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Learning Statistics with R - A tutorial for psychology students and other beginners, 2018a

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Grade<br />

66 68 70 72 74 76 78 80<br />

Anastasia<br />

Bernadette<br />

Class<br />

Figure 13.7: Plots showing the mean grade <strong>for</strong> the <strong>students</strong> in Anastasia’s <strong>and</strong> Bernadette’s <strong>tutorial</strong>s.<br />

Error bars depict 95% confidence intervals around the mean. On the basis of visual inspection, it does<br />

look like there’s a real difference between the groups, though it’s hard to say <strong>for</strong> sure.<br />

.......................................................................................................<br />

about exactly how close to zero this difference should be. And the solution to the problem is more or<br />

less the same one: we calculate a st<strong>and</strong>ard error estimate (SE), just like last time, <strong>and</strong> then divide the<br />

difference between means by this estimate. So our t-statistic will be of the <strong>for</strong>m<br />

t “ ¯X 1 ´ ¯X 2<br />

SE<br />

We just need to figure out what this st<strong>and</strong>ard error estimate actually is. This is a bit trickier than was<br />

the case <strong>for</strong> either of the two tests we’ve looked at so far, so we need to go through it a lot more carefully<br />

to underst<strong>and</strong> how it works.<br />

13.3.3 A “pooled estimate” of the st<strong>and</strong>ard deviation<br />

In the original “Student t-test”, we make the assumption that the two groups have the same population<br />

st<strong>and</strong>ard deviation: that is, regardless of whether the population means are the same, we assume that<br />

the population st<strong>and</strong>ard deviations are identical, σ 1 “ σ 2 . Since we’re assuming that the two st<strong>and</strong>ard<br />

deviations are the same, we drop the subscripts <strong>and</strong> refer to both of them as σ. How should we estimate<br />

this? How should we construct a single estimate of a st<strong>and</strong>ard deviation when we have two samples? The<br />

answer is, basically, we average them. Well, sort of. Actually, what we do is take a weighed average of<br />

the variance estimates, which we use as our pooled estimate of the variance. The weight assigned<br />

to each sample is equal to the number of observations in that sample, minus 1. Mathematically, we can<br />

write this as<br />

w 1 “ N 1 ´ 1<br />

w 2 “ N 2 ´ 1<br />

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