Question

you are conducting a test of homogeneity for the claim that two different populations have the same proportions of the following two characteristics. use a significance level of 5%. here is the sample data.

find the expected observations. round answers accurate to three decimal places.

what is the chi - square test - statistic for this data? round answer accurate to three decimal places.

what is the p - value for this test - statistic? round answer accurate to four decimal places.

what conclusion can you make about the hypothesis?
we cannot conclude that at least one of the proportions is different.
we can conclude that at least one of the proportions is different.

Explanation:

Step1: Calculate row totals and column totals

Row total for \(A\): \(28 + 51=79\)
Row total for \(B\): \(45+98 = 143\)
Column total for Population \(\#1\): \(28 + 45=73\)
Column total for Population \(\#2\): \(51+98 = 149\)
Grand total: \(79+143=222\)

Step2: Calculate expected values

The formula for expected value \(E_{ij}=\frac{\text{Row total}_i\times\text{Column total}_j}{\text{Grand total}}\)
For \(E_{A1}\) (Category \(A\), Population \(\#1\)): \(E_{A1}=\frac{79\times73}{222}\approx25.995\)
For \(E_{A2}\) (Category \(A\), Population \(\#2\)): \(E_{A2}=\frac{79\times149}{222}\approx53.005\)
For \(E_{B1}\) (Category \(B\), Population \(\#1\)): \(E_{B1}=\frac{143\times73}{222}\approx47.005\)
For \(E_{B2}\) (Category \(B\), Population \(\#2\)): \(E_{B2}=\frac{143\times149}{222}\approx95.995\)

Step3: Calculate chi - square test statistic

The formula for \(\chi^{2}=\sum\frac{(O - E)^{2}}{E}\)
\((O_{A1}-E_{A1})^{2}=(28 - 25.995)^{2}=4.02\)
\(\frac{(O_{A1}-E_{A1})^{2}}{E_{A1}}=\frac{4.02}{25.995}\approx0.155\)
\((O_{A2}-E_{A2})^{2}=(51 - 53.005)^{2}=4.02\)
\(\frac{(O_{A2}-E_{A2})^{2}}{E_{A2}}=\frac{4.02}{53.005}\approx0.076\)
\((O_{B1}-E_{B1})^{2}=(45 - 47.005)^{2}=4.02\)
\(\frac{(O_{B1}-E_{B1})^{2}}{E_{B1}}=\frac{4.02}{47.005}\approx0.086\)
\((O_{B2}-E_{B2})^{2}=(98 - 95.995)^{2}=4.02\)
\(\frac{(O_{B2}-E_{B2})^{2}}{E_{B2}}=\frac{4.02}{95.995}\approx0.042\)
\(\chi^{2}=0.155 + 0.076+0.086 + 0.042=0.359\)

Step4: Calculate degrees of freedom and P - value

Degrees of freedom \(df=(r - 1)(c - 1)=(2 - 1)(2 - 1)=1\)
Using a chi - square table or calculator, for \(\chi^{2}=0.359\) and \(df = 1\), the \(P-\text{value}\approx0.5493\)

Step5: Make a conclusion

Since \(P-\text{value}=0.5493>0.05\) (significance level), we fail to reject the null hypothesis.

Answer:

Expected values:

Category	Population #1	Population #2
B	47.005	95.995

\(\chi^{2}=0.359\)
\(P - \text{value}=0.5493\)
We cannot conclude that at least one of the proportions is different.