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Chapter 17. Inference about a Population Mean. 1. σ not known. In practice, we do not usually know population standard deviation σ Therefore, we cannot calculate σ x-bar Instead, we calculate this standard error of the mean:. 2. t Procedures. - PowerPoint PPT Presentation
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204/20/23 Inference about µ 2
σ not known
nsSEx
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:
304/20/23 Inference about µ 3
t Procedures
ns
μxt
nσ
μxz 00
Because σ is now known, we do NOT use z statistics. Instead, we use this t statistic
T procedures are based on Student’s t distribution
404/20/23 Inference about µ 4
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
604/20/23 Inference about µ 6
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
704/20/23 Inference about µ 7
Using Table CQuestion: What t critical value should I use for
95% confidence when df = 7?
Answer: t* = 2.365
804/20/23 Inference about µ 8
Confidence Interval for μ
n
stx
t* is the critical value with df = n−1 and C level of confidence
Lookup in Table C
9
ExampleStatement : 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.
10
Example (Solution)
n
stx * for CI 95%
C) (Table 262.2 :confidence 95%For ;91101 * tndf
grams 3406) to(2375 =
515.1 ±5.2890
We are 95% confident population mean µ is between 2375 and 3406 gms.
10
720262.25.2890
grams 720 grams 5.2890 sx
1104/20/23 Inference about µ 11
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)
1204/20/23 Inference about µ 12
One-Sample t Test
1 with
ndfns
μxt 0One-sample t statistic:
P-value = tail beyond tstat (use Table C)
13Basics of Significance Testing 13
P-value: Interpretation • P-value (interpretation) Smaller-and-smaller P-
values 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
1404/20/23 Inference about µ 14
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).
Example: “Weight Gain”
2.0, 0.4, 0.7, 2.0, −0.4, 2.2, −1.3, 1.2, 1.1, 2.3
lbs. 196.1 lbs.; 020.1 sxCalculate:
Do data provide significant evidence for a weight change?
1504/20/23 Inference about µ 15
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
1704/20/23 Inference about µ 17
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
18
Two-tailed P-value• For two-sided Ha,
P-value = 2 × one-sided 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
19
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
2004/20/23 Inference about µ 20
Paired SamplesResponses in matched pairs
Parameter μ now represents the population mean difference
2104/20/23 Inference about µ 21
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 A B A – B
1 2.92 1.84 1.08
2 1.88 0.95 0.93
3 5.35 4.26 1.09
4 3.81 3.18 0.63
5 4.69 3.44 1.25
6 4.86 3.69 1.17
7 5.81 4.95 0.86
8 5.55 4.47 1.08
1960.0 and 0113.1 sx
2204/20/23 Inference about µ 22
Hypotheses:
H0: μ = 0 (note: µ0 = 0, representing no mean difference)
Ha: μ > 0 (one-sided)
Ha: μ ≠ 0 (two-sided)
Test Statistic:
ns
μxt
0
Example: Matched Pairs
7181 n df80.1960
01.0113 14.59
2304/20/23 Inference about µ 23
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
Illustration (cont.)
2404/20/23 Inference about µ 24
0.16391.0113
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)
95% Confidence Interval for µ
8
0.19602.3651.0113
n
stx
1.1752 to 0.847495% confidence population mean difference µ is between 0.847 and 1.175
2504/20/23 Inference about µ 25
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
Interpreting the Confidence Interval
2604/20/23 Inference about µ 26
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)
2704/20/23 Inference about µ 27
Can we use a t procedure?
Moderately sized dataset (n = 20) w/strong skew. t procedures cannot be trusted
2804/20/23 Inference about µ 28
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.