Download Document

Chapter 17
Inference about a
Population Mean
5/24/2017
Inference about µ
1
σ not known
In practice, we do not usually know
population standard deviation σ
Therefore, we cannot calculate σx-bar
Instead, we calculate this standard error
of the mean:
SE x  s
5/24/2017
n
Inference about µ
2
t Procedures
Because σ is now known, we do NOT use z
statistics. Instead, we use this t statistic
x  μ0
z
σ
n

x  μ0
t
s
n
T procedures are based on
Student’s t distribution
5/24/2017
Inference about µ
3
Student’s t Distributions
• A “family” of distributions
• Each family member has different
degrees of freedom (df)
• More area in their tails than Normal
distributions (fatter tails)
• As df increases, s becomes a better
estimate of σ and the t distributions
becomes more Normal
• t with more than 30 df  very similar to z
5/24/2017
Inference about µ
4
t Distributions
5/24/2017
Inference about µ
5
Table C “t Table”
Table entries = t* critical values
Rows = df; Columns = probability levels
Familiarize yourself with the t table in the
“Tables and Formulas for Moore” handout
5/24/2017
Inference about µ
6
Using Table C
Question: What t critical value should I use for
95% confidence when df = 7?
Answer: t* = 2.365
5/24/2017
Inference about µ
7
Confidence Interval for μ
s
x t 
n

t* is the critical value with df = n−1
and C level of confidence
Lookup in Table C
5/24/2017
Inference about µ
8
Example
Statement : What is the population mean µ birth
weight of the SIDS population?
Data: We take an SRS of n = 10 from the population of
SIDS babies and retrieve their birth certificates. This was
their birth weights (grams): 2998, 3740, 2031, 2804,
2454, 2780, 2203, 3803, 3948, 2144
Plan: We will calculate the sample mean and
standard deviation. We will then calculate and
interpret the 95% CI for µ.
9
Example (Solution)
x  2890.5 grams s  720 grams
df  n  1  10  1  9; For 95% confidence : t *  2.262 (Table C)
s
95% CI for   x  t 
n
720
 2890.5  2.262 
10
 2890.5 ± 515.1
*
= (2375 to 3406) grams
We are 95% confident population mean µ is
between 2375 and 3406 gms.
10
One-Sample t Test (Hypotheses)
• Draw simple random sample of size n from a
large population having unknown mean µ
• Test null hypothesis H0: μ = μ0
where μ0 ≡ stated value for the population mean
– μ0 changes from problem to problem
– μ0 is NOT based on the data
– μ0 IS based on the research question
• The alternative hypothesis is:
– Ha: μ > μ0 (one-sided looking for a larger value) OR
– Ha: μ < μ0 (one-sided looking for a smaller value) OR
– Ha: μ ≠ μ0 (two-sided)
5/24/2017
Inference about µ
11
One-Sample t Test
One-sample t statistic: t 
x  μ0
s
n
with df  n  1
P-value = tail
beyond tstat
(use Table C)
5/24/2017
Inference about µ
12
P-value: Interpretation
• P-value (interpretation) Smaller-and-smaller Pvalues indicate stronger-and-stronger evidence
against H0
• Conventions:
.10 < P < 1.0  evidence against H0 not significant
.05 < P ≤ .10  evidence against H0 marginally signif.
.01 < P ≤ .05  evidence against H0 significant
P ≤ .01  evidence against H0 highly significant
Basics of Significance Testing
13
Example: “Weight Gain”
Statement: We want to know whether there is good
evidence for weight change in a particular population. We
take an SRS on n = 10 from this population and find the
following changes in weight (lbs).
2.0, 0.4, 0.7, 2.0, −0.4, 2.2, −1.3, 1.2, 1.1, 2.3
Calculate: x  1.020 lbs.; s  1.196 lbs.
Do data provide significant evidence for a
weight change?
5/24/2017
Inference about µ
14
Example “Weight Gain” (Hypotheses)
• Under null hypothesis, no weight gain in population
H0: μ = 0
Note: µ0 = 0 in this particular example
• One-sided alternative, weight gain in population.
Ha: μ > 0
• Two-sided alternative hypothesis, weight change:
Ha: μ ≠ 0
5/24/2017
Inference about µ
15
Example (Test Statistic)
t 
x  μ0
s

