The following table shows the Myers-Briggs personality preferences for a random sample of 407 people in the listed professions. Occupation | Extroverted | Introverted | Row Total Clergy (all denominations) | 63 | 43 | 106 M.D. | 70 | 91 | 161 Lawyer | 56 | 84 | 140 Column total | 189 | 218 | 407 Use the chi-square test to determine if the listed occupations and personality preferences are independent at the 0.01 level of significance. Find (or estimate) the P-value of the sample test statistic. P-Value < 0.005 0.01 < P-Value < 0.025 P-Value > 0.5 0.005 < P-Value < 0.01 0.10 < P-Value < 0.25
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For Extroverted Clergy: (106 * 189) / 407 = 49.22 For Introverted Clergy: (106 * 218) / 407 = 56.78 For Extroverted M.D: (161 * 189) / 407 = 74.76 For Introverted M.D: (161 * 218) / 407 = 86.24 For Extroverted Lawyer: (140 * 189) / 407 = 65.01 For Introverted Show more…
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The following table shows the Myers-Briggs personality preferences for a random sample of 409 people in the listed professions: Occupation | Extroverted | Introverted | Row Total Clergy (all denominations) | 65 | 43 | 108 M.D. | 68 | 94 | 162 Lawyer | 57 | 82 | 139 Column total | 190 | 219 | 409 Use the chi-square test to determine if the listed occupations and personality preferences are independent at α = 0.1. What is the level of significance? 0.05 0.10 0.90 0.01 0.95
Adi S.
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}$$Use the chi-square test to determine if the listed occupations and personality preferences are independent at the 0.05 level of significance.
Chi-Square and F Distributions
Chi-Square: Tests of Independence and of Homogeneity
Please provide the following information. (a) What is the level of significance? State the null and alternate hypotheses. (b) Find the value of the chi-square statistic for the sample. Are all the expected frequencies greater than $5 ?$ What sampling distribution will you use? What are the degrees of freedom? (c) Find or estimate the $P$ -value of the sample test statistic. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence? (e) Interpret your conclusion in the context of the application. Use the expected values $E$ to the hundredths place. Psychology: Myers-Briggs The following table shows the Myers-Briggs personality preferences for a random sample of 519 people in the listed professions (Atlas of Type Tables by Macdaid, McCaulley, and Kainz). T refers to thinking and $\mathrm{F}$ refers to feeling. $$ \begin{array}{l|c|c|c} \hline & \multicolumn{2}{c} {\text { Personality Preference Type }} & \\ \cline { 2 - 3 } \text { Occupation } & \multicolumn{1}{c} {\text { T }} & \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.
Chi-Square and $F$ Distributions
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