Question

A government health organization publishes information on AIDS in a weekly report. During one​ year, the number of provisional cases of AIDS for each of 50 regions is as shown below. Solve parts​ (a) through​ (d) below.

5656	2424	1212	590590	107107	650650	64676467	10251025	15531553	778778
599599	11561156	652652	193193	131131	7070	320320	55	1212	4444
9797	193193	14341434	894894	8686	989989	720720	13471347	47914791	195195
158158	464464	465465	190190	577577	220220	48714871	1313	2323	88
360360	118118	686686	7979	265265	534534	233233	61906190	1818	8989

a. Obtain the standard deviation of the variable​ "number of provisional AIDS​ cases" for the population of 50 regions.

sigmaσequals=nothing

cases

​(Round to two decimal places as​ needed.)

b. Consider simple random samples without replacement from the population of 50 regions. Strictly​ speaking, what is the correct formula for obtaining the standard deviation of the sample​ mean? Let n be the sample size and N be the population size. Fill in the correct answers below.

The formula is

▼

because the sampling is done

without

replacement from​ a(n)

finite

population.c. Referring to part​ (b), obtain

sigma Subscript x overbarσx

for simple random samples of size

2525

by using both formulas.

sigma Subscript x overbarσx	equals=	StartFraction sigma Over StartRoot n EndRoot EndFractionσn
	equals=	2525
		​(Round to two decimal places as​ needed.)

sigma Subscript x overbarσx	equals=	StartRoot StartFraction Upper N minus n Over Upper N minus 1 EndFraction EndRoot times StartFraction sigma Over StartRoot n EndRoot EndFractionN−nN−1•σn
	equals=	nothing
		​(Round to two decimal places as​ needed.)

Why does

sigma Subscript x overbar Baseline equals StartFraction sigma Over StartRoot n EndRoot EndFractionσx=σn

provide such a poor estimate of the true value given by

sigma Subscript x overbar Baseline equals StartRoot StartFraction Upper N minus n Over Upper N minus 1 EndFraction EndRoot times StartFraction sigma Over StartRoot n EndRoot EndFractionσx=N−nN−1•σn​?

As the value of n gets farther from​ 1, the value of

StartRoot StartFraction Upper N minus n Over Upper N minus 1 EndFraction EndRootN−nN−1

gets

▼

​1, so the value of

StartFraction sigma Over StartRoot n EndRoot EndFractionσn

gets

▼

the true value of

StartRoot StartFraction Upper N minus n Over Upper N minus 1 EndFraction EndRoot times StartFraction sigma Over StartRoot n EndRoot EndFractionN−nN−1•σn.

d. Referring to part​ (b), obtain

sigma Subscript x overbarσx

for simple random samples of size

44

by using both formulas.

sigma Subscript x overbarσx	equals=	StartFraction sigma Over StartRoot n EndRoot EndFractionσn
	equals=	nothing
		​(Round to two decimal places as​ needed.)

sigma Subscript x overbarσx	equals=	StartRoot StartFraction Upper N minus n Over Upper N minus 1 EndFraction EndRoot times StartFraction sigma Over StartRoot n EndRoot EndFractionN−nN−1•σn
	equals=	nothing
		​(Round to two decimal places as​ needed.)

Why does

sigma Subscript x overbar Baseline equals StartFraction sigma Over StartRoot n EndRoot EndFractionσx=σn

provide a somewhat reasonable estimate of the true value given by

sigma Subscript x overbar Baseline equals StartRoot StartFraction Upper N minus n Over Upper N minus 1 EndFraction EndRoot times StartFraction sigma Over StartRoot n EndRoot EndFractionσx=N−nN−1•σn​?

As the value of n gets closer to​ 1, the value of

StartRoot StartFraction Upper N minus n Over Upper N minus 1 EndFraction EndRootN−nN−1

gets

▼

​1, so the value of  

StartFraction sigma Over StartRoot n EndRoot EndFractionσn

gets

▼

the true value of

StartRoot StartFraction Upper N minus n Over Upper N minus 1 EndFraction EndRoot times StartFraction sigma Over StartRoot n EndRoot EndFractionN−nN−1•σn.

        More

Enter your answer in each of the answer boxes.

Save for Later

Answer 1