1.020  0
 2.70
1.196
n
10
df  10  1  9
5/24/2017
Inference about µ
16
Example (P-value)
• Table C, row for 9 df
• t statistic (2.70) is between
t* = 2.398 (P = 0.02) and t* = 2.821 (P = 0.01)
• One-sided P-value is between .01 and .02:
.01 < P < .02
5/24/2017
Inference about µ
17
Two-tailed P-value
• For two-sided Ha,
P-value = 2 × onesided P
• In our example, the
one-tailed P-value was
between .01 and .02
• Thus, the two-tailed P
value is between .02
and .04
18
Interpretation
• Interpret P-value in
context of claim made
by H0
• In our example,
H0: µ = 0
(no weight gain)
• Two-tailed P-value
between .02 and .04
• Conclude: significant
evidence against H0
19
Paired Samples
Responses in matched pairs
Parameter μ now represents the
population mean difference
5/24/2017
Inference about µ
20
Example: Matched Pairs
• Pollution levels in
two regions (A & B)
on 8 successive days
• Do regions differ
significantly?
• Subtract B from A =
last column
• Analyze differences
Day
1
A
2.92
B
1.84
A–B
1.08
2
3
4
1.88
5.35
3.81
0.95
4.26
3.18
0.93
1.09
0.63
5
6
7
4.69
4.86
5.81
3.44
3.69
4.95
1.25
1.17
0.86
8
5.55
4.47
1.08
x  1.0113 and s  0.1960
5/24/2017
Inference about µ
21
Example: Matched Pairs
Hypotheses:
H0: μ = 0 (note: µ0 = 0, representing no mean
difference)
Ha: μ > 0 (one-sided)
Ha: μ ≠ 0 (two-sided)
Test Statistic:
t 
x  μ0
s
 14.59
0.1960
n
5/24/2017

1.0113  0
8
Inference about µ
df  n  1  8  1  7
22
Illustration (cont.)
P-value:
• Table C  7 df row
• t statistic is greater than largest value in
table: t* = 5.408 (upper p = 0.0005).
• Thus, one-tailed P < 0.0005
• Two-tailed P = 2 × one-tailed P-value:
P < 0.001
• Conclude: highly significant evidence
against H0
5/24/2017
Inference about µ
23
95% Confidence Interval for µ
Air pollution data:
n = 8, x-bar = 1.0113, s = 0.1960
df = 8  1 = 7
For 95% confidence, use t* = 2.365 (Table C)
x t

s
n
 1.0113  2.365
0.1960
 1.0113  0.1639
8
 0.8474 to 1.1752
95% confidence population mean difference µ is between 0.847 and 1.175
5/24/2017
Inference about µ
24
Interpreting the Confidence Interval



The confidence interval seeks population mean
difference µ (IMPORTANT)
Recall the meaning of “confidence,” i.e., the
ability of the interval to capture µ upon
repetition
Recall from the prior chapter that the
confidence interval can be used to address a
null hypothesis
5/24/2017
Inference about µ
25
Normality Assumption
• t procedures require Normality, but they are
robust when n is “large”
• Sample size less than 15: Use t procedures if
data are symmetric, have a single peak with no
outliers. If data are highly skewed, avoid t.
• Sample size at least 15: Use t procedures
except in the presence of strong skewness.
• Large samples: Use t procedures even for
skewed distributions when the sample is large (n
≥ ~40)
5/24/2017
Inference about µ
26
Can we use a t procedure?
Moderately sized dataset (n = 20) w/strong skew.
t procedures cannot be trusted
5/24/2017
Inference about µ
27
Word lengths in Shakespeare’s
plays (n ≈ 1000)
The data has a strong positive skew but since the
sample is large, we can use t procedures.
5/24/2017
Inference about µ
28
Can we use t?
The distribution has no clear violations of
Normality. Therefore, we trust the t procedure.
5/24/2017
Inference about µ
29