STT 200 – LECTURE 1, SECTION 2,4
RECITATION 13
(11/27/2012)
TA: Zhen (Alan) Zhang
zhangz19@stt.msu.edu
Office hour: (C500 WH) 1:45 – 2:45PM Tuesday
(office tel.: 432-3342)
Help-room: (A102 WH) 11:20AM-12:30PM, Monday, Friday
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Class meet on Tuesday:
3:00 – 3:50PM A122 WH, Section 02
12:40 – 1:30PM A322 WH, Section 04
OVERVIEW
 We
will discuss following problems:
 Chapter 19 “Confidence Intervals for Proportions”
(Page 504)
Nos. 7, 8, 14, 22, 27, 28
 All
recitation PowerPoint slides available at here
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 Review
Confidence interval for proportion:
1 − 𝛼 % 𝑐𝑜𝑛𝑓𝑖𝑑𝑒𝑛𝑐𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝑝:
𝑝(1 − 𝑝)
𝑛
𝑝 ± 𝑧𝛼
2
The half-width of the interval, is the margin of error:
𝑀𝐸 = 𝑧𝛼
2
𝑝(1 − 𝑝)
𝑛
𝑧𝛼 will increase as the confidence level 1 − 𝛼 % increases (or as 𝛼
2
decreases).
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margin of error
Margin of error
0.08
0.08
0.10
0.10
Margin of error versus proportion
0.14
0.16
(cont’d)
0.12
 Review
0.18
Margin of error versus sample size
60
80
100
120
140
0.04
0.06
sample size
for fixed confidence level = 95%, proportion = 0.5
0.6
0.8
proportion
for fixed confidence level = 95%, sample size = 100
1.0
0.10
0.09
0.4
0.08
0.2
0.07
0.0
margin of error
0.11
0.02
0.12
0.13
Margin of error versus confidence level
0.00
margin of error
40
4
0.80
0.85
0.90
0.95
confidence level
for fixed sample size = 100, proportion = 0.5
 Review
(cont’d)
From 𝑀𝐸 = 𝑧𝛼
2
𝑝(1−𝑝)
,
𝑛
solve for sample size:
𝑧𝛼 2 𝑝(1 − 𝑝)
𝑛=
2
𝑀𝐸 2
So given margin of error, confidence level, we can calculate required
sample size. When is 𝑝 unknown since we have not collected any
samples yet, we use a conservative guess under the “worst” scenario,
that is, when 𝑝(1 − 𝑝) reaches its maximal (when 𝑝 = 0.5), it
corresponds to the largest required sample size.
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 Chapter
19 (Page 504): #7:
Which statements are true?
a)
For a given sample size, higher confidence means a smaller margin
of error.
b)
For a specified confidence level, larger samples provides smaller
margins of error.
c)
For a fixed margin of error, larger samples provide greater
confidence.
d)
For a given confidence level, halving the margin of error requires a
sample twice as large.
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 Chapter
19 (Page 504): #8:
Which statements are true?
a)
For a given sample size, reducing the margin of error will mean
lower confidence.
b)
For a certain confidence level, you can get a smaller margin of
error by selecting a bigger sample.
c)
For a fixed margin of error, smaller samples will mean lower
confidence.
d)
For a given confidence level, a sample 9 times as large will make a
margin of error one third as big.
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
Chapter 19 (Page 505): #14:
11% of a random sample of 1003 adults approved of attempts to clone a human.
a)
Find the margin of error if we want 95% confidence.
1.96 × 0.11 × 0.89 1003 = 0.0194
b)
Explain what that margin of error means.
The pollsters are 95% confident that the true population of adults who approve of
attempts to clone humans is within 1.9% of the estimated 11%.
c)
If we only need to be 90% confident, will the margin of error be larger or
smaller? Explain.
Smaller, since the critical value 𝑧𝛼 2 decreases.
d)
Find that margin of error.
1.645 × 0.11 × 0.89 1003 = 0.0163
a)
In general, if all other aspects of the situation remain the same, would smaller
samples produce smaller or larger margin of error?
Larger.
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 Chapter
19 (Page 506): #22:
A national health organization warns that 30% of middle school students have
been drunk. A local health agency randomly surveys 110 of 1212 students.
Only 21 of them report having been drunk.
a)
What proportion of the sample reported having been drunk?
𝑝 = 21 110 = 0.1909
b)
Does this mean that this city’s youth are not drinking as much as the
national data would indicate? Explain.
Maybe, since the estimation is much lower than 0.3, but this is just one
estimate.
c)
Create 95% confidence interval for the proportion of local students who
have been drunk.
95% confidence interval is 0.1909±0.0734 = (0.1175, 0.2643)
d)
Is there any reason to believe that the national level of 30% is not true of
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the middle school students in this city?
The 95% confidence interval does not cover 0.3
 Chapter
19 (Page 506): #27:
In a random survey of 226 college students, 20 reported being “only” children.
Estimate the proportion of students nationwide.
a)
Check conditions for constructing a confidence interval.
The students’ birth orders are likely to be independent. The sample was
random and consisted of less than 10% of the population. There were 20
successes and 206 failures (both greater than 10).
b)
Construct 95% confidence interval.
0.0885 ± 0.0370 = (0.0515, 0.1255)
c)
Interpret your interval.
We are 95% confident that between 5.15% and 12.55% of all college students
are “only” children.
d)
Explain what “95% confidence” means in this context.
If we were to select repeated samples like this we’d expect about 95% of the
confidence intervals we created to contain the true proportion of all college
students who are “only” children.
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 Chapter
19 (Page 506): #28:
74% of 1644 randomly selected college freshmen returned to college the next
year. Estimate the national freshman-to-sophomore retention rate.
a)
Verify that the conditions are met.
It’s a random sample; both 74% and 26% of 1644 are greater than 10.
b)
Construct a 98% confidence interval.
The critical value is 2.326. The interval is 0.74 ± 0.0252 = (0.7148,0.7652)
c)
Interpret your interval.
We’re 98% confident that between 71.48% and 76.52% of colleges freshman
return to college their sophomore years.
d)
Explain what “98% confidence” means in this context.
If we were to select repeated samples like this we’d expect about 98% of the
confidence intervals we created to contain the true proportion of all college
freshmen who return to be sophomores.
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 Closing
remark :

The true proportion 𝑝 is fixed point (not random), but unknown. It is
the confidence interval that you constructed that’s random. So saying
“𝑝 has 95% chance of falling between the 95% confidence interval”
makes no sense unless you are a Bayesian statistician. On the contrary,
we can only say “the 95% confidence interval has 95% chance of
covering the true proportion 𝑝”, or routinely, we say “If we draw
samples of this size many times and construct confidence interval in
this way, we’d expect to see approximately 95% of them cover 𝑝”.

If I tell you the confidence interval is (0.7148,0.7652), will you be able
to recover what is 𝑝 and what is the margin of error?
Yes you can. 𝑝 is always the middle point of this interval in this case, so
0.7148 + 0.7652
𝑝=
= 0.74
2
and the margin of error is half of the width, or |endpoint-middle point|
𝑀𝐸 = 0.7652 − 0.74 𝑜𝑟 = 0.74 − 0.7148 = 0.0252
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Thank you.
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