Questions: Question 2 (1 point) A color code personality test categorizes people into four colors - Red (Power), Blue (Intimacy), Green (Peace), and Yellow (Fun). In general, 25% of people are Red, 35% Blue, 20% Green, and 20% Yellow. An art class of 45 students is tested at a university and 7 are found to be Red, 18 Blue, 9 Green, and 11 Yellow. Can it be concluded that personality type has an impact on students' areas of interest and talents, such as artistic students and state the p-value? Test at a 0.05 level of significance. Red Blue Green Yellow Observed Counts 7 18 9 11 Expected Counts 11.25 15.75 9 9 Yes the p-value =0.501025 No. the p-value =0.501025 No, the p-value =0.498975 Yes, the p-value =0.498975

Question 2 (1 point) A color code personality test categorizes people into four colors - Red (Power), Blue (Intimacy), Green (Peace), and Yellow (Fun). In general, 25% of people are Red, 35% Blue, 20% Green, and 20% Yellow. An art class of 45 students is tested at a university and 7 are found to be Red, 18 Blue, 9 Green, and 11 Yellow.

Can it be concluded that personality type has an impact on students' areas of interest and talents, such as artistic students and state the p-value? Test at a 0.05 level of significance.

Red Blue Green Yellow
Observed Counts 7 18 9 11
Expected Counts 11.25 15.75 9 9

Yes the p-value =0.501025
No. the p-value =0.501025
No, the p-value =0.498975
Yes, the p-value =0.498975

Transcript text: Question 2 (1 point) A color code personality test categorizes people into four colors - Red (Power), Blue (Intimacy), Green (Peace), and Yellow (Fun). In general, 25% of people are Red, 35% Blue, 20% Green, and 20% Yellow. An art class of 45 students is tested at a university and 7 are found to be Red, 18 Blue, 9 Green, and 11 Yellow. Can it be concluded that personality type has an impact on students' areas of interest and talents, such as artistic students and state the p-value? Test at a 0.05 level of significance. \begin{tabular}{|l|l|l|l|l|} \hline & Red & Blue & Green & Yellow \\ \hline \begin{tabular}{c} Observed \\ Counts \end{tabular} & 7 & 18 & 9 & 11 \\ \hline \begin{tabular}{c} Expected \\ Counts \end{tabular} & 11.25 & 15.75 & 9 & 9 \\ \hline \end{tabular} Yes the p-value $=0.501025$ No. the $p$-value $=0.501025$ No, the p-value $=0.498975$ Yes, the $p$-value $=0.498975$

Solution

Solution Steps

Step 1: Define the Hypotheses

We are conducting a Chi-Square Goodness-of-Fit Test to determine if the distribution of personality types in an art class differs from the expected distribution based on the general population. The hypotheses are:

Null Hypothesis ($H_0$): The observed distribution of personality types matches the expected distribution.
Alternative Hypothesis ($H_a$): The observed distribution of personality types does not match the expected distribution.

Step 2: Calculate the Chi-Square Test Statistic

The Chi-Square test statistic is calculated using the formula:

\[ \chi^2 = \sum_i \frac{(O_i - E_i)^2}{E_i} \]

where $O_i$ are the observed frequencies and $E_i$ are the expected frequencies. For our data:

Observed counts: $O = [7, 18, 9, 11]$
Expected counts: $E = [11.25, 15.75, 9, 9]$

Calculating the test statistic:

\[ \chi^2 = \frac{(7 - 11.25)^2}{11.25} + \frac{(18 - 15.75)^2}{15.75} + \frac{(9 - 9)^2}{9} + \frac{(11 - 9)^2}{9} = 2.3714 \]

Step 3: Determine the Degrees of Freedom

The degrees of freedom ($df$) for a Chi-Square Goodness-of-Fit Test is given by:

\[ df = k - 1 \]

where $k$ is the number of categories. Here, $k = 4$, so:

\[ df = 4 - 1 = 3 \]

Step 4: Find the Critical Value and P-Value

For a significance level $\alpha = 0.05$ and $df = 3$, the critical value from the Chi-Square distribution table is:

\[ \chi^2(0.95, 3) = 7.8147 \]

The p-value is calculated as:

\[ P(\chi^2 > 2.3714) = 0.499 \]

Step 5: Make a Decision

Since the p-value $0.499$ is greater than the significance level $\alpha = 0.05$, we fail to reject the null hypothesis. This means there is not enough evidence to conclude that the distribution of personality types in the art class is different from the expected distribution.