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.
Understand the Problem
The question involves a hypothesis test to determine if there's a relationship between the residence place of UAE youth (Abu Dhabi or Dubai) and their opinion (favor, indifferent, opposes) on the speed limit increase on the Abu Dhabi-Dubai Road. Given a contingency table with observed frequencies, the task is to calculate the appropriate test statistic to make a determination at a 5% significance level to evaluate the dependence between the two categorical variables.
Answer
$\chi^2 = 22.151$
Answer for screen readers
The test statistic is $\chi^2 = 22.151$.
Steps to Solve
- Calculate row and column totals
First, find the row totals (for Abu Dhabi and Dubai) and column totals (for Favor, Indifferent, and 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: $64 + 84 = 148$
- Calculate the grand total
The grand total is the sum of all observations, which is also the sum of row totals or column totals.
Grand total = $285 + 215 = 500$. This matches the problem statement, so it is likely our values are correct.
- Calculate expected frequencies
The expected frequency for each cell is calculated as:
$E_{ij} = \frac{(\text{Row Total}_i) \times (\text{Column Total}_j)}{\text{Grand Total}}$
$E_{11} = \frac{285 \times 202}{500}= 115.14$ (Abu Dhabi, Favor)
$E_{12} = \frac{285 \times 150}{500}= 85.5$ (Abu Dhabi, Indifferent)
$E_{13} = \frac{285 \times 148}{500}= 84.36$ (Abu Dhabi, Opposes)
$E_{21} = \frac{215 \times 202}{500}= 86.86$ (Dubai, Favor)
$E_{22} = \frac{215 \times 150}{500}= 64.5$ (Dubai, Indifferent)
$E_{23} = \frac{215 \times 148}{500}= 63.64$ (Dubai, Opposes)
- Calculate the Chi-square test statistic
The Chi-square test statistic is calculated as:
$$ \chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}} $$
Where $O_{ij}$ is the observed frequency and $E_{ij}$ is the expected frequency.
$$ \chi^2 = \frac{(138 - 115.14)^2}{115.14} + \frac{(83 - 85.5)^2}{85.5} + \frac{(64 - 84.36)^2}{84.36} + \frac{(64 - 86.86)^2}{86.86} + \frac{(67 - 64.5)^2}{64.5} + \frac{(84 - 63.64)^2}{63.64} $$
$$ \chi^2 = \frac{(22.86)^2}{115.14} + \frac{(-2.5)^2}{85.5} + \frac{(-20.36)^2}{84.36} + \frac{(-22.86)^2}{86.86} + \frac{(2.5)^2}{64.5} + \frac{(20.36)^2}{63.64} $$
$$ \chi^2 = \frac{522.5796}{115.14} + \frac{6.25}{85.5} + \frac{414.5296}{84.36} + \frac{522.5796}{86.86} + \frac{6.25}{64.5} + \frac{414.5296}{63.64} $$
$$ \chi^2 = 4.538 + 0.073 + 4.914 + 6.016 + 0.097 + 6.513 = 22.151 $$
Therefore, the test statistic is approximately $22.151$.
The test statistic is $\chi^2 = 22.151$.
More Information
The Chi-square test statistic measures the discrepancy between observed and expected frequencies under the assumption of independence. A higher value indicates a larger difference, suggesting a stronger association between the variables. To complete the hypothesis test, this value would be compared to a critical value from the Chi-square distribution with $(r-1)(c-1)$ degrees of freedom, where $r$ is the number of rows and $c$ is the number of columns. In this case, the degrees of freedom would be $(2-1)(3-1) = 2$.
Tips
A common mistake is in calculating the expected frequencies. It's crucial to use the correct row and column totals and the grand total. Another common mistake involves errors in the Chi-square formula. Specifically, errors include subtracting in the wrong order, forgetting to square the difference, or dividing by the observed frequency instead of dividing by the expected frequency.
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