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The following table shows the Myers-Briggs personality preferences for a random sample of 406 people in the listed professions (Atlas of Type Tables by Macdaid, McCaulley, and Kainz). E refers to extroverted and I refers to introverted. $$\begin{array}{l|c|c|c} \hline \multirow{2}{*} {\text { Occupation }} & \multicolumn{3}{|c} {\text { Personality Preference Type }} \\ \\)\cline { 2 - 5 } & \\( \mathrm{E} & \mathrm{I} & \text { Row Total } \\ \hline \text { Clergy (all denominations) } & 62 & 45 & 107 \\ \hline \text { M.D. } & 68 & 94 & 162 \\ \hline \text { Lawyer } & 56 & 81 & 137 \\ \hline \text { Column Total } & 186 & 220 & 406 \\ \hline \end{array}$$Use the chi-square test to determine if the listed occupations and personality preferences are independent at the 0.05 level of significance.

Step by step solution

01

Hypotheses Formulation

First, we formulate the null and alternative hypotheses. **Null Hypothesis (H_0)**: The listed occupations and personality preferences are independent. **Alternative Hypothesis (H_1)**: The listed occupations and personality preferences are not independent.

02

Observed Frequencies Table

Identify the observed frequency counts from the provided table. These are the actual counts given: - Clergy E = 62, I = 45 - M.D. E = 68, I = 94 - Lawyer E = 56, I = 81 - Column totals: E = 186, I = 220.

03

Calculate Expected Frequencies

Use the formula for expected frequency: \[ E_{ij} = \frac{( ext{Row Total})( ext{Column Total})}{ ext{Grand Total}} \] Calculate for each cell. For example, for Clergy E: \[ E_{11} = \frac{(107)(186)}{406} \approx 49.14 \] Continue similarly for the other cells.

04

Chi-square Test Statistic Calculation

Use the formula: \[ \chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \] Calculate for each observed and expected pairing. Sum all the values to get \( \chi^2 \).

05

Degrees of Freedom

Calculate the degrees of freedom using the formula: \[ \text{d.f.} = (r-1)(c-1) \] where \( r \) is the number of rows, \( c \) is the number of columns. Substituting gives: \( (3-1)(2-1) = 2 \) degrees of freedom.

06

Determine the Critical Value

Look up the critical value from the chi-square distribution table, using \( \alpha = 0.05 \) and 2 degrees of freedom. The critical value is approximately 5.99.

07

Test Decision

Compare your calculated \( \chi^2 \) value with the critical value from the chi-square table. If \( \chi^2_{calculated} > 5.99 \), reject the null hypothesis. Otherwise, do not reject the null hypothesis.

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Key Concepts

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

Null Hypothesis

The null hypothesis is a fundamental starting point in statistical testing. It is a statement you make when beginning a hypothesis test. In our chi-square test, the null hypothesis states that the two variables in question, such as occupation and personality preferences, are independent.
This means that your null hypothesis assumes no relationship exists between the variables you are testing. In our example, the null hypothesis can be expressed as: 'The listed occupations and personality preferences are independent.'

The purpose of the null hypothesis is to provide a base or a status quo that your data can be tested against. If the test results significantly diverge from what is expected under the null hypothesis, this suggests that the hypothesis may not hold true.

Alternative Hypothesis

The alternative hypothesis offers a different perspective from the null hypothesis. It is what you consider to be true if the null hypothesis is rejected.
In our case, the alternative hypothesis would be that an association does exist between the two variables, occupations and personality preferences. This is articulated as: 'The listed occupations and personality preferences are not independent.'

The alternative hypothesis explores the possibility of a dependency or relationship between the variables. This hypothesis is critical to scientific testing as it represents new findings or results that contradict the status quo.

Expected Frequencies

Expected frequencies are a crucial component of performing a chi-square test. They represent what we expect to find if the null hypothesis were true. To calculate them, you use the formula: \[ E_{ij} = \frac{(\text{Row Total})(\text{Column Total})}{\text{Grand Total}} \]
In our example, for the Clergy with an Extroverted preference, the expected frequency is determined by plugging in the appropriate row and column totals into the formula, yielding approximately 49.14.

Calculating expected frequencies enables you to measure how observed data deviates from what is expected under the null hypothesis. Large deviations suggest that the null hypothesis may not be valid.

Degrees of Freedom

Degrees of freedom is an important concept in statistical calculations, affecting how the chi-square test statistic is interpreted. It reflects the number of values that are free to vary in the final calculation of a statistic.

For our chi-square test, degrees of freedom are calculated using the formula: \[ \text{d.f.} = (r-1)(c-1) \] where \( r \) is the number of rows, and \( c \) is the number of columns in the table of data. In our case with 3 occupations and 2 personality preferences, the degrees of freedom is \( (3-1)(2-1) = 2 \).

This metric is essential because it helps identify the appropriate distribution of the chi-square statistic, which in turn is used to determine if the test evidence is sufficient to reject the null hypothesis.