1. STAT 312 Chapter 7 - Statistical Intervals Based on a Single Sample
7.1 - Basic Properties of Confidence Intervals
7.2 - Large-Sample Confidence Intervals for a Population Mean and Proportion
7.3 - Intervals Based on a Normal Population Distribution
7.4 - Confidence Intervals for the Variance and Standard Deviation of a Normal Pop
Chapter 8 - Tests of Hypotheses Based on a Single Sample
8.1 - Hypotheses and Test Procedures
8.2 - Z-Tests for Hypotheses about a Population Mean
8.3 - The One-Sample T-Test
8.4 - Tests Concerning a Population Proportion
8.5 - Further Aspects of Hypothesis Testing

2. “Statistical Inference”
POPULATION
via… “Hypothesis Testing”
Study Question:
Has “Mean (i.e., average) Age at First Birth” of women in the U.S. changed since 2010 (25.4 yrs old)?
Present Day: Assume “Mean Age at First Birth” follows a normal distribution (i.e., “bell curve”) in the population.
Assume
Population Distribution X
H0: pop mean age  = 25.4
(i.e., no change since 2010)
Random Sample
size n = 400 ages
The reasonableness of the normality assumption is empirically verifiable (e.g., histogram, Q-Q plot) and in fact formally testable from the sample data If violated (e.g., skewed) or inconclusive (e.g., small sample size), then a transformation (e.g. logarithm) or “distribution-free” nonparametric tests should be used instead…
Examples: Sign Test, Wilcoxon Signed Rank Test (= Mann-Whitney U Test) x4 x1 x3 x2 x5
… etc…
x400

3. “Statistical Inference”
POPULATION
via… “Hypothesis Testing”
Study Question:
Has “Mean (i.e., average) Age at First Birth” of women in the U.S. changed since 2010 (25.4 yrs old)?
Present Day: Assume “Mean Age at First Birth” follows a normal distribution (i.e., “bell curve”) in the population.
Population Distribution X
H0: pop mean age  = 25.4
(i.e., no change since 2010)
Random Sample
size n = 400 ages x4 x1
Sample size n partially depends on the power of the test, i.e., the desired probability of correctly rejecting a false null hypothesis (80% or more). Coming up next! x3 x2 x5
… etc…
x400

4. P(Accept H0 | H0 is true) = 1 – α. P(Reject H0 | H0 is true) = α.
Power and Sample Size
IF the null hypothesis H0: μ = μ0 is true, then we should expect a random sample mean to lie in its “acceptance region” with probability 1 – α, the “confidence level.”
That is,
P(Accept H0 | H0 is true) = 1 – α.
Therefore, we should expect a random sample mean to lie in its “rejection region” with probability α, the “significance level.”
P(Reject H0 | H0 is true) = α.
1  
H0:  = 0
Acceptance Region for H0
Rejection Region
/2
“Null Distribution”
“Type 1 Error”
μ0 + zα/2 (σ /

5. P(Reject H0 | H0 is false) = 1 – . P(Accept H0 | H0 is false) = .
Power and Sample Size
1  
H0:  = 0
Acceptance Region for H0
Rejection Region
/2
“Null Distribution”
“Alternative Distribution”
IF the null hypothesis H0: μ = μ0 is false,
then the “power” to correctly reject it in favor of a particular alternative HA: μ = μ1 is
P(Reject H0 | H0 is false) = 1 – .
Thus,
P(Accept H0 | H0 is false) = .
1 – 
“Type 2 Error” 
HA: μ = μ1
Set them equal to each other, and solve for n…
μ0 + zα/2 (σ /
μ1 – z (σ /

6. Given:
X ~ N(μ , σ ) Normally-distributed population random variable,
with unknown mean, but known standard deviation
H0: μ = μ0 Null Hypothesis value
HA: μ = μ1 Alternative Hypothesis specific value
 significance level (or equivalently, confidence level 1 – )
1 –  power (or equivalently, Type 2 error rate  )
Then the minimum required sample size is:
N(0, 1)
1    z
Example: σ = 1.5 yrs, μ0 = 25.4 yrs,  = .05
 z.025 = 1.96
Suppose it is suspected that currently, μ1 = 26 yrs.
Want more power!
Want 90% power of correctly rejecting H0 in favor of HA, if it is false
 1 –  = .90
  = .10
 z.10 = 1.28
 = |26 – 25.4| / 1.5 = 0.4
qnorm(.9)
So… minimum sample size required is
n  66

