Lektsii (1) / Lecture 18

ICEF, 2012/2013 STATISTICS 1 year LECTURES

Лекция 18	29.01.13
	Confidence interval for a difference of population means
	(matched samples)

Example. Is it true that in average the right hand is more powerful than the left hand? Let µr , µl be the mean power of the right and left hand, respectively, and let ∆ = µr −µl . Suppose that there is a random sample of n people, and let xi be the right hand’s power, and yi be the left hand’s power for the ith person. Thus the sample consists of the pairs {(xi , yi ), i =1,..., n}/ It should emphasized that xi and yi are dependent, while (xi , yi ) and (x j , y j ) are independent for i ≠ j . Such samples are called matched or paired samples. In order to construct the confidence

interval for ∆ = µr

−µl let’s denote di = xi − yi and note that E(di ) =∆. If we assume that the

differences di = xi

− yi

are normally distributed then the problem is reduced to the problem of

calculating confidence interval for the mean of a normal population. Namely,

(n −1)

sd

,

∆ = d

±t

α / 2

n

, sd

are sample mean and sample standard deviation for the observations d1,..., dn .

where d

= 0.9, sd =1.1, α = 0.05. Then

Let, for example, n = 60, d

s

= 0.9 ±2.00

1.1

= 0.9 ±0.284, ∆ [0.615, 1.184].

∆ = d

±t

(n −1)

α / 2

n

60

So, we have statistical evidence in favor of statement that the right hand is more powerful than the left hand.

Confidence interval for the difference of two proportions

Let there are two populations with population proportions π1, π2 respectively (the proportions of

voting for democrats in two different regions, the shares of blue-eyes people among women and men, etc.) Taking two (independent) samples of sizes n1, n2 , respectively from these populations

we get the corresponding sample proportions p1, p2 . Suppose that the normal approximation for the sample proportions can be used (remind the conditions!), then

p		π	,	π	(1−π	)	, p			π		,	π		(1−π		)
	≈ N			1	1			2	≈ N		2			2		2
1		1			n1										n2
and (due to independency)

p	− p		π		−π		,	π	(1	−π	)	+	π		(1	−π		)
		≈ N		1		2		1		1				2			2	.
1	2								n1						n2

Then the approximate confidence interval for the difference is as follows:

π	1	−π	2	= p − p ± z		p1(1 − p1 )	+	p2 (1 − p2 )	.
				1 2	α / 2	n1		n2

Пример. Перед теледебатами Буша поддерживали 630 из 1200 опрошенных, после − 490 из 1000. Можно ли утверждать, что симпатии уменьшились?

p1 = 0.525, p2 = 0.49, π1 −π2 = 0.0325 ±0.0419, π1 −π2 [−0.0069, 0.0769] нет оснований утверждать, что симпатии уменьшились.