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8/3/2019 Stats Lecture 09 Small Samples
http://slidepdf.com/reader/full/stats-lecture-09-small-samples 1/44
Inference From Small Samples
Quantitative Methods for EconomicsDr. Katherine SauerMetropolitan State College of Denver
8/3/2019 Stats Lecture 09 Small Samples
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Chapter Overview:
I. Normal Population, σ is known
II. The t -distribution (aka Student’s t -distribution)III. Difference Between Means from Small, Independent SamplesIV. The F-test for equality of two variancesV. Difference between Means, Paired Samples
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I. Normal Population, σ is known
For n < 30:
When the population is Normal and the population standard deviationis known, then the sampling distribution for sample means is
n N x
,~
The confidence interval is
The Test Statistic is
x Z x 2 /
x
H x Z
0
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Example: The temperature (degrees C) of a cooled storage unit istaken on 8 consecutive days.
4.5 4.8 5.2 4.7 3.8 3.7 4.1 3.9
Temperatures for this type of storage unit are known to beNormally distributed with a standard deviation of σ =0.35.
Construct a 90% confidence interval for the true mean temperature.
3375.4 xCalculate the sample mean:
For α = 0.10, Z α /2 = 1.6449
Calculate the standard error:
x Z x 2 /
1237.0
8
35.0
n x
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4.3375 + 1.6449(0.1237)4.3375 + 0.2035
4.1340 to 4.5410
We are 90% sure that the true population mean is in thisinterval.
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Test the hypothesis that the mean temperature is 4 degrees.
H0: µ = 4H1: µ ≠ 4
For α = 0.10, α / 2 = 0.05 Z = 1.6449
µ= 4
Z = -1.6449
RejectH0
Accept H0
Z = 1.6449
RejectH0
Reject the null if Z > 1.6449 or Z < -1.6449
x
H x
x Z
0
n x
73.21237.0
43375.4 x Z
Z = 2.73
Reject the null and conclude that the
average temperature is not 4 degrees.
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p-value = pr (Z > 2.73) + pr(Z < -2.73)= 0.0032 + 0.0032= 0.0064
There is only a 0.64% chance of selecting the given sample if the true mean is 4.
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Often, we don’t know the population standard deviation.
We can no longer use the Z table.
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II. The t -distribution (aka Student’s t -distribution)
Fun origin: A chemist at the Guinness brewery in Dublin inventedthe t-distribution in order to monitor quality in brewing, usingsmall samples from Normal populations with σ unknown.
If random samples of size n are selected from a Normal populationwith mean µ and σ unknown, then the distribution of sample meansis a t-distribution.
xn st x ,~ 1 n
ss
x
(n-1) refers to the degrees of freedom
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The t-distribution is similar to the Normal distribution in severalways:
it is bell shapedit is symmetrical about the mean
is the number of standard errors between thesample mean and population mean x
s
xt
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Ex: find the tailarea equal to5% when the
sample size is10.
10-1 =9 degreesof freedom
Tail area = 0.05
Critical t-valueis 1.8331
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In large samples, when σ is unknown, we often use Z instead of t.
When samples are large, Z and t are close.
Statistical software always uses t when σ is unknown, even forlarge samples.
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The confidence interval for a small sample from a Normalpopulation with unknown σ is
xn st x 2 / ,1
The test statistic for a small sample from a Normal populationwith unknown σ is
xs
xt
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Example: The waiting time at an airline check in counter isknown to be Normally distributed. A random sample of 5passengers were interviewed. They reported the following waittimes: 15.5 21.2 12.6 18.4 22.9 minutes.Construct a 90% confidence interval for the average wait time.
Calculate the sample average wait time: 12.18 x
n
ss
xCalculate the standard error:
Remember todivide by n-1
for thevariance!!!
xi xi - mean (xi - mean)^2 mean 18.1215.5 -2.62 6.8644 variance 17.43721.2 3.08 9.4864 st dev 4.17576312.6 -5.52 30.470418.4 0.28 0.078422.9 4.78 22.8484
69.748 sum
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8675.15
1758.4
n
ss x
xn st x 2 / ,1
Find the critical value for t :tn-1, α /2 = t 4, 0.05 = 2.1318
Construct the interval:
18.12 + (2.1318)(1.8675)18.12 + 3.981114.1389 to 22.1011
We are 90% confident that the average wait time is inthis range.
