(E=\frac{15\times40}{80}=7.5\)
For row \(D\), column \(Small\): \(E=\frac{10\times40}{80}=5\)
For row \(D\), column \(Large\): \(E=\frac{10\times40}{80}=5\)
For row \(E\), column \(Small\): \(E=\frac{10\times40}{80}=5\)
For row \(E\), column \(Large\): \(E=\frac{10\times40}{80}=5\)

The smallest expected count is \(5\)

Answer:

Yes, the smallest expected count is \(5\), so all expected counts are at least \(5\).