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This variation of t-test is used for Paired Data

Two set of observations are *paired* if each observation in one set has exactly one corresponding observation is another set.

Examples:

- pre- and post-test scores on the same person
- measures in pairs at the same time or place
- outcome with or without a treatment - on same subject (cross-over study)

library(openintro) data(textbooks) t.test(textbooks$diff, mu=x.bar.nul, alternative='two.sided')

or

t.test(textbooks$uclaNew, textbooks$amazNew, paired=T, alternative='two.sided')

- two samples: local bookshop and amazon
- $\mu_\text{dif} = \mu_l - \mu_a$ - the mean of difference in the price

Test

- $H_0: \mu_\text{dif} = 0$ - there's no difference in the price
- $H_A: \mu_\text{dif} \ne 0$ - there's some difference

Calculations

- $\bar{X}_\text{dif} = 12.76$
- Standard Error: $\text{SE}_{\bar{X}_\text{dif}} = \cfrac{s_\text{dif}}{\sqrt{n_\text{dif}}} = 1.67$
- $T = \cfrac{\bar{X}_\text{dif}}{\text{SE}_{\bar{X}_\text{dif}}} = \cfrac{12.76}{1.67} = 7.59$
- $p = 6 \cdot 10^{-11}$, less than $\alpha = 0.05$, so we reject $H_0$

library(openintro) data(textbooks) hist(textbooks$diff, col='yellow') n = length(textbooks$diff) s = sd(textbooks$diff) se = s / sqrt(n) x.bar.nul = 0 x.bar.dif = mean(textbooks$diff) t = (x.bar.dif - x.bar.nul) / se t p = pt(t, df=n-1, lower.tail=F) * 2 p

or

t.test(textbooks$diff, mu=x.bar.nul, alternative='two.sided')

Let $\mu_d = \mu_0 - \mu_1$ be the difference between two methods

Our test:

- $H_0: \mu_d = 0, H_A: \mu_d \neq 0$

Say, we have:

- $\bar{X}_d = 6.854$
- $s_d = 11.056$
- $n = 398$

Test statistics:

- $\cfrac{\bar{X}_d - 0}{s_d / \sqrt{n}} = \cfrac{6.854}{11.056 / \sqrt{398}} \approx 12.37$

Then we compare it with $t_{397}$

- $p$-value is $2.9 \cdot 10^{29}$

And we conclude that the difference between the two methods is not 0