Answer :
To perform a hypothesis test of the independence of safety ratings and industry classification, we can use the Chi-Square Test of Independence. Here is the step-by-step solution:
### Step 1: State the Hypotheses
- Null Hypothesis ([tex]\(H_0\)[/tex]): There is no association between safety ratings and industry classification. In other words, they are independent.
- Alternative Hypothesis ([tex]\(H_1\)[/tex]): There is an association between safety ratings and industry classification. In other words, they are not independent.
### Step 2: Organize the Data
The given data can be summarized in a contingency table:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Stock Rating} & \text{Industry I} & \text{Industry II} & \text{Industry III} & \text{Industry IV} & \text{Total} \\ \hline \text{High} & 25 & 30 & 42 & 20 & 117 \\ \hline \text{Average} & 32 & 32 & 14 & 33 & 111 \\ \hline \text{Low} & 18 & 29 & 12 & 13 & 72 \\ \hline \text{Total} & 75 & 91 & 68 & 66 & 300 \\ \hline \end{array} \][/tex]
### Step 3: Compute the Expected Frequencies
The expected frequency for each cell in a contingency table is calculated using the formula:
[tex]\[ E_{ij} = \frac{( \text{Row Total} \times \text{Column Total} )}{ \text{Overall Total} } \][/tex]
For example, for the cell corresponding to "High" rating and "Industry I":
[tex]\[ E_{11} = \frac{( \text{Total High} \times \text{Total Industry I} )}{ \text{Overall Total} } = \frac{(117 \times 75)}{300} = 29.25 \][/tex]
Calculate the expected frequencies for all cells.
### Step 4: Compute the Test Statistic
The test statistic for the Chi-Square Test of Independence is computed using the formula:
[tex]\[ \chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \][/tex]
where [tex]\(O_{ij}\)[/tex] is the observed frequency and [tex]\(E_{ij}\)[/tex] is the expected frequency.
### Step 5: Find the Degrees of Freedom
The degrees of freedom for the Chi-Square Test of Independence is given by:
[tex]\[ ( \text{Number of Rows} - 1 ) \times ( \text{Number of Columns} - 1 ) \][/tex]
For our table:
[tex]\[ (3 - 1) \times (4 - 1) = 2 \times 3 = 6 \][/tex]
### Step 6: Determine the Critical Value and Make a Decision
Using a significance level of [tex]\(\alpha = 0.01\)[/tex] and 6 degrees of freedom, find the critical value from the Chi-Square distribution table. Let's denote this critical value by [tex]\(\chi^2_{\alpha}\)[/tex].
If the computed [tex]\(\chi^2\)[/tex] statistic is greater than [tex]\(\chi^2_{\alpha}\)[/tex], we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
### Conclusion
If the hypothesis test results in rejecting the null hypothesis, we conclude that there is an association between safety ratings and industry classification. If we fail to reject the null hypothesis, we conclude that there is no significant association between safety ratings and industry classification.
### Step 1: State the Hypotheses
- Null Hypothesis ([tex]\(H_0\)[/tex]): There is no association between safety ratings and industry classification. In other words, they are independent.
- Alternative Hypothesis ([tex]\(H_1\)[/tex]): There is an association between safety ratings and industry classification. In other words, they are not independent.
### Step 2: Organize the Data
The given data can be summarized in a contingency table:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Stock Rating} & \text{Industry I} & \text{Industry II} & \text{Industry III} & \text{Industry IV} & \text{Total} \\ \hline \text{High} & 25 & 30 & 42 & 20 & 117 \\ \hline \text{Average} & 32 & 32 & 14 & 33 & 111 \\ \hline \text{Low} & 18 & 29 & 12 & 13 & 72 \\ \hline \text{Total} & 75 & 91 & 68 & 66 & 300 \\ \hline \end{array} \][/tex]
### Step 3: Compute the Expected Frequencies
The expected frequency for each cell in a contingency table is calculated using the formula:
[tex]\[ E_{ij} = \frac{( \text{Row Total} \times \text{Column Total} )}{ \text{Overall Total} } \][/tex]
For example, for the cell corresponding to "High" rating and "Industry I":
[tex]\[ E_{11} = \frac{( \text{Total High} \times \text{Total Industry I} )}{ \text{Overall Total} } = \frac{(117 \times 75)}{300} = 29.25 \][/tex]
Calculate the expected frequencies for all cells.
### Step 4: Compute the Test Statistic
The test statistic for the Chi-Square Test of Independence is computed using the formula:
[tex]\[ \chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \][/tex]
where [tex]\(O_{ij}\)[/tex] is the observed frequency and [tex]\(E_{ij}\)[/tex] is the expected frequency.
### Step 5: Find the Degrees of Freedom
The degrees of freedom for the Chi-Square Test of Independence is given by:
[tex]\[ ( \text{Number of Rows} - 1 ) \times ( \text{Number of Columns} - 1 ) \][/tex]
For our table:
[tex]\[ (3 - 1) \times (4 - 1) = 2 \times 3 = 6 \][/tex]
### Step 6: Determine the Critical Value and Make a Decision
Using a significance level of [tex]\(\alpha = 0.01\)[/tex] and 6 degrees of freedom, find the critical value from the Chi-Square distribution table. Let's denote this critical value by [tex]\(\chi^2_{\alpha}\)[/tex].
If the computed [tex]\(\chi^2\)[/tex] statistic is greater than [tex]\(\chi^2_{\alpha}\)[/tex], we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
### Conclusion
If the hypothesis test results in rejecting the null hypothesis, we conclude that there is an association between safety ratings and industry classification. If we fail to reject the null hypothesis, we conclude that there is no significant association between safety ratings and industry classification.