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Example: Test the hypothesis that the average wait time is atmost 20 minutes.
1. State the null and alternative hypothesesH0: µ < 20H1: µ > 20
one-sided test, upper tail
2. Sketch the graph and identify the critical regionα =0.10 t 4, 0.1 = 1.5332
µ =20
t =1.5332
RejectH0
Accept H0 Accept H 0 if 1.5332 < t
Reject H 0 if t > 1.5332
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3. Calculate t: x
s
xt
n
ss
x
8675.15
1758.4n
ss x 0067.18675.1
2012.18t
12.18 x
µ =20
t =1.5332
RejectH0
Accept H0
t = -1.0067
Accept H 0 because-1.0067 < 1.5332
Accept the null and validate the claim that at most the averagewait time is 20 minutes.
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4. p-value is the area to the right of -1.0067
(rarely look up in t- distribution table… software)
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Example: The temperature (degrees C) of a cooled storage unit istaken on 8 consecutive days.
4.5 4.8 5.2 4.7 3.8 3.7 4.1 3.9
At the 90% level, test the hypothesis that the mean temperature is4 degrees.
H0: µ = 4H1: µ ≠ 4
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xi xi - mean (xi - mean)^2 mean 4.33754.5 0.1625 0.0264 variance 0.2941074.8 0.4625 0.2139 st dev 0.542316
5.2 0.8625 0.74394.7 0.3625 0.13143.8 -0.5375 0.28893.7 -0.6375 0.40644.1 -0.2375 0.05643.9 -0.4375 0.1914
2.0588 sum
Let’s verify the output:
1917.08
542316.0
n
ss x
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xn st x 2 / ,1 t 7, 0.05 = 1.8946
4.3375 + (1.8946)(0.19174)4.3375 + 0.3633
3.9742 to 4.7008
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xs
xt
76.1
19174.0
43375.4t
This is a two-tail test.
- 1.8946 < 1.76 < 1.8946
Accept the null.
If we had rejected the null, the p-value would have told us thelevel of significance.
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III. Difference Between Means from Small, Independent Samples
Example: Promoters of e-learning software design a test foreffectiveness of an online course based on typing tutor software.Two groups are randomly selected. Group 1 consists of 10 subjectswho have completed a course that did not use supporting software.Group 2 consists of 8 subjects who used the online software.
The typing speeds (wpm) are as follows.Group 1: 23, 35, 37, 12, 26, 60, 13, 24, 27, 53
Group 2: 56, 30, 55, 48, 35, 40, 33, 23
Construct a 90% confidence interval for the difference in meantyping speed between the two groups. Can you conclude that thosewho used the online software can type faster?
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xi xi - mean ( xi - mean)^2 xi xi - mean ( xi - mean)^223 -8 64 56 16 25635 4 16 30 -10 10037 6 36 55 15 22512 -19 361 48 8 6426 -5 25 35 -5 2560 29 841 40 0 013 -18 324 33 -7 4924 -7 49 23 -17 28927 -4 16 sum 100853 22 484
sum 2216
mean 31 mean 40variance 246.2222 variance 144
st dev 15.69147 st dev 12
Group 1 Group 2
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We’ll need to construct a pooled estimate of variance .
2
)1()1(
21
222
2112
nn
snsns
p
5.2012810
)12)(18()69147.15)(110( 222 ps
Use the pooled estimate of variance to find the standard error.
21
2 1121 nn
ss p x x
7333.68
1
10
15.201
21
x xs
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Find the critical t value:degrees of freedom = n1 + n2 – 2
= 16
α / 2 = 0.05
t 16, 0.05 = 1.7459
Construct the interval:40 – 31 + 1.7459(6.7333)9 + 11.7557
-2.7557 to 20.7557The interval contains 0. We can conclude that thedifference between means is zero.
Typing speeds between the 2 groups are the same.
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At the 95% level, test the hypotheses that the mean typing speed isfaster for those who used the software.
