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The following table shows the Myers-Briggs personality preference and area of study for a random sample of 519 college students (Applications of the Myers- Briggs Type Indicator in Higher Education, edited by Provost and Anchors). In the table, IN refers to introvert, intuitive; EN refers to extrovert, intuitive; IS refers to introvert, sensing; and ES refers to extrovert, sensing. $$\begin{array}{l|c|c|c|c} \hline \begin{array}{l} \text { Myers-Briggs } \\ \text { Preference } \end{array} & \begin{array}{c} \text { Arts \& } \\ \text { Science } \end{array} & \text { Business } & \begin{array}{c} \text { Allied } \\ \text { Health } \end{array} & \text { Row Total } \\ \hline \text { IN } & 64 & 15 & 17 & 96 \\ \hline \text { EN } & 82 & 42 & 30 & 154 \\ \hline \text { IS } & 68 & 35 & 12 & 115 \\ \hline \text { ES } & 75 & 42 & 37 & 154 \\ \hline \text { Column Total } & 289 & 134 & 96 & 519 \\ \hline \end{array}$$ Use a chi-square test to determine if Myers-Briggs preference type is independent of area of study at the 0.05 level of significance.

Step by step solution

01

Understand the Hypotheses

We are conducting a chi-square test for independence. The null hypothesis \(H_0\) states that Myers-Briggs preference type is independent of the area of study. The alternative hypothesis \(H_a\) asserts that there is a dependency between Myers-Briggs preference type and area of study.

02

Set Up the Table and Calculate Expected Frequencies

To test these hypotheses, we create a contingency table from the given data. The expected frequency for each cell is calculated using the formula: \(E = \frac{\text{(Row Total)} \times \text{(Column Total)}}{\text{(Grand Total)}}\). We will calculate the expected frequencies for each cell in the table.

03

Compute the Expected Frequencies

For instance, the expected frequency for IN and Arts & Science is \(E = \frac{96 \times 289}{519} \approx 53.47\). Repeat this calculation for all cells to fill out the expected frequency table.

04

Calculate the Chi-Square Statistic

Use the formula: \[\chi^2 = \sum \frac{(O - E)^2}{E}\] where \(O\) is the observed frequency and \(E\) is the expected frequency. Compute this for each cell and sum all values to find the chi-square statistic.

05

Determine Degrees of Freedom

The degrees of freedom for this test is calculated using \((r-1)(c-1)\), where \(r\) is the number of rows and \(c\) is the number of columns in the table. Here, \( (4-1)(3-1) = 6 \) degrees of freedom.

06

Compare with Critical Value and Make Decision

Refer to a chi-square distribution table to find the critical value at 0.05 level of significance for 6 degrees of freedom. The critical value is approximately 12.592. Compare the computed \(\chi^2\) statistic to this critical value to either reject or fail to reject \(H_0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Myers-Briggs Type Indicator

The Myers-Briggs Type Indicator (MBTI) is a popular psychological tool used to assess people's personality types. This indicator identifies four key preferences that combine to form one of 16 personality types.
The preferences revolve around how individuals engage with the world:

• Introversion vs. Extroversion
• Sensing vs. Intuition
• Thinking vs. Feeling
• Judging vs. Perceiving

In the context of the exercise, categories such as "IN" (introvert, intuitive) and "EN" (extrovert, intuitive) are used to identify different personality types among students. Understanding these types helps to discern behavioral patterns that might influence one's area of study. The test can often reveal insights into whether certain types are naturally drawn to specific fields, making it a valuable tool in both education and career counseling.

Contingency Table

A contingency table is a statistical tool that organizes data into a matrix format, facilitating the analysis of the relationship between two categorical variables.
In this scenario, the table showcases the relationship between Myers-Briggs personality preferences and students' chosen areas of study.
Each cell in the table contains the count of observations for each combination of categories – for example, how many 'IN' personality types chose Arts & Science.

• Row Totals: Sum of observations across different study areas for each Myers-Briggs type.
• Column Totals: Sum of observations across personality types for each study area.
• Grand Total: The total number of observations across all categories, here being 519 students.

Contingency tables are critical for conducting chi-square tests, as they help establish expected versus observed counts, which is essential for testing hypotheses concerning independence or association.

Degrees of Freedom

In statistical analysis, degrees of freedom (df) are critical for determining the validity and significance of computed test statistics. It essentially refers to the number of values in a calculation that are free to vary.
For a chi-square test within a contingency table, degrees of freedom depend on the number of categories within the variables. The formula is \[(r-1)(c-1)\]where \(r\) is the number of rows, and \(c\) is the number of columns.
This lets us know the number of comparisons we have in our data set.

• For this case, with 4 personality preferences (rows) and 3 areas of study (columns), the degrees of freedom are \[(4-1)(3-1) = 6\]

Knowing the degrees of freedom is vital as it allows us to look up the necessary critical value in a chi-square distribution table to interpret our results correctly.

Expected Frequencies

Expected frequencies form the backbone of a chi-square test. They are the predicted counts we expect to see in each contingency table cell if the null hypothesis were true.
These are calculated using the formula:\[E = \frac{\text{(Row Total)} \times \text{(Column Total)}}{\text{(Grand Total)}}\]For each cell:

• Example: The expected frequency for 'IN' preferences in Arts & Science is calculated as:\[E = \frac{96 \times 289}{519} \approx 53.47\]
• This is repeated for all entries in the table, allowing a thorough comparison between observed data and what is expected under the assumption of independence.

By comparing observed frequencies (what we actually see) with these expected frequencies, the chi-square statistic can reveal whether the differences are due to mere chance or indicate a real association between the variables.