Required Sample Size, Type II Error Probabilities
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Transcript of Required Sample Size, Type II Error Probabilities
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Required Sample Size, Type II Error Probabilities
Required Sample Size, Type II Error Probabilities
Chapter 23 Inference for Means: Part 2
Required Sample Size To Estimate a Population Mean
• If you desire a C% confidence interval for a population mean with an accuracy specified by you, how large does the sample size need to be?
• We will denote the accuracy by ME, which stands for Margin of Error.
Example: Sample Size to Estimate a Population Mean
• Suppose we want to estimate the unknown mean height of male students at NC State with a confidence interval.
• We want to be 95% confident that our estimate is within .5 inch of
• How large does our sample size need to be?
Confidence Interval for
*1
*1
In terms of the margin of error ME,
the CI for can be expressed as
The confidence interval for is
so ME
n
n
x ME
sx t
n
st
n
• Good news: we have an equation• Bad news:
1. Need to know s2. We don’t know n so we don’t know the degrees of
freedom to find t*n-1
*1
2*1
So we can find the sample size by solving
this equation for n:
ME
which gives
n
n
st
n
t sn
ME
A Way Around this Problem: Approximate by Using the
Standard Normal*
*
2*
Use the corresponding z from the standard normal
to form the equation
Solve for n:
sME z
n
z sn
ME
1.96n
1.96n
.95
Confidence level
Sampling distribution of x
ME ME
2
set M 1.96 and solve for
1.96 (estimate with s)
E nn
nME
Estimating s• Previously collected data or prior knowledge
of the population• If the population is normal or near-normal,
then s can be conservatively estimated by
s range
6• 99.7% of obs. within 3 of the mean
Example: sample size to estimate mean height µ of NCSU undergrad. male students
We want to be 95% confident that we are within .5 inch of so ME = .5; z*=1.96
• Suppose previous data indicates that s is about 2 inches.
• n= [(1.96)(2)/(.5)]2 = 61.47• We should sample 62 male students
2*z sn
ME
Example: Sample Size to Estimate a Population Mean -Textbooks
• Suppose the financial aid office wants to estimate the mean NCSU semester textbook cost within ME=$25 with 98% confidence. How many students should be sampled? Previous data shows is about $85.
2 2z * σ (2.33)(85)
n 62.76ME 25
round up to n = 63
Example: Sample Size to Estimate a Population Mean -NFL footballs
• The manufacturer of NFL footballs uses a machine to inflate new footballs
• The mean inflation pressure is 13.5 psi, but uncontrollable factors cause the pressures of individual footballs to vary from 13.3 psi to 13.7 psi
• After throwing 6 interceptions in a recent game, Peyton Manning complains that the balls are not properly inflated.
The manufacturer wishes to estimate the mean inflation pressure to within .025
psi with a 99% confidence interval. How many footballs should be sampled?
Example: Sample Size to Estimate a Population Mean
• The manufacturer wishes to estimate the mean inflation pressure to within .025 pound with a 99% confidence interval. How may footballs should be sampled?
• 99% confidence z* = 2.576; ME = .025 = ? Inflation pressures range from 13.3 to 13.7 psi• So range =13.7 – 13.3 = .4; range/6 = .4/6 = .067
2*z
nME
22.58 .067
47.66 48.025
n
1 2 3 48
. . .
Significance Levels and Rejections Regions
Hypothesis Tests for
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Levels and RejectionRegions, Right-Tail; n=26 (df=25)
If HA: > 0 and =.10then RR={t: t > 1.316}
If HA: > 0 and =.05 then RR={t: t > 1.708}
If HA: > 0 and =.01then RR={t: t > 2.485}
Rej Region
.10 t > 1.316
.05 t > 1.708
.01 t > 2.485
0 0
0
:H
yt
sn
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Hypothesis Testing for , Type II Error Probabilities (Right-tail example)
• Example– A new billing system for a department store will be cost-
effective only if the mean monthly account is more than $170.
– A sample of 401 accounts has a mean of $174 and s = $65.
– Can we conclude that the new system will be cost effective?
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• Hypotheses– The population of interest is the credit accounts at
the store.– We want to know whether the mean account for all
customers is greater than $170.
