Question

1) Use the following data to run an independent t-test. Does the result reach significance? х 12 Y 16 16 19 14 14 15 20 11 13 12 15 2) Think of an experiment you would like to run and describe it to me here. After your description, tell me which of the three t-tests we have learned you would use to test the result of your experiment.

Answer 1

The null and alternative hypothesis

$H0: \mu _{x}=\mu _{y}, Ha: \mu _{x}\neq \mu _{y}$

Test statistic

$t= \frac{\bar{x}-\bar{y}}{s\sqrt{(1/n1+1/n2)}}$

where

$s=\sqrt{((n1-1)sx^{2}+(n2-1)sy^{2})/(n1+n2-2)}$

	X	(x-13.33)^2			Y	(y-16.17)^2
	12	1.7689			16	0.0289
	16	7.1289			19	8.0089
	14	0.4489			14	4.7089
	15	2.7889			20	14.6689
	11	5.4289			13	10.0489
	12	1.7689			15	1.3689
sum/ss	80	19.3334		sum/ss	97	38.8334
mean =sum/24	13.33333			mean =sum/15	16.16667
sx^2=ss/23		3.86668		sy^2=ss/14		7.76668
sx		1.9663876		sy		2.78687639

$\bar{x}=13.33 ,\bar{y}=16.17, sx^{2}=3.87, sy^{2}=7.77 , n1=6, n2= 6$

$s=\sqrt{(5*3.87+5*7.77/10}=2.41$

Therefore

$t= \frac{13.33-16.17}{2.41\sqrt{(1/6+1/6)}}$

= -2.03

df =6+6-2 =10

P value =0.0698

Since P value > 0.05 , the result is not significant

We fail to reject H0

There is not sufficient evidence to conclude that population means are different .

Note : P value using excel"=T.DIST.2T(2.03,10)"