Confidence Interval for a Proportion

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Confidence Interval for a Proportion
Since the sampling distribution is a distribution
of all possible samples of size n, one selected
at random has 95 chance of being within 2
standarddeviation unit of the population
proportion p.
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Confidence Interval for a Proportion
The True Mean (Unknown)
Sample 1
Sample 2
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Confidence Interval for a Proportion
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Confidence Interval for a Proportion
Notice that the standard deviation
has been replaced by
This latter quantity is called the standard error
of thesampling distribution and is an estimator
for the standard deviation when the value of p
is unknown and is estimated by p. The notation
for thestandard error of p is SE(p). The
quantity zSE(P) is called the margin of
error. The confidence interval is p ME or (p
ME, p ME). Also, ME .5(length of the
confidence interval)
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Confidence Interval Problem Using TI-83 Add-in
Program
Problem A survey was done to see what
proportionof people supported affirmative
action. 1013 people weresurveyed and 537 were
in support. p 537/1013 .530. Find a 93
confidence interval for p.
Confidence interval p ME p z SE(P) PRGM
CRITVAL ENTER1 NORMAL DIST ENTER (ENTER
CONF LEVEL AS A DECIMAL? .93 ENTER OUTPUT
CRITICAL VALUE 1.8119 PRGM 8STDERROR
ENTER1 1-PROP ENTERp .530n 1013OUTPUT
SE(p) .0157 CALCULATE CONFIDENCE INTERVAL
(C.I) C.I .530 1.8119 .0157 .530
.0284 (.530 - .0284, .530 .0284) (.5016,
.5584) We say 93 of 100 confidence intervals
constructed inthis way will contain the true
proportion of supporters.
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Confidence Interval Problem Using TI-83 Black
Box Program
Problem A survey was done to see what
proportionof people supported affirmative
action. 1013 people weresurveyed and 537 were
in support. p 537/1013 .530. Find a 93
confidence interval for p. STAT TESTS - A 1
PROPZInt ENTERx 537 (The number of successes
in p, np rounded to an integer, or
the numerator of p) ENTERn 1013 (Sample size)
- ENTER C- LEVEL .93 (as a decimal)
ENTER CALCULATE ENTER OUTPUT (.5017, .5585)
The slight difference in answers is due to the
roundingof the components using the add-in
program outputs. The black box program is most
accurate.
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Problems
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Required Sample Size on TI-83
Often we want to determine a sample size to
satisfya certain confidence level and certain
margin of error.
If we do not have and estimate for p, we estimate
it to be 0.5. Problem The margin of error in
the previous affirmative action support poll was
.0284, and p .530. How big does the sample
have to be to have a confidence level of 95 with
a ME .01 PRGM 7 SAMPLSIZ ENTER1
(PROPORTION) ENTERCONFIDENCE LEVEL .95
ENTERME .01p EST .530OUTPUT n (sample
size) 9570
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Problems
What is the margin of error for the confidence
interval? How large a sample is required to have
a margin of error of 0.02?
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Problems
d. If you wanted to have a margin of error of 1
for the respondents to this poll, and a 95
confidence interval, how many respondents would
be needed?
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Confidence Interval for Mean with Known s
For confidence intervals for means, we
distinguish two cases 1) we know the population
standard deviation, and 2) we have to estimate
the population deviation from the sample. For
case 1 we use the normalsampling distribution
when conditions are met.
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Confidence Interval for Mean with Known s TI-83
Problem A survey was made of cosmic radiation
exposure by Flight Personnel of Xingjian
Airlines. The sample mean was 219 mrems, based on
a sample size of 100 flight crew. Assume that s
isknown to be 35 mrems. Find the 95 confidence
interval. STAT 7TESTS ZInterval
ENTER Input Stats DOWN ARROW s 35 219n
100C-Level .95Calculate ENTER OUTPUT
(212.14, 225.86)
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Confidence Interval for Mean with Unknown s
When s is unknown, additional variability is
introduced Into the sampling distribution,
spreading it out. We cannotuse the normal
model, we must use the t-distribution, which
like the normal model, is unimodal, symmetric,
and bell shaped. While normal model shapes are
determined µ and s, t-distributions require µ,
s, and df (degrees of freedom n-1).
z critical value for 95 confidence 1.96 t
critical value for 95 confidence and 4df 2.78
t critical value for 95 confidence and 12df
2.18
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One Sample Confidence Interval for a Mean Unknown
s
s standard deviation of sample and thestandard
error of the sampling distribution is
When the sample size is less than 30, we make a
probability plot of the sample. It the plot is
reasonably close to a straight line, we consider
the population normal, and conclude condition 2
abovesatisfied for the sampling distribution to
be a t-distribution. The smaller the sample,
the closer the probability plot pattern must be
to a straight line. Samples as small as 5 may be
suitable for t-distribution procedures.
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Probability Plot
Problem A study was done to determine the
selfishness of Chimpanzees. The could pull one
rope and only get food for themselves. The other
rope would provide food for both Chimps. The
following numbers represent the number of times
each of the 7 chimps pulled both ropes in 36
trials 23, 22, 21, 24, 19, 20, 20.
Probability Plot
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Probability Plot Continued-TI-83
Problem Continued. The data is 23, 22, 21, 24,
19, 20, 20. The mean of this sample is 21.29 and
the standard deviation of this sample is 1.80.
Construct the probability plot and find a 99
confidence interval. STAT 1EDIT ENTER(Enter
the data in L1) 2nd STAT PLOT ENTERONTYPE
(Right arrow to 6th symbol)DATA LIST L1DATA
AXIS XZOOM 9 ZOOMSTAT PRGM 1 CRITVAL
ENTER2 T DIST ENTERCONFIDENCE LEVEL? .99df
(INTEGER)? 6 (n-1)CR VALUE 3.7074C.I 21.29
3.7074 C.I (18.77, 23.81)
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One Sample Confidence Interval TI-83
Since we do not have the population s, we must
use a t-distribution. Since the sample size is
greater than 30, conditions are satisfied.STAT
TESTS 8Tinterval ENTERInput Stats
ENTERx28.5 Sx24.2n500C-Level.90Calculate
EnterOutput (26.717, 30.283)We are 90
confident the interval contains the population
mean. 90 of 100 confidence intervals constructed
this way will contain the population mean.
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One Sample Confidence Interval TI-83
Since the sample size is less than 30, we needto
check a probability plot for normality. STAT-1EDI
T ENTER(Enter the data in List 1) 2nd STAT
PLOT (Examine Plot for linearity. Sincethe plot
is close to linear, conditions are ok)STAT
TESTS 8Tinterval ENTERInput DataList
L1Freq 1 (This entry is always 1)C-Level
.90Calculate ENTEROUTPUT (19.504, 24.296)
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Required Sample Size TI-83
We need program inputs of the confidence level,
the margin of error, and an estimate of s. If
wedont have a good estimate of s, we estimate
it to be ¼ the range of sample data or ¼(700-50)
162.50. PRGM 7SAMPLSIZ ENTERSELECT? 2
(Mean) CONF LEVEL .95 ME 10s EST
162.50 OUTPUT n 1017
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Problems
21
Problems
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Problems
Sketch and interpret a probability plot for this
data. If you wanted to cut the margin of error
in half, how big would the sample have to be?