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StatisticsQ&A Library36) 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 < 7.815Question

Asked Dec 2, 2019

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Step 1

The null and alternative hypotheses can be written as follows:

*Null hypothesis:*

H_{0}: The student’s plan is distributed same as expected.

*Alternative hypothesis:*

H_{1}: 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 *p*_{1 }= 0.50, *p*_{2 }= 0.20, *p*_{3 }= 0.30, are respectively

The formula for expected count is,

Expected count = *n*×*p*_{i}

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. - Click the
**Shade...**

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