{"id":126,"date":"2018-10-31T18:05:15","date_gmt":"2018-10-31T22:05:15","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/simplestats\/?post_type=chapter&p=126"},"modified":"2019-11-02T21:14:17","modified_gmt":"2019-11-03T01:14:17","slug":"9-3-the-chi-square","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/simplestats\/chapter\/9-3-the-chi-square\/","title":{"raw":"9.3 Between Two Discrete Variables: The \u03c72, Part 1","rendered":"9.3 Between Two Discrete Variables: The \u03c72, Part 1"},"content":{"raw":"[latexpage]\r\n\r\nAs in the previous section, here you need to recall how we examine potential association between two variables both treated as discrete (Section 7.2.2, https:\/\/pressbooks.bccampus.ca\/simplestats\/chapter\/7-2-2-between-two-discrete-variables\/<\/a>). We described such associations through contingency tables, reporting differences of proportions as appropriate.\r\n\r\n \r\n\r\nWe can start with the simplest, binary case: when the discrete variables have two groups each. Then we compare the groups of interest (categories of one variable) on one of the categories of the other variable. (The example in Chapter 7 we used was to compare the percentage of first-year students who like the campus cafeteria to the percentage of second-year students who do.)\r\n\r\n \r\n\r\nThe t<\/em>-test for testing difference of two<\/em> proportions.<\/strong> When we have only two proportions (or percentages) to compare, we can actually use the same t<\/em>-test we used for testing differences of means, again treating the difference<\/em> as a single, normally distributed statistic. Since we have categorical variables, however, and no standard deviations\/variances, we resort to measuring population variability by \u03c0(1-\u03c0) and sample variability by\u00a0p<\/em>(1-p<\/em>)[footnote]Do not forget that p<\/em> here stands for proportion<\/em>, not probability\/p<\/em>-value<\/em>.[\/footnote]. (See Section 6.7.2, https:\/\/pressbooks.bccampus.ca\/simplestats\/chapter\/6-7-2-confidence-intervals-for-proportions\/<\/a>.) We can thus simply substitute that into the formula for z<\/em>:\r\n\r\n \r\n\r\n$z=\\frac{(p_1 -p_2)-(\\pi_1 -\\pi_2 )}{\\sqrt{\\frac{\\pi_1(1-\\pi_1)}{N_1}+\\frac{\\pi_2(1-\\pi_2)}{N_2}}}$\r\n\r\n \r\n\r\nwhere, of course, under the null hypothesis $(\\pi_1 -\\pi_2 )=0$. Then, using the sample proportions leaves us with t<\/em>:\r\n\r\n \r\n\r\n$t=\\frac{(p_1 -p_2)}{\\sqrt{\\frac{p_1(1-p_1)}{N_1}+\\frac{p_2(1-p_2)}{N_2}}}$\r\n\r\n \r\n\r\nAgain, under the null hypothesis the two groups' proportions are assumed to be the same so effectively we have:\r\n\r\n \r\n\r\n$t=\\frac{(p_1 -p_2)}{\\sqrt{p(1-p)(\\frac{1}{N_1}+\\frac{2}{N_2})}}$\r\n\r\n \r\n\r\nLet's revisit the cafeteria-preferences example from Section 7.2.2 to see how the t<\/em>-test for testing difference of proportions works.\r\n\r\n \r\n
\r\n

Example 9.3 Do You Like the Campus Cafeteria? (A t-Test)<\/em><\/p>\r\n\r\n<\/header>\r\n

\r\n\r\n \r\n\r\nIn Chapter 7 we imagined that you asked 35 students in your class[footnote]Note that this of course is not a random sample; we are using it here only for illustrating how hypothesis testing works so we are effectively pretending it is random. In a real-life study, you should not use non-probability samples for statistical inference.[\/footnote] whether they liked the campus cafeteria: 12 of your classmates said yes<\/em> (i.e., 34.3 percent), 7 (out of 15) first-years and 5 (out of 20) second-years (46.7 percent of all first-years and 25 percent of all second-years, respectively).\r\n\r\n \r\n\r\nWe want to know whether the observed in the sample difference in proportions (0.467-0.25=0.217) is statistically significant: can it be generlized to a larger student population, or is it due to a regular sampling variability?\r\n