anN.49 Abstainers The Harvard study mentioned in the previ-fous exercise estimated that 19 % of college students ab-stain from drinking alcohol. To estimate this proportion inyour school, how large a random sample would you needto estimate it to within 0.05 with probability 0.95, if beforeconducting the studyC8.55 Vunwilling to guess the proportion value ata. You areyour school?h. You use the Harvard study as a guideline?Use the results from parts a and b to explain whystrategy (a) is inefficient if you are quite sure you'll geta sample proportion that is far from 0.50.SS0).ityS8.50 How many businesses fail?mate the proportion of businesses started in the year 2006that had failed within five years of their start-up. Howlarge a sample size is needed to guarantee estimating thisproportion correct to withina. 0.10 with probability 0.95?b. 0.05 with probability 0.95?c. 0.05 with probability 0.99?d. Compare sample sizes for parts a and b, and b and c,and summarize the effects of decreasing the margin oferror and incre asing the confidence level.851 Canada and the death penalty1998 indicated that 48% of Canadians favor imposingthe death penalty (Canada does not have it). A report byAmnesty International on this and related polls (www.amnesty.ca) did not report the sample size but stated,"Polls of this size are considered to be accurate within2.5 percentage points 95% of the time." About how largewas the sample size?8.52 Farm size An estimate is needed of the mean acreageof farms in Ontario, Canada. A 95% confidence intervalshould have a margin of error of 25 acres. A study 10heA study is planned to esti-onheеа8.56ntsTRYeninf aor"plehatelyataledA poll in Canada inen8.57anvelofyearsagoor 200 acres for farm size.a. About houin this province had a sample standard deviation8.58mmnle of farms is needed?

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Asked Nov 8, 2019
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8.50 A,B,C

an
N.49 Abstainers The Harvard study mentioned in the previ-
f
ous exercise estimated that 19 % of college students ab-
stain from drinking alcohol. To estimate this proportion in
your school, how large a random sample would you need
to estimate it to within 0.05 with probability 0.95, if before
conducting the study
C
8.55 V
unwilling to guess the proportion value at
a. You are
your school?
h. You use the Harvard study as a guideline?
Use the results from parts a and b to explain why
strategy (a) is inefficient if you are quite sure you'll get
a sample proportion that is far from 0.50.
S
S
0).
ity
S
8.50 How many businesses fail?
mate the proportion of businesses started in the year 2006
that had failed within five years of their start-up. How
large a sample size is needed to guarantee estimating this
proportion correct to within
a. 0.10 with probability 0.95?
b. 0.05 with probability 0.95?
c. 0.05 with probability 0.99?
d. Compare sample sizes for parts a and b, and b and c,
and summarize the effects of decreasing the margin of
error and incre asing the confidence level.
851 Canada and the death penalty
1998 indicated that 48% of Canadians favor imposing
the death penalty (Canada does not have it). A report by
Amnesty International on this and related polls (www.
amnesty.ca) did not report the sample size but stated,
"Polls of this size are considered to be accurate within
2.5 percentage points 95% of the time." About how large
was the sample size?
8.52 Farm size An estimate is needed of the mean acreage
of farms in Ontario, Canada. A 95% confidence interval
should have a margin of error of 25 acres. A study 10
he
A study is planned to esti-
on
he
еа
8.56
nts
TRY
en
in
f a
or"
ple
hat
ely
ata
led
A poll in Canada in
en
8.57
an
vel
of
years
ago
or 200 acres for farm size.
a. About hou
in this province had a sample standard deviation
8.58
mmnle of farms is needed?
help_outline

