Step 1

From the provided information,

The hypotheses can be constructed as:

\(H_0\): There is no association between gender and preferred award.

\(H_1\): There is association between gender and preferred award.

Assume the level of significance \((\alpha) = 0.05\)

Step 2

The expected frequencies can be obtained as:

\(E_11=\frac{\text{Row total} \times \text{Column total}}{\text{Total}} =\frac{169 \times 31}{362}\)

\(=14.47\)

\(E_12=\frac{169 \times 149}{362}\)

\(=69.56\)

\(E_13=\frac{169 \times182}{362}\)

\(=84.97\)

\(E_21=\frac{193 \times 31}{362}\)

\(=16.53\)

\(E_22=\frac{193 \times 149}{362}\)

\(=79.44\)

\(E_23=\frac{193 \times 182}{362}\)

\(=97.03\) Step 3 The value of chi square test statistic can be obtained as: \(x^2=\sum \frac{(O-E)^2}{E}\)

\(=\left[ \frac{(20-14.47)^2}{14.47}+\frac{(76-69.56)^2}{69.56}+\frac{(73-84.97)^2}{84.97}+\frac{(11-16.53)^2}{16.53}+\frac{(73-79.44)^2}{79.44}+\frac{(109-97.03)^2}{97.03}\right]\)

the chi square test statistic is 8.245

Step 4

The degree of freedom \(=(r-1) \cdot (c-1) = (2-1) \cdot (3-1) = 2\) The p value of the test statistic with 2 degree of freedom from the chi square table is 0.0162 which is less than level of significance, therefore, the null hypothesis would reject and it can be concluded that there is sufficient evidence to support the claim that there is some association between gender and preferred award.

Yes, the test indicates an association between gender and preferred award.

From the provided information,

The hypotheses can be constructed as:

\(H_0\): There is no association between gender and preferred award.

\(H_1\): There is association between gender and preferred award.

Assume the level of significance \((\alpha) = 0.05\)

Step 2

The expected frequencies can be obtained as:

\(E_11=\frac{\text{Row total} \times \text{Column total}}{\text{Total}} =\frac{169 \times 31}{362}\)

\(=14.47\)

\(E_12=\frac{169 \times 149}{362}\)

\(=69.56\)

\(E_13=\frac{169 \times182}{362}\)

\(=84.97\)

\(E_21=\frac{193 \times 31}{362}\)

\(=16.53\)

\(E_22=\frac{193 \times 149}{362}\)

\(=79.44\)

\(E_23=\frac{193 \times 182}{362}\)

\(=97.03\) Step 3 The value of chi square test statistic can be obtained as: \(x^2=\sum \frac{(O-E)^2}{E}\)

\(=\left[ \frac{(20-14.47)^2}{14.47}+\frac{(76-69.56)^2}{69.56}+\frac{(73-84.97)^2}{84.97}+\frac{(11-16.53)^2}{16.53}+\frac{(73-79.44)^2}{79.44}+\frac{(109-97.03)^2}{97.03}\right]\)

the chi square test statistic is 8.245

Step 4

The degree of freedom \(=(r-1) \cdot (c-1) = (2-1) \cdot (3-1) = 2\) The p value of the test statistic with 2 degree of freedom from the chi square table is 0.0162 which is less than level of significance, therefore, the null hypothesis would reject and it can be concluded that there is sufficient evidence to support the claim that there is some association between gender and preferred award.

Yes, the test indicates an association between gender and preferred award.