The following table shows the Myers-Briggs personality preference and area of study for a random sample of 519 college students. In the table, IN refers to introvert, intuitive; EN refers to extrovert, intuitive; IS refers to introvert, sensing; and ES refers to extrovert, sensing. Myers-Briggs Preference Arts & Science Business Allied Health Row Total IN 66 13 17 96 EN 77 47 30 154 IS 62 32 21 115 ES 76 37 41 154 Column Total 281 129 109 519 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. (a) What is the level of significance? State the null and alternate hypotheses. H0: Myers-Briggs type and area of study are not independent. H1: Myers-Briggs type and area of study are not independent.H0: Myers-Briggs type and area of study are independent. H1: Myers-Briggs type and area of study are independent. H0: Myers-Briggs type and area of study are independent. H1: Myers-Briggs type and area of study are not independent.H0: Myers-Briggs type and area of study are not independent. H1: Myers-Briggs type and area of study 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 5? YesNo What sampling distribution will you use? chi-square uniform normal Student's tbinomial What are the degrees of freedom? (c) Find or estimate the P-value of the sample test statistic. (Round your answer to three decimal places.) p-value > 0.1000 .050 < p-value < 0.100 0.025 < p-value < 0.0500 .010 < p-value < 0.0250 .005 < p-value < 0.010 p-value < 0.005 (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence? Since the P-value > ?, we fail to reject the null hypothesis. Since the P-value > ?, we reject the null hypothesis. Since the P-value ≤ ?, we reject the null hypothesis. Since the P-value ≤ ?, we fail to reject the null hypothesis. (e) Interpret your conclusion in the context of the application. At the 5% level of significance, there is sufficient evidence to conclude that Myers-Briggs type and area of study are not independent. At the 5% level of significance, there is insufficient evidence to conclude that Myers-Briggs type and area of study are not in

The following table shows the Myers-Briggs personality preference and area of study for a random sample of 519 college students. In the table, IN refers to introvert, intuitive; EN refers to extrovert, intuitive; IS refers to introvert, sensing; and ES refers to extrovert, sensing. Myers-Briggs Preference Arts & Science Business Allied Health Row Total IN 66 13 17 96 EN 77 47 30 154 IS 62 32 21 115 ES 76 37 41 154 Column Total 281 129 109 519 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. (a) What is the level of significance? State the null and alternate hypotheses. H0: Myers-Briggs type and area of study are not independent. H1: Myers-Briggs type and area of study are not independent.H0: Myers-Briggs type and area of study are independent. H1: Myers-Briggs type and area of study are independent. H0: Myers-Briggs type and area of study are independent. H1: Myers-Briggs type and area of study are not independent.H0: Myers-Briggs type and area of study are not independent. H1: Myers-Briggs type and area of study 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 5? YesNo What sampling distribution will you use? chi-square uniform normal Student's tbinomial What are the degrees of freedom? (c) Find or estimate the P-value of the sample test statistic. (Round your answer to three decimal places.) p-value > 0.1000 .050 < p-value < 0.100 0.025 < p-value < 0.0500 .010 < p-value < 0.0250 .005 < p-value < 0.010 p-value < 0.005 (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence? Since the P-value > ?, we fail to reject the null hypothesis. Since the P-value > ?, we reject the null hypothesis. Since the P-value ≤ ?, we reject the null hypothesis. Since the P-value ≤ ?, we fail to reject the null hypothesis. (e) Interpret your conclusion in the context of the application. At the 5% level of significance, there is sufficient evidence to conclude that Myers-Briggs type and area of study are not independent. At the 5% level of significance, there is insufficient evidence to conclude that Myers-Briggs type and area of study are not in

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Description: The following table shows the Myers-Briggs personality preference and area of study for a random sample of 519 college students. In the table, IN refers to introvert, intuitive; EN refers to extrovert, intuitive; IS refers to introvert, sensing; and

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Myers-Briggs Preference	Arts & Science	Business	Allied Health	Row Total
IN	66	13	17	96
EN	77	47	30	154
IS	62	32	21	115
ES	76	37	41	154
Column Total	281	129	109	519

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.

(a) What is the level of significance?

State the null and alternate hypotheses.

H₀: Myers-Briggs type and area of study are not independent.
H₁: Myers-Briggs type and area of study are not independent.H₀: Myers-Briggs type and area of study are independent.
H₁: Myers-Briggs type and area of study are independent. H₀: Myers-Briggs type and area of study are independent.
H₁: Myers-Briggs type and area of study are not independent.H₀: Myers-Briggs type and area of study are not independent.
H₁: Myers-Briggs type and area of study 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 5?

YesNo

What sampling distribution will you use?

chi-square

uniform

normal

Student's tbinomial

What are the degrees of freedom?

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

p-value > 0.1000

.050 < p-value < 0.100

0.025 < p-value < 0.0500

.010 < p-value < 0.0250

.005 < p-value < 0.010

p-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence?

Since the P-value > ?, we fail to reject the null hypothesis.

Since the P-value > ?, we reject the null hypothesis.

Since the P-value ≤ ?, we reject the null hypothesis.

Since the P-value ≤ ?, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, there is sufficient evidence to conclude that Myers-Briggs type and area of study are not independent.

At the 5% level of significance, there is insufficient evidence to conclude that Myers-Briggs type and area of study are not in

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