HA : > 170
– Where is the mean account value for all customers
– We will choose significance level = .05
Right-tail example: hypotheses, significance level
H0 : = 170
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• The rejection region: reject H0 if the test statistic t satisfies t > t.05,n-1 = t.05,400 = 1.649
• We will reject H0 if the value of the test statistic t is greater than 1.649• Results from the n = 401 randomly selected customers:
A Right - Tail Test: Rejection Region
$174, $65x s
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– Hypotheses:H0 : = 170HA : > 170
174 170test statistic: 1.23
65 401
xts n
Right-tail example: test statistic and conclusion
data: 174, 65x s
, 1 .05,400 1.649nt t t Recall that the rejection region is
Since the test statistic t = 1.23, and 1.23 < 1.649,We do not reject the null hypothesis H0: = 170.
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400-value ( 1.23) .1097P P t
01.23t
P-value: The probability of observing a value of the test statistic as extreme or more extreme then t = 1.23, given that = 170 is…
Right-tail example: P-value and conclusion
t400
Since the P-value > .05, we conclude that there is not sufficient evidence to reject H0 : =170.
Type II error is possible
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Calculating , the Probability of aType II Error
• Calculating for the t test is not at all straightforward and is beyond the level of this course– The distribution of the test statistic t is quite
complicated when H0 is false and HA is true– However, we can obtain very good approximate
values for using z (the standard normal) in place of t.
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Calculating , the Probability of aType II Error (cont.)
• We need to1. specify an appropriate significance level ;2. Determine the rejection region in terms of z3. Then calculate the probability of not being in the
rejection when = 1, where 1 is a value of that makes HA true.
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– Test statistic:H0 : = 170HA : > 170
Choose = .05Rejection region in terms of z: z > z.05 = 1.645
Example (cont.) calculating
rejection region in terms of :
1701.645
65
40065
170 1.645 175.34.400
x
xz
x
= 0.05
175.34170
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Express the rejection region directly, not in standardized
terms
175.34
=.05
= 170
Example (cont.) calculating
– The rejection region with = .05.
34.175x
Do not reject H0
180
HA: = 180
H0: = 170
Specify the alternative value under HA.
– Let the alternative value be = 180 (rather than just >170)
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175.34
=.05
= 170
Example (cont.) calculating
34.175x 180
H1: = 180
H0: = 170
– A Type II error occurs when a false H0 is not rejected. Suppose =180, that is H0 is false.
A false H0……is not rejected
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175.34= 170
Example (cont.) calculating
180
H1: = 180
H0: = 170
0(180) ( 175.34 )
( 175.34 180)
P x given that H is false
P x given that
0764.)40065
18034.175z(P
Power when =180 =
1-(180)=.9236
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• Increasing the significance level decreases the value of and vice versa
Effects on of changing
= 170 180
2 >2 <
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• A hypothesis test is effectively defined by the significance level and by the sample size n.
• If the probability of a Type II error is judged to be too large, we can reduce it by– increasing , and/or– increasing the sample size.
Judging the Test
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• Increasing the sample size reduces
Judging the Test
By increasing the sample size the standard deviation of the sampling distribution of the mean decreases. Thus, the cutoff value of for the rejection region decreases.
Recall : ,x s
RR z z or x zs n n
29Lx 180= 170
Judging the Test
Lx
Note what happens when n increases:
Lx LxLx Lx
does not change,but becomes smaller
• Increasing the sample size reduces
Recall : ,x s
RR z z or x zs n n
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• Increasing the sample size reduces • In the example, suppose n increases from 400 to
1000.65
170 1.645 173.381000
173.38 180( ) ( 3.22) 0
65 1000
sx z
n
P Z P Z
Judging the Test
• remains 5%, but the probability of a Type II drops dramatically.
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A Left - Tail Test• Self-Addressed Stamped Envelopes.
– The chief financial officer in FedEx believes that including a stamped self-addressed (SSA) envelop in the monthly invoice sent to customers will decrease the amount of time it take for customers to pay their monthly bills.
– Currently, customers return their payments in 24 days on the average, with a standard deviation of 6 days.
– Stamped self-addressed envelopes are included with the bills for 76 randomly selected customers. The number of days until they return their payment is recorded.
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• The parameter tested is the population mean payment period () for customers who receive self-addressed stamped envelopes with their bill
• The hypotheses are:H0: = 24H1: < 24
• Use = .05; n = 76.
A Left - Tail Test: Hypotheses
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• The rejection region: reject H0 if the test statistic t satisfies t < t.05,75 = 1.665
• We will reject H0 if the value of the test statistic t is less than 1.665• Results from the 76 randomly selected customers:
A Left - Tail Test: Rejection Region
22.95 days, 6 daysx s
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• The value of the test statistic t is:
A Left -Tail Test: Test Statistic
.05 1.665t t t
22.95 241.52
6 76
xts n
Since the test statistic t = 1.52, and 1.52 > 1.665,We do not reject the null hypothesis. Note that the P-value = P(t75 < -1.52) = .066 > .05.
