The following table shows the Myers-Briggs personality preferences for a random sample of 406 people in the listed professions. E refers to extroverted and I refers to introverted.Personality TypeОссирationClergy (al denominations)ERow Total5948107M.D.6498162Lawyer5285137Column Total175231406A USE SALTUse the chi-square test to determine it 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.OH,: Myers-Briggs preference and profession are not independentH,: Myers-Briggs preference and profession are independent.O Hg: Myers-Briggs preference and profession are independentH: Myers-Briggs preference and profession are independent.OH: Myers-Briggs preference and profession are not independentH: Myers-Briggs preference and profession are not independent.O Ho: Myers-Briggs preference and profession are independentH: Myers-Briggs preference and profession are not 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 frequendies greater than 5?O YesO NoWhat sampling distribution will you use?O binomialO normalO uniformO chi-squareO Student'sEWhat are the degrees of freedom?(c) Find or estimate the Pvalue of the sample test statistic.O p-value > 0.100O 0.050 < p-value < 0.100O 0.025 < p-value < 0.050O 0.010 < p-value < 0.025O 0.005 < p-value < 0.010O prvalue < 0.005(d) Based on your answers in parts (a) to (C), wil you reject or fail to reject the null hypothesis of independence?O Since the Pvalue > a, we fail to reject the nul hypothesis.O Since the Pvalue > a, we reject the nul hypothesis.O Since the Pvalue s a, we reject the null hypothesis.O Since the Pvalue sa, we fail to reject the nul hypothesis.(e) Interpret your condusion in the context of the application.O At the 5% level of significance, there is insufficient evidence to conclude that Myers-Briggs preference and the profession are not independent.O At the 5% level of significance, there is sufficient evidence to conclude that Myers-Briggs preference and the profession are not independent.

The chi-square test of independence can be used to determine whether the Myers-Briggs preferences and profession are independent or not. In this test, there will be two types of counts, one is expected counts (denoted by E) and other is observed or actual counts (denoted by O). The test will determine whether the actual counts are significantly different from the expected count or not. If the test shows that there are evidence of significant difference in between them, then it can be concluded that the test is significant.(a): The level of significance is 5%, that is, α=0.05. The null hypothesis for the test can be given as, H0: Myers-Briggs preferences and profession are independent, and the alternate hypothesis for the test can be given as, Ha: Myers-Briggs preferences and profession are not independent. Hence, the last option is the correct option.(b): The observed counts O for each category or cell are mentioned in the table. To calculate the expected count E for all the 6 categories, use the formula, Expected counts=(Row total)(Column total)Sample size. The expected count/frequency for each category is mentioned in the table below after calculation. Extroverted Introverted Clergy 46.121 60.879 M. D. 69.828 92.172 Lawyer 59.052 77.948 The value of test statistics can be calculated by using the formula, χ2=∑(Observed-Expected)2Expected. Substitute the expected and the observed frequencies for each category in the above mentioned formula for χ2 and calculate the test statistic. Calculation gives χ2≈8.656. It can observe from the table that all the expected frequencies are greater than 5. The sampling distribution will be chi-square distribution with degrees of freedom d.f.=(r-1)(c-1). where, r is number of rows of contingency table. c is number of columns of contingency table. Calculate the required d.f. by substituting r=3 and c=2 in the above-mentioned formula for d.f. Calculation gives d.f.=2. d.f.=(r-1)(c-1)=3-12-1=2(c): To find the p-value, use the chi-square distribution with d.f.=2 and evaluate the proportion of data that lies beyond test statistic χ2≈8.656. It gives p-value=0.013. Hence, the correct option is 0.010<p-value<0.025. (d): The level of significance is 5%, that is, α=0.05. Since, p-value<α, this indicates that there are sufficient evidences against the null hypothesis. Therefore, the null hypothesis is rejected. Hence, the correct option is the third option. (e): As the null hypothesis gets rejected, it can be concluded that at a significance level of 5%, it can be said that Myers-Briggs preferences and profession are not independent. Hence, the correct option is the second option.

Math

Statistics

The following table shows the Myers-Briggs personality preferences for a random sample of 406 people in the listed professions. E refers to extroverted and I refers to introverted. Personality Tvee Occupation Clery fal denominatians) M.D. Lawyer Column Tetal Row Total 107 162 137 406 48 59 64 52 175 98 231 A USE SALT Use the chi-square test to determine r the listed occupations and persenality preferences are independent at the 0.05 level of significance. a) what is the level of significance? State the null and alternate hypotheses. OH Myers-Briggs preference and profession are not indepandent H,: Myers-Briggs preference and profession are independent. O Hg: Myers-Briggs preference and profession are independent H: Myers-riggs preference and pratession are independent. OH Myers-Briggs praference and profession are nat independent H Myers-riggs preference and pratession are not independent. OH Myers-Briggs preference and profession are independent H: Myers-Briggs prference and profession are not independent. (B) Find the value of the chi-square statistic for the sample. (Round the expectad frequencies to at least three decimal places. Round the test statistic to three decimal places.) Are all the expected frequendes greater than 57 O ves O No What sampling distribution will you use? O binomial O narmal O unitorm O d-sqare O Student's t What are the degrees of freedom? (C) Find or estimate the Prvalue of the sample test statistic. Opvalue>.100 O 0.oso pvalue <0.100 O 0.025 4, we fal to reject the null hypothesis. O Since the Pvalue >4, we reject the nul hypothesis. O Since the Pvaluesa, we reject the nul hypothesis. O Since the Avaluesa, we fal to reject the null hypothesis. (e) Interpret your condusion in the context of the application. O At the S% level of significance, there is insuficient evidence to conclude that Myers-Briggs preference and the profession are not independent. O At the S% level of significance, there is sufficient evidence to conclude that Myers-Briggs preference and the profession are not independent.