$ (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**Â 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,**Â **provisional,**Â or **acceptable**. A **high-quality candidate**Â is expected to obtain $80\%$ acceptable, $15\%$ provisional, and $5\%$ unacceptable based on experience.

A **quality candidate**Â was evaluated by $100$ companies and received $60$ **acceptabl****e**, $25$ **provisional**, and $15$**Â unacceptable ratings**.

The **formula for test statistics**Â 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**Â 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**Â 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,**Â **provisional,**Â or **acceptable**. A **high-quality candidate**Â is expected to obtain $80\%$ acceptable, $15\%$ provisional, and $5\%$ unacceptable based on experience.

A **quality candidate**Â was evaluated by $100$ companies and received $60$ **acceptabl**e, $25$ **provisional**, and $15$**Â unacceptable ratings**.

The **formula for test statistics**Â 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.**