36) A random group of seniors was selected from a university and asked about their plans forthe following year. The school advising office claims that 50% of the students plan to36)work, 20% of the students plan to continue in school, and 30% of the students plan totake some time off. Is there evidence to reject this hypothesis at a 0.05?1PlansWorkSchoolTime offNumber of students2113A) There is not evidence to reject the claim that the students' plans are distributed asclaimed because the test value 4.198 < 5.991B) There is evidence to reject the claim that the students' plans are distributed asclaimed because the test value 4.198 < 5.991) There is not evidence to reject the claim that the students' plans are distributed asclaimed because the test value 4.198 < 7.815D) There is evidence to reject the claim that the students' plans are distributed asclaimed because the test value 4.198 < 7.815

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Asked Dec 2, 2019
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36) A random group of seniors was selected from a university and asked about their plans for
the following year. The school advising office claims that 50% of the students plan to
36)
work, 20% of the students plan to continue in school, and 30% of the students plan to
take some time off. Is there evidence to reject this hypothesis at a 0.05?
1
Plans
Work
School
Time off
Number of students
21
13
A) There is not evidence to reject the claim that the students' plans are distributed as
claimed because the test value 4.198 < 5.991
B) There is evidence to reject the claim that the students' plans are distributed as
claimed because the test value 4.198 < 5.991
) There is not evidence to reject the claim that the students' plans are distributed as
claimed because the test value 4.198 < 7.815
D) There is evidence to reject the claim that the students' plans are distributed as
claimed because the test value 4.198 < 7.815
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36) A random group of seniors was selected from a university and asked about their plans for the following year. The school advising office claims that 50% of the students plan to 36) work, 20% of the students plan to continue in school, and 30% of the students plan to take some time off. Is there evidence to reject this hypothesis at a 0.05? 1 Plans Work School Time off Number of students 21 13 A) There is not evidence to reject the claim that the students' plans are distributed as claimed because the test value 4.198 < 5.991 B) There is evidence to reject the claim that the students' plans are distributed as claimed because the test value 4.198 < 5.991 ) There is not evidence to reject the claim that the students' plans are distributed as claimed because the test value 4.198 < 7.815 D) There is evidence to reject the claim that the students' plans are distributed as claimed because the test value 4.198 < 7.815

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Expert Answer

Step 1

The null and alternative hypotheses can be written as follows:

Null hypothesis:

H0: The student’s plan is distributed same as expected.

Alternative hypothesis:

H1: The student’s plan is distributed different from expected.

The expected frequencies are calculated as follows:.

From the given information the observed frequency is 21, 13 and 8 for the specific proportions p1 = 0.50, p2 = 0.20, p3 = 0.30, are respectively

The formula for expected count is,

Expected count = n×pi

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Frequency Expected count = n»xpi 42(0.50) 21 42(0.20) 8.4 Plans Work 21 13 School 42(0.30) 12.6 Time off 8 Total n= 42

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Step 2

The test statistics for the given data is given and calculated as follows:

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Observed-Expected) Expected (21-21) (13-8.4)(8-12.6) 21 8.4 12.6 0.000+2.519 1.679 =4.198

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Step 3

The test statistics X02= 4.198.

The degrees of freedom is obtained below:

Df = n–1

     = (3–1)

     = 2

Step-by-step procedure to conduct chi-square test using MINITAB software:

  • Choose Graph > Probability Distribution Plot choose View Probability > OK.
  • From Distribution, choose ‘Chi-square’ distribution.
  • In Degrees of freedom, enter 2.
  • Click the Shade...
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Distribution Plot Chi-Square, df2 0.5 0.4 0.3 0.2 0.1 0.05 0.0 5.991 0 х Density

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