{"id":1137,"date":"2019-04-06T02:17:08","date_gmt":"2019-04-06T06:17:08","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/simplestats\/?post_type=chapter&p=1137"},"modified":"2019-11-02T19:46:06","modified_gmt":"2019-11-02T23:46:06","slug":"9-1-between-a-discrete-and-a-continuous-variable","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/simplestats\/chapter\/9-1-between-a-discrete-and-a-continuous-variable\/","title":{"raw":"9.1 Between a Discrete and a Continuous Variable: The t-test","rendered":"9.1 Between a Discrete and a Continuous Variable: The t-test"},"content":{"raw":"[latexpage]\r\n\r\nFor this part, you need to recall (from Section 7.2.1, https:\/\/pressbooks.bccampus.ca\/simplestats\/chapter\/7-2-1-between-a-discrete-and-a-continuous-variable\/<\/a>) how we described bivariate associations between two variables, one of which is treated as discrete and one as continuous. In this case we essentially compared the groups (categories of the discrete variable) by their mean (or median) value on the continuous variable.\u00a0 We examine the potential association between such variables visually through boxplots and numerically through a difference of means.\r\n\r\n \r\n\r\nNow the question in front of us is: even if we do see a difference in the means of the different groups in sample data<\/em>, how certain can we be that this association is real and reflective of the population? As we learned in Chapter 8, to answer this question, we need to test the difference for statistical significance.\r\n\r\n \r\n\r\nWe start with a few theoretical notes, which we will then apply to the example I used in Chapter 7 about the potential gender difference in average income. In this way we will be able to test whether the difference observed in the NHS 2011<\/em> data (\\$16,401 in favour of men to be precise) is statistically significant or not. In the latter half of this section we will see what happens when there are more than two groups' means to compare.\r\n\r\n \r\n\r\nTesting the difference of two means.<\/strong> Recall from Section 8.3 (https:\/\/pressbooks.bccampus.ca\/simplestats\/chapter\/8-3-hypothesis-testing\/<\/a>) that we tested whether the employees who took a training course indeed had a higher average productivity by simply calculating the z<\/em>-value (or, using the estimated standard error, the t<\/em>-value with a given df<\/em>) for the mean and then finding its associated p<\/em>-value. We could then compare the p<\/em>-value to the preselected\u00a0\u03b1<\/em>-level and make a conclusion regarding the null hypothesis.\r\n\r\n \r\n\r\nYou will be happy to know that testing a difference of means follows the same principle: obtain the\u00a0z <\/em>(or rather, the\u00a0t<\/em>-value), get the associated\u00a0p<\/em>-value, compare to the\u00a0\u03b1<\/em>. What is not the same is that now we are testing expressly a difference of two means -- so we need the\u00a0t<\/em>-value for the difference<\/em>. It turns out, we can calculate one as easily as ever, as long as we had the standard error of the difference<\/em>[footnote]I hope you have not forgotten that $z=\\frac{\\overline{x}-\\mu}{\\sigma_\\overline{x}}$, where the standard error $\\sigma_\\overline{x}$ $=\\frac{\\sigma}{N}$.[\/footnote].\r\n\r\n \r\n\r\nThe standard error of a difference of two means is a combination of their separate standard errors:<\/strong>\r\n\r\n \r\n\r\n$\\sigma_(\\overline{x}_1-\\overline{x}_2)$ $=\\sqrt{\\frac{\\sigma_1^2}{N_1}+\\frac{\\sigma_2^2}{N_2}}$ = standard error of the difference of two means<\/em>\r\n\r\n \r\n\r\nwhere the subscripts refer to the first and second group being compared.