chapter 11: inference about a mean
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Chapter 11: Inference About a Mean. In Chapter 11:. 11.1 Estimated Standard Error of the Mean 11.2 Student’s t Distribution 11.3 One-Sample t Test 11.4 Confidence Interval for μ 11.5 Paired Samples 11.6 Conditions for Inference 11.7 Sample Size and Power. - PowerPoint PPT PresentationTRANSCRIPT
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Apr 20, 2023
Chapter 11: Chapter 11: Inference About a MeanInference About a Mean
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In Chapter 11:
11.1 Estimated Standard Error of the Mean
11.2 Student’s t Distribution
11.3 One-Sample t Test
11.4 Confidence Interval for μ
11.5 Paired Samples
11.6 Conditions for Inference
11.7 Sample Size and Power
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§11.1 Estimated Standard Error of the Mean
• We rarely know population standard deviation σ instead, we calculate sample standard deviations s and use this as an estimate of σ
• We then use s to calculate this estimated standard error of the mean:
• Using s instead of σ adds a source of uncertainty z procedures no longer apply use t procedures instead
n
sSEx
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§11.2 Student’s t distributions • A family of distributions identified by “Student”
(William Sealy Gosset) in 1908
• t family members are identified by their degrees of freedom, df.
• t distributions are similar to z distributions but with broader tails
• As df increases → t tails get skinnier → t become more like z
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t table (Table C)
Table C:Entries t values Rows dfColumns probabilities
Use Table C to look up t values and probabilities
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Left tail:Pr(T9 < -1.383) = 0.10
Right tail:Pr(T9 > 1.383) = 0.10
Understanding Table CLet tdf,p ≡ a t value with df degrees of freedom and cumulative probability p. For example, t9,.90 = 1.383
Table C. Traditional t tableCumulative p 0.75 0.80 0.85 0.90 0.95 0.975
Upper-tail p 0.25 0.20 0.15 0.10 0.05 0.025
df = 9 0.703 0.883 1.100 1.383 1.833 2.262
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§11.3 One-Sample t TestA. Hypotheses. H0: µ = µ0 vs. Ha: µ ≠ µ0 (two-sided)
[ Ha: µ < µ0 (left-sided) or Ha: µ > µ0 (right-sided)]
B. Test statistic.
C. P-value. Convert tstat to P-value [table C or software]. Small P strong evidence against H0
D. Significance level (optional). See Ch 9 for guidelines.
1 with 0stat
ndf
ns
xt
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One-Sample t Test: ExampleStatement of the problem: • Do SIDS babies have lower than average birth
weights?• We know from prior research that the mean birth
weight of the non-SIDs babies in this population is 3300 grams
• We study n = 10 SIDS babies, determine their birth weights, and calculate x-bar = 2890.5 and s = 720.
• Do these data provide significant evidence that SIDs babies have different birth weights than the rest of the population?
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A. H0: µ = 3300 versus Ha: µ ≠ 3300 (two-sided)
B. Test statistic
C. P = 0.1054 [next slide]Weak evidence against H0
D. (optional) Data are not significant at α = .10
91101
80.110720
33005.2890 0
stat
ndf
SE
xt
x
One-Sample t Test: Example
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Converting the tstat to a P-value
tstat P-value via Table C. Wedge |tstat| between critical value landmarks on Table C. One-tailed 0.05 < P < 0.10 and two-tailed 0.10 < P < 0.20.
tstat P-value via software. Use a software utility to determine that a t of −1.80 with 9 df has two-tails of 0.1054.
Table C. Traditional t tableCumulative p 0.75 0.80 0.85 0.90 0.95 0.975
Upper-tail p 0.25 0.20 0.15 0.10 0.05 0.025
df = 9 0.703 0.883 1.100 1.383 1.833 2.262
|tstat| = 1.80
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Two-sided P-value associated with a t statistic of -1.80 and 9 df
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§11.4 Confidence Interval for µ
n
stx n 21,1for CI %100)1(
• Typical point “estimate ± margin of error” formula • tn-1,1-α/2 is from t table (see bottom row for conf. level)• Similar to z procedure except uses s instead of σ • Similar to z procedure except uses t instead of z• Alternative formula:
n
sSESEtx xxn where
21,1
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Confidence Interval: Example 1
grams 3405.6) to(2375.4 =
515.1 ±5.2890 10
720262.25.2890
for CI 95%
10 0.720 5.2890
205.1,110
n
stx
nsx
Let us calculate a 95% confidence interval for μ for the birth weight of SIDS babies.
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Confidence Interval: Example 2
Data are “% of ideal body weight” in 18 diabetics: {107, 119, 99, 114, 120, 104, 88, 114, 124, 116, 101, 121, 152, 100, 125, 114, 95, 117}. Based on these data we calculate a 95% CI for μ.
120.0) (105.6, = 7.17 ±112.778
)44.3)(110.2(778.112))((
table) (from 110.2
400.318
242.14
18 424.14 778.112
2
205.
