The following table shows the Myers-Briggs personality...

The following table shows the Myers-Briggs personality preferences for a random sample of 519 people in the listed professions (Atlas of Type Tables by Macdaid, McCaulley, and Kainz). T refers to thinking and F refers to feeling. $$\begin{array}{l|c|c|c} \hline \multirow{2}{*} {\text { Occupation }} & \multicolumn{2}{c} {\text { Personality Preference Type }} \\ \)\cline { 2 - 5 } & \( \mathrm{T} & \mathrm{F} & \text { Row Total } \\ \hline \text { Clergy (all denominations) } & 57 & 91 & 148 \\ \hline \text { M.D. } & 77 & 82 & 159 \\ \hline \text { Lawyer } & 118 & 94 & 212 \\ \hline \text { Column Total } & 252 & 267 & 519 \\ \hline \end{array}$$ Use the chi-square test to determine if the listed occupations and personality preferences are independent at the 0.01 level of significance.

The following table shows the Myers-Briggs personality preferences for a random sample of 519 people in the listed professions (Atlas of Type Tables by Macdaid, McCaulley, and Kainz). T refers to thinking and F refers to feeling.
$$\begin{array}{l|c|c|c}
\hline \multirow{2}{*} {\text { Occupation }} & \multicolumn{2}{c} {\text { Personality Preference Type }} \\
\)\cline { 2 - 5 } & \( \mathrm{T} & \mathrm{F} & \text { Row Total } \\
\hline \text { Clergy (all denominations) } & 57 & 91 & 148 \\
\hline \text { M.D. } & 77 & 82 & 159 \\
\hline \text { Lawyer } & 118 & 94 & 212 \\
\hline \text { Column Total } & 252 & 267 & 519 \\
\hline
\end{array}$$
Use the chi-square test to determine if the listed occupations and personality preferences are independent at the 0.01 level of significance.

Key Concepts

Chi-Square Test of Independence

The chi-square test of independence is a statistical method used to determine whether there is a significant association between two categorical variables. It compares the observed frequencies in a contingency table to the frequencies expected if the variables were independent, providing a p-value to assess the strength of the evidence against the null hypothesis of no association.

Contingency Table

A contingency table, also known as a cross-tabulation, is a matrix format that displays the frequency distribution of variables. It is used in the chi-square test to organize and visually summarize the counts of observations across the different categories of two categorical variables, which is essential for calculating observed and expected frequencies.

Expected Frequency Calculation

Expected frequencies in a contingency table are calculated by multiplying the marginal totals of the corresponding row and column and then dividing by the overall total number of observations. These expected counts represent the frequencies we would anticipate if there were no association between the variables, serving as a baseline to compare with the observed data.

Degrees of Freedom

Degrees of freedom in the context of the chi-square test of independence are calculated based on the size of the contingency table, typically using the formula (number of rows - 1) multiplied by (number of columns - 1). This value is necessary for determining the appropriate chi-square distribution against which the test statistic is compared.

Significance Level and Hypothesis Testing

The significance level, such as 0.01 in this case, sets the threshold for rejecting the null hypothesis in a statistical test. In hypothesis testing, the null hypothesis states that there is no relationship between the two categorical variables, while the alternative hypothesis suggests an association. The significance level is used to decide whether the observed chi-square statistic indicates enough evidence to reject the null hypothesis.

Transcript

00:01 Okay, so let us get down to business now.

00:03 Now, in all the questions from now on, we have some data, and we have to perform the kai square analysis by ourselves, okay, manually.

00:13 So what do we have over here? over here, we have a table that shows mb personality preferences for a random sample of around 500 people, and they have different professions, different occupations.

00:28 Now, what we have to do is we have to use the kai square test to determine if the listed occupations and personality preferences are independent at 0 .01 level of significance.

00:38 So, our alpha, just a moment, if our alpha is 0 .01.

00:48 All right.

00:52 So let me just draw the table because this is a big question and we are going to need the entire table.

01:01 With us.

01:03 So just a moment.

01:08 All right.

01:09 We have the table with us.

01:10 So we have occupation.

01:13 We have occupation.

01:17 Then we have personality preference.

01:19 Over here we have t.

01:21 Over here we have f.

01:24 Right.

01:25 And let me just write this occupation over here.

01:31 Okay.

01:32 First one is clergy.

01:37 Then we have md.

01:39 Then we have lawyer.

01:43 And then there will be the column.

01:46 Total okay 50 this is 57 this is 91 this is 77 this is 77 this is 82 118 over here we have 94 right over here we will have the row totals the row total and over here we will have the column total what are the row totals if i add them this is 148 this is 159 this is 212 over here we have 252 this is 267 and the overall total is 519 this should be 519 okay now what is the first step in this analysis the first step is finding what we call the expected values okay how do we find e e is going to be, okay, we have e over here, e is going to be row column totals multiplied, a row column totals divided by the grand total.

03:15 So, this is row total multiplied by the column total, multiplied by the column total.

03:30 Okay, this is row total multiplied by the column total upon the grand total upon the grand total.

03:40 Okay, this is going to give us the expected.

03:42 For all of these cells so if i take my calculator and get to work this is going to be let's say for now i'm not going to write the labels let me just write these columns away okay the first row the second row the third row okay the labels will be the same over here we are going to have the occupations these are going to be the personality traits tnf right yeah tnf okay these are to be the personality traits.

04:18 Now i just have to write in the expected values in these middle cells.

04:23 We are going to use this formula.

04:25 So for clergy and t, that is the first cell, my row total is 148, multiplied by column total is 252 divided by 519.

04:37 Okay.

04:41 So my expected value is 71 .86...

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