\r\n\r\n \r\n\r\nThe z<\/em>-value for a difference of two means follows the ordinary z<\/em>-value formula, but with the difference<\/em> taking the place of the single mean:\r\n\r\n \r\n\r\n$z=\\frac{(\\overline{x_1} -\\overline{x_2})-(\\mu_1 -\\mu_2 )}{\\sigma_(\\overline{x}_1-\\overline{x}_2)}$\r\n\r\n \r\n\r\nHowever, under the null hypothesis we hypothesize there is no difference in the population means, as such $\\mu_1=\\mu_2$, and thus $\\mu_1-\\mu_2=0$. Accounting for that in the formula, along with substituting the standard error with its own formula from above, we get:\r\n\r\n \r\n\r\n$z=\\frac{\\overline{x_1} -\\overline{x_2}}{\\sqrt{\\frac{\\sigma_1^2}{N_1}+\\frac{\\sigma_2^2}{N_2}}}$\r\n\r\n \r\n\r\nFinally, since we generally don't know the population parameters but work with sample data, we estimate the standard error\u00a0\u03c3<\/em> with the sample standard error s<\/em>, thus moving to the\u00a0t<\/em>-value\u00a0through which we test the difference for statistical significance:<\/strong>\r\n\r\n \r\n\r\n$t=\\frac{\\overline{x_1} -\\overline{x_2}}{\\sqrt{\\frac{s_1^2}{N_1}+\\frac{s_2^2}{N_2}}}$ = t-test for the difference of means<\/em>[footnote]The more observant of you would notice that the squared standard deviations of the two groups, i.e., the\u00a0s1<\/sub>2<\/sup><\/em>\u00a0and s2<\/sub>2<\/sup><\/em> here are of course the groups' variances (which we need if we are to have them under the square root). In this version of the formula, the groups are taken to have unequal<\/em> variances, which is a more conservative assumption than assuming the variances of the two groups are equal. If we have a good reason to assume equal<\/em> variances, then s1<\/sub>2\u00a0<\/sup><\/em>and s2<\/sub>2<\/sup><\/em>\u00a0will just be the same (combined, or pooled) variance s2<\/sup><\/em>, and the formula will look like this:\r\n\r\n \r\n\r\n$t=\\frac{\\overline{x_1} -\\overline{x_2}}{s\\sqrt{\\frac{1}{N_1}+\\frac{1}{N_2}}}$\u00a0[\/footnote]\r\n\r\n \r\n\r\nNote than unlike the single value case where the df=N<\/em>-1, when working with a difference of means of two groups the\u00a0<\/span>df=N<\/span><\/em>-2.<\/span>\r\n\r\n \r\n\r\nBefore you eyes glaze over (completely), rest assured that SPSS calculates this for you; I only provide it here to show you that the logic of hypothesis testing is the same, only the formulas change to accommodate the testing of a difference of means<\/em> rather than a single mean.\r\n\r\n \r\n\r\nFrom this point on, it's easy: you only need to check the p<\/em>-value of the t<\/em>-value you have obtained (given the specific df<\/em>)[footnote]You can do that through an online p<\/em>-value calculator for the t<\/em>-distribution like this one here: https:\/\/www.socscistatistics.com\/pvalues\/tdistribution.aspx<\/a>.[\/footnote], and compare it to the significance level, and voila<\/em> -- you have yourself a significance test!\r\n\r\n \r\n\r\nLet's see how this all works out in an example.\u00a0 A few sections back I promised you to test the gender differences in average income, didn't I?\r\n\r\n \r\n
\r\n

Example 9.1 Testing Gender Differences in Average Income, NHS 2011\u00a0<\/em><\/p>\r\n\r\n<\/header>\r\n

\r\n\r\n \r\n\r\nAs in Example 7.2 in Section 7.2.1, I use a random sample of about 3 percent of the entire NHS 2011<\/em> data, this time resulting in N<\/em>=21,902[footnote]Since I use a new random sub-sample of the data, you can consider this an indirect illustration of sampling variation. For comparison of sample statistics as well as variable description, refer back to Example 7.2[\/footnote].\r\n\r\n \r\n\r\nWe are still interested in whether women and men on average earn differently per year, i.e., whether gender<\/em> affects income<\/em>:\r\n