Apr 19, 2023

Chapter 11: Chapter 11: Inference About a MeanInference 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

§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

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sSEx

§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

t table (Table C)

Table C:Entries t values Rows dfColumns probabilities

Use Table C to look up t values and probabilities

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

§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

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?

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

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

Two-sided P-value associated with a t statistic of -1.80 and 9 df

§11.4 Confidence Interval for µ

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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

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.

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

§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”

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

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

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 ↓ ↓ ↓ ↓

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.

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

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:

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

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

SPSS Output: Oat Bran data

§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)

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)

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

§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

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]

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z||

121

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

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21

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