# 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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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

Step 2

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

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.

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