 # 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.

$$\begin{array}{ccc} & Age \ Response & {\begin{array}{l|r|r|r|r|r} & 18-34 & 35-49 & 50-64 & 65+ & Total \\ \hline Support & 91 & 161 & 272 & 332 & 856 \\ \hline Oppose & 25 & 74 & 211 & 255 & 565 \\ \hline Don't know & 4 & 13 & 20 & 51 & 88 \\ \hline Total & 120 & 248 & 503 & 638 & 1509 \end{array}} \ \end{array}$$

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:

$$\begin{array}{l|r|r|r|r|r} & 18-34 & 35-49 & 50-64 & 65+ & Total \\ \hline Support & 91 & 161 & 272 & 332 & 856 \\ \hline Oppose & 25 & 74 & 211 & 255 & 565 \\ \hline Don't know & 4 & 13 & 20 & 51 & 88 \\ \hline Total & 120 & 248 & 503 & 638 & 1509 \end{array}$$
$$\alpha=\text{Sign if icence level}=0.05\ \text{as sumption}$$

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. $$\displaystyle{H}_{{0}}$$ : There is no association between age and response. $$\displaystyle{H}_{\alpha}$$: 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. $$\displaystyle{E}_{{{11}}}={\frac{{{r}_{{1}}\times{c}_{{1}}}}{{{n}}}}={\frac{{{856}\times{120}}}{{{1509}}}}\approx{68.07}$$
$$\displaystyle{E}_{{{12}}}={\frac{{{r}_{{1}}\times{c}_{{2}}}}{{{n}}}}={\frac{{{856}\times{248}}}{{{1509}}}}\approx{140.68}$$
$$\displaystyle{E}_{{{13}}}={\frac{{{r}_{{1}}\times{c}_{{3}}}}{{{n}}}}={\frac{{{856}\times{503}}}{{{1509}}}}\approx{285.33}$$
$$\displaystyle{E}_{{{14}}}={\frac{{{r}_{{1}}\times{c}_{{4}}}}{{{n}}}}={\frac{{{856}\times{638}}}{{{1509}}}}\approx{361.91}$$
$$\displaystyle{E}_{{{21}}}={\frac{{{r}_{{2}}\times{c}_{{1}}}}{{{n}}}}={\frac{{{565}\times{120}}}{{{1509}}}}\approx{44.93}$$
$$\displaystyle{E}_{{{22}}}={\frac{{{r}_{{2}}\times{c}_{{2}}}}{{{n}}}}={\frac{{{565}\times{248}}}{{{1509}}}}\approx{92.86}$$
$$\displaystyle{E}_{{{23}}}={\frac{{{r}_{{2}}\times{c}_{{3}}}}{{{n}}}}={\frac{{{565}\times{503}}}{{{1509}}}}\approx{188.33}$$
$$\displaystyle{E}_{{{24}}}={\frac{{{r}_{{2}}\times{c}_{{4}}}}{{{n}}}}={\frac{{{565}\times{638}}}{{{1509}}}}\approx{238.88}$$
$$\displaystyle{E}_{{{31}}}={\frac{{{r}_{{3}}\times{c}_{{1}}}}{{{n}}}}={\frac{{{88}\times{120}}}{{{1509}}}}\approx{7.00}$$
$$\displaystyle{E}_{{{32}}}={\frac{{{r}_{{3}}\times{c}_{{2}}}}{{{n}}}}={\frac{{{88}\times{248}}}{{{1509}}}}\approx{14.46}$$
$$\displaystyle{E}_{{{33}}}={\frac{{{r}_{{3}}\times{c}_{{3}}}}{{{n}}}}={\frac{{{88}\times{503}}}{{{1509}}}}\approx{29.33}$$
$$\displaystyle{E}_{{{34}}}={\frac{{{r}_{{3}}\times{c}_{{4}}}}{{{n}}}}={\frac{{{88}\times{638}}}{{{1509}}}}\approx{37.21}$$

Conditions The conditions for performing a chi-square test of homogeneity /independence are: Random, Independent (10%), Large counts. Random: Satisfied, because the sample is a random sample. Independent: Satisfied, because the 1509 U.S. adults are less than 10% of all U.S. adults (since there are more than 15090 U.S. adults) Large counts: Satisfied, because all expected counts are at least 5. Since all conditions are satisfied, it is appropriate to carry out a test of homogeneity /independence. Hypothesis test The chi-square subtotals are the squared differences between the observed and expected frequencies, divided by the expected frequency. The value of the test-statistic is then the sum of the chi-square subtotals: $$\chi^2 = \sum \frac{(O - E)^2}{E}$$
$$\displaystyle={\frac{{{\left({91}-{68.07}\right)}^{{2}}}}{{{68.07}}}}+{\frac{{{\left({161}-{140.68}\right)}^{{2}}}}{{{140.68}}}}+{\frac{{{\left({272}-{285.33}\right)}^{{2}}}}{{{285.33}}}}+{\frac{{{\left({332}-{361.91}\right)}^{{2}}}}{{{361.91}}}}+{\frac{{{\left({25}-{44.93}\right)}^{{2}}}}{{{44.93}}}}+{\frac{{{\left({74}-{92.86}\right)}^{{2}}}}{{{92.86}}}}+{\frac{{{\left({211}-{188.33}\right)}^{{2}}}}{{{188.33}}}}+{\frac{{{\left({255}-{238.88}\right)}^{{2}}}}{{{238.88}}}}+{\frac{{{\left({4}-{7.00}\right)}^{{2}}}}{{{7.00}}}}+{\frac{{{\left({13}-{14.46}\right)}^{{2}}}}{{{14.46}}}}+{\frac{{{\left({20}-{29.33}\right)}^{{2}}}}{{{29.33}}}}+{\frac{{{\left({51}-{37.21}\right)}^{{2}}}}{{{37.21}}}}$$
$$\displaystyle\approx{39.7549}$$ The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of the chi-square distribution table in the appendix containing the $$\displaystyle\chi^{{2}}$$ -value in the row
$$\displaystyle{d}{f}={\left({r}-{1}\right)}{\left({c}-{1}\right)}={\left({3}-{1}\right)}{\left({4}-{1}\right)}={6}:$$
$$\displaystyle{P}{<}{0.0005}$$ Command $$\displaystyle{T}{I}\frac{{83}}{{84}}-{c}{a}{l}{c}\underline{{a}}\to{r}:\chi^{{2}}{c}{d}{f{{\left({39.7549},{1}{E}{99},{6}\right)}}}$$ which results in a P-value of 0. Note: You can replace 1E99 by any other very large number. If the P-value is less than or equal to the significance level, then the null hypothesis is rejected: $$\displaystyle{P}{<}{0.05}\Rightarrow{R}{e}{j}{e}{c}{t}{H}_{{0}}$$ There is convincing evidence that there is an association between age and response.