Use the chi-squared test to test the significance of differences between observed and expected results (the formula for the chi-squared test will be provided, as shown in the Mathematical requirements)

Sourav PanNovember 1, 2024

Answered

Where:
O_i = \text{observed frequency for category } i
E_i = \text{expected frequency for category } i
\text{The summation is over all categories.}

1. \text{State the Hypotheses:}
- \text{Null Hypothesis } (H_0): \text{ There is no significant difference between observed and expected frequencies.}
- \text{Alternative Hypothesis } (H_a): \text{ There is a significant difference between observed and expected frequencies.}

2. \text{Collect Data:}
- \text{Obtain the observed frequencies } (O) \text{ for each category.}
- \text{Determine the expected frequencies } (E) \text{ based on the theoretical distribution or proportions.}

3. \text{Calculate the Chi-Squared Statistic:}
- \text{Use the formula provided to compute } \chi^2.

4. \text{Determine the Degrees of Freedom:}
- \text{Degrees of freedom } (df) \text{ are calculated as:}
df = k - 1
\text{where } k \text{ is the number of categories.}

5. \text{Find the Critical Value:}
- \text{Using a chi-squared distribution table, find the critical value for } \chi^2 \text{ at a specific significance level (commonly } \alpha = 0.05) \text{ and the calculated degrees of freedom.}

6. \text{Make a Decision:}
- \text{If the calculated } \chi^2 \text{ is greater than the critical value from the table, reject the null hypothesis } (H_0). \text{ Otherwise, fail to reject it.}

\begin{array}{|c|c|c|}
\hline
\text{Category} & \text{Observed (O)} & \text{Expected (E)} \\
\hline
\text{Category 1} & 30 & 25 \\
\text{Category 2} & 15 & 20 \\
\text{Category 3} & 25 & 30 \\
\hline
\end{array}

1. \text{Calculate } \chi^2:
\chi^2 = \frac{(30-25)^2}{25} + \frac{(15-20)^2}{20} + \frac{(25-30)^2}{30}

2. \text{Degrees of Freedom:}
df = 3 - 1 = 2

3. \text{Find the Critical Value:}
- \text{For } \alpha = 0.05 \text{ and } df = 2, \text{ the critical value is approximately } 5.991.

4. \text{Decision:}
- \text{Since } 3.0833 < 5.991, \text{ we fail to reject the null hypothesis } (H_0). \text{ There is no significant difference between the observed and expected frequencies.}