$ (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.