Since our decision is to not reject the null hypothesis,A Type II error is possible.
Since the rejection region is
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• The CFO thinks that a decrease of one day in the average payment return time will cover the costs of the envelopes since customer checks can be deposited earlier.
• What is (23), the probability of a Type II error when the true mean payment return time is 23 days?
Left-Tail Test: Calculating , the Probability of a Type II Error
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– Test statistic:H0 : = 24HA : < 24
Choose = .05Rejection region in terms of z: z < -z.05 = -1.645
Left-tail test: calculating (cont.)
rejection region in terms of :
241.645
6
756
24 1.645 22.86.75
x
xz
x
= 0.05
22.86 24
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Express the rejection region directly, not in standardized
terms
22.86
=.05
= 23
Left-tail test: calculating (cont.)
– The rejection region with = .05.
22.86x
Do not reject H0
24
HA: = 23H0: = 24
Specify the alternative value under HA.
– Let the alternative value be = 23 (rather than just < 24)
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22.86 = 23
Left-tail test: calculating (cont.)
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H1: = 23
H0: = 24
0(23) ( 22.86 )
( 22.86 23)
P x given that H is false
P x given that
22.86 23.718
6 75P z
Power when =23 = 1-(23)=.282
=.05
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A Two - Tail Test for • The Federal Communications Commission
(FCC) wants competition between phone companies. The FCC wants to investigate if AT&T rates differ from their competitor’s rates.
• According to data from the (FCC) the mean monthly long-distance bills for all AT&T residential customers is $17.09.
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A Two - Tail Test (cont.)• A random sample of 100 AT&T customers is
selected and their bills are recalculated using a leading competitor’s rates.
• The mean and standard deviation of the bills using the competitor’s rates are
• Can we infer that there is a difference between AT&T’s bills and the competitor’s bills (on the average)?
$17.55, $3.87x s
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• Is the mean different from 17.09?
• n = 100; use = .05
H0: = 17.09
: 17.09AH
A Two - Tail Test (cont.)
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0
20.025 20.025
17.55 17.09
3.87.1
1001 9
xts n
-t= -1.9842 t= 1.9842
Rejection region
Rejection region
A Two – Tail Test (cont.)
.025,99 .025,99
1.9842 1.9842
t t or t t
t or t
t99
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20.025 20.02517.55 17.09
1.193.87 100
xts n
-t= -1.9842 t= 1.9842
There is insufficient evidence to conclude that there is a difference between the bills of AT&T and the competitor.
-1.19
Also, by the P-value approach:The P-value = P(t < -1.19) + P(t > 1.19) = 2(.1184) = .2368 > .05
1.190
A Two – Tail Test: Conclusion
A Type II error is possible
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• The FCC would like to detect a decrease of $1.50 in the average competitor’s bill. (17.09-1.50=15.59)
• What is (15.59), the probability of a Type II error when the true mean competitor’s bill is $15.59?
Two-Tail Test: Calculating , the Probability of a Type II Error
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20.025
17.0916.33 17.85
20.025
Rejection regionTwo – Tail Test: Calculating (cont.)
.025 .025
1.96 1.96
z z or z z
z or z
rejection region in terms of :
17.091.96
3.87100
3.8717.09 1.96
10016.33
17.091.96
3.87100
3.8717.09 1.96
10017.85
x
xz
x
x
xz
x
x
Do not reject H0
Reject H0
4616.63
= 15.59
Two – Tail Test: Calculating (cont.)
17.09
HA: = 15.59
H0: = 17.09
(15.59) (16.33 17.85 given that 15.59)P x 16.33 15.59 15.59 17.85 15.59
3.87 100 3.87 100 3.87 100
(1.912 5.84) .028
xP
P z
Power when =15.59 = 1-(15.59)=.972
=.05
17.85
General formula: Type II Error Probability (A) for a Level Test
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00
00
0 00 /2 /2
:
: 1
:
AA
AA
A AA
H P z zn
H P z zn
H P z z P z zn n
Sample Size n for which a level test also has (A) =
48
2
0
2
/2
0
( )for a 1-tailed (right or left) test
( ) for a 2-tailed test (approx. solution)
A
A
z z
nz z