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A job candidate at a large job fair can be classified as unacceptable, provisional, or acceptable. Based on past experience, a high-quality candidate is expected to get 80 percent acceptable ratings, 15 percent provisional ratings, and 5 percent unacceptable ratings. A high-quality candidate was evaluated by 100 companies and received 60 acceptable, 25 provisional, and 15 unacceptable ratings. A chi-square goodness-of-fit-test was conducted to investigate whether the evaluation of the candidate is consistent with past experience. What is the value of the chi-square test statistic and number of degrees of freedom for the test?

$ (a) \chi ^{2} = \dfrac{(15-5)^{2}}{5} + \dfrac{(25-15)^{2}}{15} +\dfrac{(60-80)^{2}}{80} with \: 2df $

$ (b) \chi ^{2} = \dfrac{(15-5)^{2}}{5} + \dfrac{(25-15)^{2}}{15} +\dfrac{(60-80)^{2}}{80} with \: 3df $

$ (c) \chi ^{2} = \dfrac{(15-5)^{2}}{5} + \dfrac{(25-15)^{2}}{15} +\dfrac{(60-80)^{2}}{80} with \: 99df $

$ (d) \chi ^{2} = \dfrac{(5-15)^{2}}{15} + \dfrac{(15-25)^{2}}{25} +\dfrac{(80-60)^{2}}{60} with \: 2df $

$ (e) \chi ^{2} = \dfrac{(5-15)^{2}}{15} + \dfrac{(15-25)^{2}}{25} +\dfrac{(80-60)^{2}}{60} with \: 3df $

This article aims to find the chi-square test statistics. This article uses the concept of chi-square test statistics. The formula for chi-square test statistics is

\[\chi _{c}^{2} = \sum \dfrac{(O_{i} – E_{i})^{2}}{E_{i}} \]

Expert Answer

It is a given that a large job fair is classified as unacceptable, provisional, or acceptable. A high-quality candidate is expected to obtain $80\%$ acceptable, $15\%$ provisional, and $5\%$ unacceptable based on experience.

A quality candidate was evaluated by $100$ companies and received $60$ acceptable, $25$ provisional, and $15$ unacceptable ratings.

The formula for test statistics is given as:

\[\chi ^{2} = \sum _{i= 1}^{n} \dfrac{(O_{i} – E_{i})^{2}}{E_{i}} \]

$ O_{i}$ is the observed frequencies, and $ E_{i}$ is the expected frequencies.

Observed frequencies

Calculate the expected frequencies

Calculate the chi-square test statistic

\[\chi ^{2} = \dfrac{(15-5)^{2}}{5} + \dfrac{(25-15)^{2}}{15} +\dfrac{(60-80)^{2}}{80} \]

\[= \dfrac{400}{80} +\dfrac{100}{15} +\dfrac{100}{5} \]

\[= 5+ 6.667 +20 \]

\[= 31.667\]

Degree of freedom

\[df = (n0.\: of \:categories) – 1\]

\[df = 3-1 =2\]

The chi-square test statistics is $ \chi ^{2} = \dfrac{(15-5)^{2}}{5} + \dfrac{(25-15)^{2}}{15} +\dfrac{(60-80)^{2}}{80} with \: 2df $.

The option $ A$ is correct.

Numerical Result

The chi-square test statistics is $ \chi ^{2} = \dfrac{(15-5)^{2}}{5} + \dfrac{(25-15)^{2}}{15} +\dfrac{(60-80)^{2}}{80} with \: 2df $.

The option $A$ is correct.

Example

A job applicant at a significant job fair may be classified as Unacceptable, Provisional, or Acceptable. Based on experience, a high-quality candidate is expected to receive 80 percent acceptable, 15 percent provisional, and 5 percent unacceptable ratings. A quality candidate was evaluated by 100 companies and received 60 acceptable, 25 provisional, and 15 unacceptable ratings. A chi square goodness-of-fit test was performed to determine whether candidate ratings were consistent with prior experience. What are the value of the chi-square test statistic and the number of degrees of freedom for the test?

$ (a) \chi ^{2} = \dfrac{(20-10)^{2}}{10} + \dfrac{(25-15)^{2}}{15} +\dfrac{(60-80)^{2}}{80} with \: 2df $

Solution

It is a given that a large job fair is classified as unacceptable, provisional, or acceptable. A high-quality candidate is expected to obtain $80\%$ acceptable, $15\%$ provisional, and $5\%$ unacceptable based on experience.

A quality candidate was evaluated by $100$ companies and received $60$ acceptable, $25$ provisional, and $15$ unacceptable ratings.

The formula for test statistics is given as

\[\chi ^{2} = \sum _{i= 1}^{n} \dfrac{(O_{i} – E_{i})^{2}}{E_{i}} \]

$ O_{i}$ is the observed frequencies, and $ E_{i}$ is the expected frequencies.

Observed frequencies

Calculate the expected frequencies

Calculate the chi-square test statistic

\[\chi ^{2} = \dfrac{(20-10)^{2}}{10} + \dfrac{(25-15)^{2}}{15} +\dfrac{(60-80)^{2}}{80} \]

\[= \dfrac{400}{80} +\dfrac{100}{15} +\dfrac{100}{10} \]

\[= 5+ 6.667 +10 \]

\[= 21.667\]

Degree of freedom

\[df = (no.\: of \:categories) – 1\]

\[df = 3-1 =2\]

The chi-square test statistics is $ \chi ^{2} = \dfrac{(20-10)^{2}}{10} + \dfrac{(25-15)^{2}}{15} +\dfrac{(60-80)^{2}}{80} with \: 2df $.

The option $A$ is correct.

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