The following table shows the Myers-Briggs personality preference and area of st
ID: 3236801 • Letter: T
Question
The following table shows the Myers-Briggs personality preference and area of study for a random sample of 519 college students (Applications of the Myers-Briggs Type Indicator in Higher Education, edited by Provost and Anchors). In the table. IN refers to introvert, intuitive; EN refers to extrovert, intuitive; IS refers to introvert, sensing; and ES refer, to extrovert, sensing. Use a chi square test to determine if Myers-Briggs preference type is independent of area of study at the 0.05 level of significance.Explanation / Answer
Solution:-
The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
H0: Myers-Briggs preference type is independent of area of study.
Ha: Myers-Briggs preference type is not independent of area of study.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square test for independence.
Analyze sample data. Applying the chi-square test for independence to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.
DF = (r - 1) * (c - 1) = (4 - 1) * (3 - 1) = 6
Er,c = (nr * nc) / n
E1,1 = (96 * 289) / 519 = 53.457
E1,2 = (96 * 134) / 519 = 24.786
E1,3 = (96 * 96) / 519 = 17.757
E2,1 = (154 * 289) / 519 = 85.75337
E2,2 = (154 * 134) / 519 = 39.761
E2,3 = (154 * 96) / 519 = 28.486
E3,1 = (115 * 289) / 519 = 64.037
E3,2 = (115 * 134) / 519 = 29.692
E3,3 = (115 * 96) / 519 = 21.272
E4,1 = (154 * 289) / 519 = 85.753
E4,2 = (154 * 134) / 519 = 39.761
E4,3 = (154 * 96) / 519 = 28.486
2 = [ (Or,c - Er,c)2 / Er,c ]
2 = (64 - 53.457)^2/53.457 + (15 - 24.786)^2/24.786 + (17 - 17.757)^2/17.757 + (82 - 85.75337)^2/85.75337 + (42 - 39.761)^2/39.761 + (30 - 28.486)^2/28.486 + (68 - 64.037)^2/64.037 + (35 - 29.692)^2/29.692 + (12 - 21.272)^2/21.272 + (75 - 85.753)^2/85.753 + (42 - 39.761)^2/39.761 + (37 - 28.486)^2/28.486
2 = 15.6009
where DF is the degrees of freedom, r is the number of rows, c is the number of columns, nr is the number of observations from level r, nc is the number of observations from level c, n is the number of observations in the sample, Er,c is the expected frequency count, and Or,c is the observed frequency count.
The P-value is the probability that a chi-square statistic having 6 degrees of freedom is more extreme than 15.6009.
We use the Chi-Square Distribution Calculator to find P(2 > 15.6009)
The P-Value is 0.016064. The result is significant at p < 0.05.
Interpret results. Since the P-value (0.016064) is less than the significance level (0.05), we cannot accept the null hypothesis. Thus, we conclude that Myers-Briggs preference type is not independent of area of study.
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