2
1,1
975,.171,1181,1
xn
n
x
SEtx
ttttn
sSE
nsx
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§11.5 Paired Samples
• Paired samples: Each point in one sample is matched to a unique point in the other sample
• Pairs be achieved via sequential samples within individuals (e.g., pre-test/post-test), cross-over trials, and match procedures
• Also called “matched-pairs” and “dependent samples”
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Example: Paired Samples
• A study addresses whether oat bran reduce LDL cholesterol with a cross-over design.
• Subjects “cross-over” from a cornflake diet to an oat bran diet. – Half subjects start on CORNFLK, half on OATBRAN– Two weeks on diet 1 – Measures LDL cholesterol– Washout period– Switch diet– Two weeks on diet 2– Measures LDL cholesterol
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Example, Data Subject CORNFLK OATBRAN
---- ------- -------
1 4.61 3.84
2 6.42 5.57
3 5.40 5.85
4 4.54 4.80
5 3.98 3.68
6 3.82 2.96
7 5.01 4.41
8 4.34 3.72
9 3.80 3.49
10 4.56 3.84
11 5.35 5.26
12 3.89 3.73
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Calculate Difference Variable “DELTA”• Step 1 is to create difference variable “DELTA”• Let DELTA = CORNFLK - OATBRAN • Order of subtraction does not materially effect results (but but does
change sign of differences)• Here are the first three observations:
Positive values represent lower LDL on oatbran
ID CORNFLK OATBRAN DELTA ---- ------- ------- ----- 1 4.61 3.84 0.77 2 6.42 5.57 0.85 3 5.40 5.85 -0.45 ↓ ↓ ↓ ↓
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Explore DELTA Values
Stemplot
|-0|42|+0|0133|+0|667788 ×1
Here are all the twelve paired differences (DELTAs): 0.77, 0.85, −0.45, −0.26, 0.30, 0.86, 0.60, 0.62, 0.31, 0.72, 0.09, 0.16
EDA shows a slight negative skew, a median of about 0.45, with results varying from −0.4 to 0.8.
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Descriptive stats for DELTA
Data (DELTAs): 0.77, 0.85, −0.45, −0.26, 0.30, 0.86, 0.60, 0.62, 0.31, 0.72, 0.09, 0.16
0.4335
0.3808
12
d
d
s
x
n
The subscript d will be used to denote statistics for difference variable DELTA
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95% Confidence Interval for µd
n
stx dndd 21,1 for CI %100)1(
A t procedure directed toward the DELTA variable calculates the confidence interval for the mean difference.
)656.0 to105.0(
2754.00.3808 12
4335.201.20.3808 for CI %95
C) Table (from 201.2 use confidence 95%For 975,.111112 205
d
, tt .
“Oat bran” data:
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Paired t Test
• Similar to one-sample t test • μ0 is usually set to 0, representing “no
mean difference”, i.e., H0: μ = 0• Test statistic:
ndf
ns
xt
d
d
1
0stat
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Paired t Test: Example“Oat bran” data
A. Hypotheses. H0: µd = 0 vs. Ha: µd 0B. Test statistic.
C. P-value. P = 0.011 (via computer). The evidence against H0 is statistically significant.
D. Significance level (optional). The evidence against H0 is significant at α = .05 but is not significant at α = .01
111121
043.312/4335.
038083.0 0
stat
ndf
ns
xt d
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SPSS Output: Oat Bran data
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§11.6 Conditions for Inference
t procedures require these conditions:
• SRS (individual observations or DELTAs)
• Valid information (no information bias)
• Normal population or large sample (central limit theorem)
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The Normality Condition
The Normality condition applies to the sampling distribution of the mean, not the population. Therefore, it is OK to use t procedures when:
• The population is Normal
• Population is not Normal but is symmetrical and n is at least 5 to 10
• The population is skewed and the n is at least 30 to 100 (depending on the extent of the skew)
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Can a t procedures be used?• Dataset A is skewed
and small: avoid t procedures
• Dataset B has a mild skew and is moderate in size: use t procedures
• Data set C is highly skewed and is small: avoid t procedure
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§11.7 Sample Size and Power
• Questions: – How big a sample is needed to limit the
margin of error to m?
– How big a sample is needed to test H0 with 1−β power at significance level α?
– What is the power of a test given certain conditions?
• In this presentation, we cover only the last question
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Power
where: • α ≡ (two-sided) alpha level of the test• Δ ≡ “the mean difference worth detecting” (i.e.,
the mean under the alternative hypothesis minus the mean under the null hypothesis)
• n ≡ sample size• σ ≡ standard deviation in the population• Φ(z) ≡ the cumulative probability of z on a
Standard Normal distribution [Table B]
n
z||
121
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Power: Illustrative Example
SIDS birth weight example. Consider the SIDS illustration in which n = 10 and σ is assumed to be 720 gms. Let α = 0.05 (two-sided). What is the power of a test under these conditions to detect a mean difference of 300 gms?
26%about ispower The
2611.064.0
720
10|300|96.1
||1
21
n
z