 # A random sample of U.S. adults was recently asked, "Would you support or oppose major Khadija Wells 2021-01-02 Answered

A random sample of U.S. adults was recently asked, "Would you support or oppose major new spending by the federal government that would help undergraduates pay tuition at public colleges without needing loans?" The two-way table shows the responses, grouped by age.

Do these data provide convincing evidence of an association between age and opinion about loan-free tuition in the population of U.S. adults?

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Given:

$\begin{array}{lrrrrr}& 18-34& 35-49& 50-64& 65+& Total\\ Support& 91& 161& 272& 332& 856\\ Oppose& 25& 74& 211& 255& 565\\ Do{n}^{\prime }tknow& 4& 13& 20& 51& 88\\ Total& 120& 248& 503& 638& 1509\end{array}$

The null hypothesis states that there is no difference in the distribution of the categorical variable for each of the populations/treatments. The alternative hypotesis states that there is a difference. ${H}_{0}$ : There is no association between age and response. ${H}_{\alpha }$: There is an association between age and response. The expected frequencies E are the product of the column and row total, divided by the table total. ${E}_{11}=\frac{{r}_{1}×{c}_{1}}{n}=\frac{856×120}{1509}\approx 68.07$
${E}_{12}=\frac{{r}_{1}×{c}_{2}}{n}=\frac{856×248}{1509}\approx 140.68$
${E}_{13}=\frac{{r}_{1}×{c}_{3}}{n}=\frac{856×503}{1509}\approx 285.33$
${E}_{14}=\frac{{r}_{1}×{c}_{4}}{n}=\frac{856×638}{1509}\approx 361.91$
${E}_{21}=\frac{{r}_{2}×{c}_{1}}{n}=\frac{565×120}{1509}\approx 44.93$
${E}_{22}=\frac{{r}_{2}×{c}_{2}}{n}=\frac{565×248}{1509}\approx 92.86$
${E}_{23}=\frac{{r}_{2}×{c}_{3}}{n}=\frac{565×503}{1509}\approx 188.33$
${E}_{24}=\frac{{r}_{2}×{c}_{4}}{n}=\frac{565×638}{1509}\approx 238.88$
${E}_{31}=\frac{{r}_{3}×{c}_{1}}{n}=\frac{88×120}{1509}\approx 7.00$
${E}_{32}=\frac{{r}_{3}×{c}_{2}}{n}=\frac{88×248}{1509}\approx 14.46$