The following table shows the MyersBriggs personality prefer

The following table shows the Myers-Briggs personality preferences for a random sample of 406 people in the listed professions. table shows the Myers-Briggs personality preferences for a random sample of 406 people in the isted professions. E refers to extroverted and I refers to introverted. Personality Type Occupation Cl Clergy (all denominations) M.D Column Total Row Total 107 162 63 59 8 137 406 Use the chi-square test to determine if the listed occupations and personality preferences are independent at the 0.05 level of significance (a) What is the level of significance? State the null and alternate hypotheses. O Ho: Myers-Briggs preference and profession are independent H1: Myers-Briggs preference and profession are not independent. O Ho: Myers-Briggs preference and profession are not independent Hy: Myers-Briggs preference and profession are independent. O Ho: Myers-Briggs preference and profession are not independent H1: Myers-Briggs preference and profession are not independent. Ho: Myers-Briggs preference and profession are independent H1: Myers-Briggs preference and profession are independent. (b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to at least three decimal places. Round the test statistic to three decimal places.) Are all the expected frequencies greater than 57 O Yes O No What sampling distribution will you use? O normal O chi-square O Student\'st O binomial O uniform What are the degrees of freedom? (c) Find or estimate the P-value of the sample test statistic. (Round your answer to yhree decimal places.) (d) Based on your answers in parts (a) to (c), will you rejoct or fail to reject the null hypothesis of independence? O Since the P-value > a, we fail to reject the nüli hypothesis. since the P-value > , we reject the null hypothesis O Since the P-value s a, we reject the null hypothesis. Sirice the P-value acr, we fail to reject the null hypothesis.

Solution

(a) What is the level of significance? = 5%

b)Find the value of the chi-square statistic for the sample. (Round the expected frequencies to at least three decimal places. Round the test statistic to three decimal places. ChiSq = 14.610

What are the degrees of freedom?
Ans: 2

c)Find or estimate the P-value of the sample test statistic. (Round your answer to three decimal places.)
Ans: 0.000672