Answer :
Let's go through the steps to complete the table and answer the questions in the most detailed manner:
### Expected Frequencies
First, we calculate the total observed frequency:
[tex]\[ 18 + 9 + 15 + 13 + 10 = 65 \][/tex]
Since there are 5 categories, the expected frequency for each category is:
[tex]\[ \frac{65}{5} = 13.00 \][/tex]
So, the expected frequencies for categories A, B, C, D, and E are:
[tex]\[ \text{Expected Frequency} = 13.00 \][/tex]
### (O-E)^2 / E Calculation
Now, we calculate [tex]\((O-E)^2 / E\)[/tex] for each category.
#### Category A:
[tex]\[ O = 18, \, E = 13.00 \][/tex]
[tex]\[ \left(\frac{(18 - 13)^2}{13}\right) = \left(\frac{25}{13}\right) = 1.923 \][/tex]
#### Category B:
[tex]\[ O = 9, \, E = 13.00 \][/tex]
[tex]\[ \left(\frac{(9 - 13)^2}{13}\right) = \left(\frac{16}{13}\right) = 1.231 \][/tex]
#### Category C:
[tex]\[ O = 15, \, E = 13.00 \][/tex]
[tex]\[ \left(\frac{(15 - 13)^2}{13}\right) = \left(\frac{4}{13}\right) = 0.308 \][/tex]
#### Category D:
[tex]\[ O = 13, \, E = 13.00 \][/tex]
[tex]\[ \left(\frac{(13 - 13)^2}{13}\right) = \left(\frac{0}{13}\right) = 0.000 \][/tex]
#### Category E:
[tex]\[ O = 10, \, E = 13.00 \][/tex]
[tex]\[ \left(\frac{(10 - 13)^2}{13}\right) = \left(\frac{9}{13}\right) = 0.692 \][/tex]
### Completing the Table
[tex]\[ \begin{array}{|c|l|l|l|} \hline \text{Category} & \text{Observed Frequency} & \text{Expected Frequency} & \left(\frac{(O-E)^2}{E}\right)\\ \hline \text{A} & 18 & 13.00 & 1.923 \\ \hline \text{B} & 9 & 13.00 & 1.231 \\ \hline \text{C} & 15 & 13.00 & 0.308 \\ \hline \text{D} & 13 & 13.00 & 0.000 \\ \hline \text{E} & 10 & 13.00 & 0.692 \\ \hline \end{array} \][/tex]
### Chi-Square Test Statistic
To find the chi-square test statistic, sum all the [tex]\(\left(\frac{(O-E)^2}{E}\right)\)[/tex] values:
[tex]\[ \chi^2 = 1.923 + 1.231 + 0.308 + 0.000 + 0.692 = 4.154 \][/tex]
### Degrees of Freedom
The degrees of freedom (df) in a chi-square test is calculated as:
[tex]\[ \text{df} = \text{number of categories} - 1 = 5 - 1 = 4 \][/tex]
### p-Value
A chi-square test with the test statistic value of 4.154 and degrees of freedom 4 yields a p-value of:
[tex]\[ p = 0.386 \][/tex]
### Decision Based on p-Value
Comparing the p-value to the significance level [tex]\(\alpha = 0.05\)[/tex]:
[tex]\[ p \text{ (0.386) is greater than } \alpha \text{ (0.05)} \][/tex]
### Conclusion
Since the p-value is greater than [tex]\(\alpha\)[/tex], we:
[tex]\[ \text{fail to reject the null} \][/tex]
This leads to the final conclusion:
[tex]\[ \text{There is not sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected.} \][/tex]
Thus, our final answers are:
[tex]\[ \begin{array}{|c|l|l|l|} \hline \text{Category} & \text{Observed Frequency} & \text{Expected Frequency} & \left(\frac{(O-E)^2}{E}\right)\\ \hline \text{A} & 18 & 13.00 & 1.923 \\ \hline \text{B} & 9 & 13.00 & 1.231 \\ \hline \text{C} & 15 & 13.00 & 0.308 \\ \hline \text{D} & 13 & 13.00 & 0.000 \\ \hline \text{E} & 10 & 13.00 & 0.692 \\ \hline \end{array} \][/tex]
[tex]\[ \chi^2 = 4.154 \][/tex]
[tex]\[ \text{df} = 4 \][/tex]
[tex]\[ p \text{-value} = 0.386 \][/tex]
The p-value is greater than [tex]\(\alpha\)[/tex].
We fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected.
### Expected Frequencies
First, we calculate the total observed frequency:
[tex]\[ 18 + 9 + 15 + 13 + 10 = 65 \][/tex]
Since there are 5 categories, the expected frequency for each category is:
[tex]\[ \frac{65}{5} = 13.00 \][/tex]
So, the expected frequencies for categories A, B, C, D, and E are:
[tex]\[ \text{Expected Frequency} = 13.00 \][/tex]
### (O-E)^2 / E Calculation
Now, we calculate [tex]\((O-E)^2 / E\)[/tex] for each category.
#### Category A:
[tex]\[ O = 18, \, E = 13.00 \][/tex]
[tex]\[ \left(\frac{(18 - 13)^2}{13}\right) = \left(\frac{25}{13}\right) = 1.923 \][/tex]
#### Category B:
[tex]\[ O = 9, \, E = 13.00 \][/tex]
[tex]\[ \left(\frac{(9 - 13)^2}{13}\right) = \left(\frac{16}{13}\right) = 1.231 \][/tex]
#### Category C:
[tex]\[ O = 15, \, E = 13.00 \][/tex]
[tex]\[ \left(\frac{(15 - 13)^2}{13}\right) = \left(\frac{4}{13}\right) = 0.308 \][/tex]
#### Category D:
[tex]\[ O = 13, \, E = 13.00 \][/tex]
[tex]\[ \left(\frac{(13 - 13)^2}{13}\right) = \left(\frac{0}{13}\right) = 0.000 \][/tex]
#### Category E:
[tex]\[ O = 10, \, E = 13.00 \][/tex]
[tex]\[ \left(\frac{(10 - 13)^2}{13}\right) = \left(\frac{9}{13}\right) = 0.692 \][/tex]
### Completing the Table
[tex]\[ \begin{array}{|c|l|l|l|} \hline \text{Category} & \text{Observed Frequency} & \text{Expected Frequency} & \left(\frac{(O-E)^2}{E}\right)\\ \hline \text{A} & 18 & 13.00 & 1.923 \\ \hline \text{B} & 9 & 13.00 & 1.231 \\ \hline \text{C} & 15 & 13.00 & 0.308 \\ \hline \text{D} & 13 & 13.00 & 0.000 \\ \hline \text{E} & 10 & 13.00 & 0.692 \\ \hline \end{array} \][/tex]
### Chi-Square Test Statistic
To find the chi-square test statistic, sum all the [tex]\(\left(\frac{(O-E)^2}{E}\right)\)[/tex] values:
[tex]\[ \chi^2 = 1.923 + 1.231 + 0.308 + 0.000 + 0.692 = 4.154 \][/tex]
### Degrees of Freedom
The degrees of freedom (df) in a chi-square test is calculated as:
[tex]\[ \text{df} = \text{number of categories} - 1 = 5 - 1 = 4 \][/tex]
### p-Value
A chi-square test with the test statistic value of 4.154 and degrees of freedom 4 yields a p-value of:
[tex]\[ p = 0.386 \][/tex]
### Decision Based on p-Value
Comparing the p-value to the significance level [tex]\(\alpha = 0.05\)[/tex]:
[tex]\[ p \text{ (0.386) is greater than } \alpha \text{ (0.05)} \][/tex]
### Conclusion
Since the p-value is greater than [tex]\(\alpha\)[/tex], we:
[tex]\[ \text{fail to reject the null} \][/tex]
This leads to the final conclusion:
[tex]\[ \text{There is not sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected.} \][/tex]
Thus, our final answers are:
[tex]\[ \begin{array}{|c|l|l|l|} \hline \text{Category} & \text{Observed Frequency} & \text{Expected Frequency} & \left(\frac{(O-E)^2}{E}\right)\\ \hline \text{A} & 18 & 13.00 & 1.923 \\ \hline \text{B} & 9 & 13.00 & 1.231 \\ \hline \text{C} & 15 & 13.00 & 0.308 \\ \hline \text{D} & 13 & 13.00 & 0.000 \\ \hline \text{E} & 10 & 13.00 & 0.692 \\ \hline \end{array} \][/tex]
[tex]\[ \chi^2 = 4.154 \][/tex]
[tex]\[ \text{df} = 4 \][/tex]
[tex]\[ p \text{-value} = 0.386 \][/tex]
The p-value is greater than [tex]\(\alpha\)[/tex].
We fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected.
