A random sample of 500 UAE youth who are questioned regarding their place of residence and opinion regarding the speed limit increase on Abu Dhabi-Dubai Road. We will test if the r... A random sample of 500 UAE youth who are questioned regarding their place of residence and opinion regarding the speed limit increase on Abu Dhabi-Dubai Road. We will test if the residence place and their opinion on speed limit are dependent at a 5% level of significance. Calculate the test statistic. Read more

Steps to Solve

1. Calculate the row and column totals First, we need to calculate the row totals (Abu Dhabi and Dubai) and column totals (Favor, Indifferent, Opposes). Row Totals: Abu Dhabi: $138 + 83 + 64 = 285$ Dubai: $64 + 67 + 84 = 215$ Column Totals: Favor: $138 + 64 = 202$ Indifferent: $83 + 67 = 150$ Opposes: $64 + 84 = 148$

2. Calculate the grand total The grand total is the sum of all observations, which is also the sum of the row totals or the column totals: $285 + 215 = 202 + 150 + 148 = 500$

3. Calculate the expected frequencies The expected frequency for each cell is calculated using the formula: $E_{ij} = \frac{(Row\ Total_i) \times (Column\ Total_j)}{Grand\ Total}$, where $E_{ij}$ is the expected frequency for the cell in the $i^{th}$ row and $j^{th}$ column. $E_{11} = \frac{285 \times 202}{500} = 115.14$ $E_{12} = \frac{285 \times 150}{500} = 85.5$ $E_{13} = \frac{285 \times 148}{500} = 84.36$ $E_{21} = \frac{215 \times 202}{500} = 86.86$ $E_{22} = \frac{215 \times 150}{500} = 64.5$ $E_{23} = \frac{215 \times 148}{500} = 63.64$

4. Calculate the chi-square statistic The chi-square test statistic is calculated using the formula: $\chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}}$, where $O_{ij}$ is the observed frequency and $E_{ij}$ is the expected frequency for each cell. $\chi^2 = \frac{(138 - 115.14)^2}{115.14} + \frac{(83 - 85.5)^2}{85.5} + \frac{(64 - 84.36)^2}{84.36} + \frac{(64 - 86.86)^2}{86.86} + \frac{(67 - 64.5)^2}{64.5} + \frac{(84 - 63.64)^2}{63.64}$ $\chi^2 = \frac{(22.86)^2}{115.14} + \frac{(-2.5)^2}{85.5} + \frac{(-20.36)^2}{84.36} + \frac{(-22.86)^2}{86.86} + \frac{(2.5)^2}{64.5} + \frac{(20.36)^2}{63.64}$ $\chi^2 = \frac{522.5796}{115.14} + \frac{6.25}{85.5} + \frac{414.5296}{84.36} + \frac{522.5796}{86.86} + \frac{6.25}{64.5} + \frac{414.5296}{63.64}$ $\chi^2 = 4.538 + 0.073 + 4.914 + 6.016 + 0.097 + 6.513$ $\chi^2 = 22.151$ Therefore, the test statistic is approximately $22.15$.