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an N.49 Abstainers The Harvard study mentioned in the previ- f ous exercise estimated that 19 % of college students ab- stain from drinking alcohol. To estimate this proportion in your school, how large a random sample would you need to estimate it to within 0.05 with probability 0.95, if before conducting the study C 8.55 V unwilling to guess the proportion value at a. You are your school? h. You use the Harvard study as a guideline? Use the results from parts a and b to explain why strategy (a) is inefficient if you are quite sure you'll get a sample proportion that is far from 0.50. S S 0). ity S 8.50 How many businesses fail? mate the proportion of businesses started in the year 2006 that had failed within five years of their start-up. How large a sample size is needed to guarantee estimating this proportion correct to within a. 0.10 with probability 0.95? b. 0.05 with probability 0.95? c. 0.05 with probability 0.99? d. Compare sample sizes for parts a and b, and b and c, and summarize the effects of decreasing the margin of error and incre asing the confidence level. 851 Canada and the death penalty 1998 indicated that 48% of Canadians favor imposing the death penalty (Canada does not have it). A report by Amnesty International on this and related polls (www. amnesty.ca) did not report the sample size but stated, "Polls of this size are considered to be accurate within 2.5 percentage points 95% of the time." About how large was the sample size? 8.52 Farm size An estimate is needed of the mean acreage of farms in Ontario, Canada. A 95% confidence interval should have a margin of error of 25 acres. A study 10 he A study is planned to esti- on he еа 8.56 nts TRY en in f a or" ple hat ely ata led A poll in Canada in en 8.57 an vel of years ago or 200 acres for farm size. a. About hou in this province had a sample standard deviation 8.58 mmnle of farms is needed?

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Expert Answer

Step 1

a)

Given data

Error =  0.10

Confidence level = 0.95

Zo.05
Zo.025
-1.95996
2
NORM.S.INV (0.025))
(From Excel =
Since, sample proportion is not given
Assuming sample proportion, p 0.5
Margin of error formula is given by
|Р(1-р)
E Za X
п
2
Simplifying the above formula
2
Za
1.95996
п %3 p(1 — р) х
= 96.036 97
0.5 x 0.5 x
E
0.1
help_outline

Image Transcriptionclose

Zo.05 Zo.025 -1.95996 2 NORM.S.INV (0.025)) (From Excel = Since, sample proportion is not given Assuming sample proportion, p 0.5 Margin of error formula is given by |Р(1-р) E Za X п 2 Simplifying the above formula 2 Za 1.95996 п %3 p(1 — р) х = 96.036 97 0.5 x 0.5 x E 0.1

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Step 2

b)

Given data

Error =  0.05

Confidence level = 0.95

Significance level = a = 1 - 0.95 = 0.05
= Zo.025
-1.95996
Zo.05
2
NORM.S.INV (0.025))
(From Excel =
Since, sample proportion is not given
Assuming sample proportion, p 0.5
Margin of error formula is given by
|Р(1-р)
E Za X
п
Simplifying the above formula
2
Za
2
п %3 p(1 — р) х
1.95996
0.5 x 0.5 x
= 384.144 2 385
0.05
Е
help_outline

Image Transcriptionclose

Significance level = a = 1 - 0.95 = 0.05 = Zo.025 -1.95996 Zo.05 2 NORM.S.INV (0.025)) (From Excel = Since, sample proportion is not given Assuming sample proportion, p 0.5 Margin of error formula is given by |Р(1-р) E Za X п Simplifying the above formula 2 Za 2 п %3 p(1 — р) х 1.95996 0.5 x 0.5 x = 384.144 2 385 0.05 Е

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Step 3

c)

Given data

Error =  0.10

Confidence ...

Significance level = a = 1 - 0.99 = 0.01
Zo.01 = Zo.005
-2.576
2
(From Excel NORM.S.INV(0.005))
Since, sample proportion is not given
Assuming sample proportion, p 0.5
Margin of error formula is given by
P(1-p)
E Za X
n
Simplifying the above formula
2
Za'
2
E
-2.576
n p(1 - p) x
663.5776 664
0.5 x 0.5 x
0.05
help_outline

Image Transcriptionclose

Significance level = a = 1 - 0.99 = 0.01 Zo.01 = Zo.005 -2.576 2 (From Excel NORM.S.INV(0.005)) Since, sample proportion is not given Assuming sample proportion, p 0.5 Margin of error formula is given by P(1-p) E Za X n Simplifying the above formula 2 Za' 2 E -2.576 n p(1 - p) x 663.5776 664 0.5 x 0.5 x 0.05

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