H0: µ 1 = µ 2H1: µ 1 > µ 2
one tailed test
α = 0.05
t 16, 0.05 = 1.7459
µ1 = µ2
Accept H0
t = 1.7459
RejectH0
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t =1.3366
The test statistic is
21
2
2121
11
)()(
nns
x xt
p
3366.17333.6
)0()3140(t
µ1 = µ2
Accept H0
t = 1.7459
RejectH0
Accept the nullhypotheses that thetyping speed of both
groups is the same.
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Assumptions made in solving this problem:1. independent samples2. random samples from Normal populations3. the variance is the same for both populations
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IV. The F-test for equality of two variances
To figure out if two populations have similar variances, we will look at the sample variances.
If the ratio of the sample variances is close to 1, then the hypothesisthat the populations have equal variance is plausible.
The sampling distribution of is an F-distribution, when thesamples are independent and selected from Normal populationswith equal variances.
2
2
2
1
s
s
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The F-distribution is not symmetrical and depends on thedegrees of freedom in each sample.
v1 = n1 – 1 v2 = n2 - 1
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Ex: Suppose sample 1 has 10 observations and sample 2 has 8observations. Find the critical F-value for the 5% level.
v1 = 9 v2 = 7
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If we wanted the 2.5% level, we’d need a different table.
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Example: Using the data from the typing example, test whetherthe sample variances are equal at the 95% level.
H0: σ21 = σ22 H1: σ2
1 ≠ σ 22
this is a 2-tail test
α /2 = 0.025F: v1 = 10-1 = 9 v2 = 8-1 = 7
F = 4.82
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Calculate the test statistic2
2
2
1
s
s
7099.1144
22.24622
21
ss
F = 1.7099
Accept the nullhypothesis andconclude that thepopulation variancesare equal.
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Instead, test the hypothesis that the variance of population 1exceeds the variance of population 2.
H0: σ21 < σ22 H1: σ2
1 > σ22
this is a 1-tail test, upper tail
α= 0.05F: v1 = 10-1 = 9 v2 = 8-1 = 7
F = 3.69
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Calculate the test statistic2
2
2
1
s
s
7099.1144
22.24622
21
ss
F = 1.7099
Accept the nullhypothesis andconclude thevariance of
population 1 is lessthan or equal to thevariance of population 2.
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V. Difference between Means, Paired Samples
Paired t-tests are used when data consists of pairs of measurementson the same subjects.
ex: before and after
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Example: The typing speeds for 7 people are recorded before andafter completing a course using typing tutor software.
Person Before After DifferenceJM 32 46 14AC 10 18 8TB 65 58 -7
AF 39 50 11AO 24 36 12PD 10 24 14FF 24 21 -3
Construct a 90% confidence interval for the difference betweenaverage typing speed before and after the course.
α /2 = 0.05degrees of freedom = 7-1 = 6
t 6, 0.05 = 1.9432
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Calculate the mean of the differences:49 / 7 = 7
Calculate the sample standard deviation:
Person Difference dif - mean (dif - mean)^2JM 14 7 49AC 8 1 1TB -7 -14 196
AF 11 4 16AO 12 5 25PD 14 7 49FF -3 -10 100
436
variance 72.6667st dev 8.5245
Calculate the sample standard error:2219.3
7
5242.8
n
ss d
d
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Construct the interval:
7 + 1.9432(3.2219)7 + 6.2608
0.7392 to 13.2608
We are 90% confident that the true difference in average typingspeeds is between 0.7392 words per minute and 13.2608 words
per minute.
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Now at the 2.5% level, test the hypothesis that typing speeds have
increased after taking the course.
H0 : µ d < 0H1: µ d > 0
one sided test
α = 0.025
degrees of freedom = 6
t 6, 0.025 = 2.447
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t =2.1726
µd = 0
Accept H0
t = 2.447
RejectH0
Calculate the test statistic:
sterror claim H estimate
t 0
1726.22219.3
07t
Accept the null hypothesis andconclude that typing speeds did
not improve during the course.
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Concepts:t-distributionF-distribution
Skills:Construct confidence interval and perform hypothesis test formeans from small, independent samples
Perform an F-test
Construct confidence interval and perform hypothesis test for the
difference between means from small, independent samples
Construct confidence interval and perform hypothesis test for thedifference between paired means from small, independent samples
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