7. Given:
X ~ N(μ , σ ) Normally-distributed population random variable,
with unknown mean, but known standard deviation
H0: μ = μ0 Null Hypothesis value
HA: μ = μ1 Alternative Hypothesis specific value
 significance level (or equivalently, confidence level 1 – )
1 –  power (or equivalently, Type 2 error rate  )
Then the minimum required sample size is:
N(0, 1)
1    z
Example: σ = 1.5 yrs, μ0 = 25.4 yrs,  = .05
 z.025 = 1.96
Change μ1
Suppose it is suspected that currently, μ1 = 26 yrs.
Want 95% power of correctly rejecting H0 in favor of HA, if it is false
Want 90% power of correctly rejecting H0 in favor of HA, if it is false
 1 –  = .90
 1 –  = .95
  = .05
  = .10
 z.05 = 1.645
 z.10 = 1.28
 = |26 – 25.4| / 1.5 = 0.4
qnorm(.9)
qnorm(.975)
So… minimum sample size required is
n  82
n  66

8. Given:
X ~ N(μ , σ ) Normally-distributed population random variable,
with unknown mean, but known standard deviation
H0: μ = μ0 Null Hypothesis value
HA: μ = μ1 Alternative Hypothesis specific value
 significance level (or equivalently, confidence level 1 – )
1 –  power (or equivalently, Type 2 error rate  )
Then the minimum required sample size is:
N(0, 1)
1    z
Example: σ = 1.5 yrs, μ0 = 25.4 yrs,  = .05
 z.025 = 1.96
Suppose it is suspected that currently, μ1 = 25.7 yrs.
Suppose it is suspected that currently, μ1 = 26 yrs.
Want 95% power of correctly rejecting H0 in favor of HA, if it is false
 1 –  = .95
  = .05
 z.05 = 1.645
 = |25.7 – 25.4| / 1.5 = 0.2
 = |26 – 25.4| / 1.5 = 0.4
qnorm(.975)
So… minimum sample size required is
n  82
n  325

9. Given:
X ~ N(μ , σ ) Normally-distributed population random variable,
with unknown mean, but known standard deviation
H0: μ = μ0 Null Hypothesis value
HA: μ = μ1 Alternative Hypothesis specific value
 significance level (or equivalently, confidence level 1 – )
1 –  power (or equivalently, Type 2 error rate  )
Then the minimum required sample size is:
N(0, 1)
1    z
Example: σ = 1.5 yrs, μ0 = 25.4 yrs,  = .05
 z.025 = 1.96
Suppose it is suspected that currently, μ1 = 25.7 yrs.
With n = 400, how much power exists to correctly reject H0 in favor of HA, if it is false?
Power = 1 –  =
= , i.e., 98%

11. Comments

12. Given:
X ~ N(μ , σ ) Normally-distributed population random variable,
with unknown mean, but known standard deviation
H0: μ = μ0 Null Hypothesis
HA: μ ≠ μ0 Alternative Hypothesis (2-sided)
 significance level (or equivalently, confidence level 1 – )
n sample size
From this, we obtain…
“standard error” s.e.
sample mean
sample standard deviation
…with which to test the null hypothesis (via CI, AR, p-value).
In practice however, it is far more common that the
true population standard deviation σ is unknown.
So we must estimate it from the sample!
(estimate)
x1, x2,…, xn
Recall that

13. Given:
X ~ N(μ , σ ) Normally-distributed population random variable,
with unknown mean, but known standard deviation
H0: μ = μ0 Null Hypothesis
HA: μ ≠ μ0 Alternative Hypothesis (2-sided)
 significance level (or equivalently, confidence level 1 – )
n sample size
From this, we obtain…
“standard error” s.e.
sample mean
sample standard deviation
…with which to test the null hypothesis (via CI, AR, p-value).
In practice however, it is far more common that the
true population standard deviation σ is unknown.
So we must estimate it from the sample!
This introduces additional variability from one sample to another… PROBLEM???
Not if n is “large”…say,  30.
(estimate)
But what if n < 30? T-test!
x1, x2,…, xn
Recall that

14. Student’s T-Distribution
… is actually a family of distributions, indexed by the degrees of freedom, labeled tdf.
William S. Gossett ( ) t2 t1
t10 t3
Z ~ N(0, 1)
As the sample size n gets large, tdf converges to the standard normal distribution Z ~ N(0, 1). So the T-test is especially useful when n < 30.

15. Student’s T-Distribution
… is actually a family of distributions, indexed by the degrees of freedom, labeled tdf.
William S. Gossett ( )
Z ~ N(0, 1) t4
.025
1.96
As the sample size n gets large, tdf converges to the standard normal distribution Z ~ N(0, 1). So the T-test is especially useful when n < 30.

16. Lecture Notes Appendix…
or…
qt(.975, 4)
qt(.025, 4, lower.tail = F)
[1]

17. Student’s T-Distribution
… is actually a family of distributions, indexed by the degrees of freedom, labeled tdf.
William S. Gossett ( )
Z ~ N(0, 1) t4
.025
.025
1.96
2.776
Because any t-distribution has heavier tails than the Z-distribution, it follows that for the same right-tailed area value, t-score > z-score.

18. If n is small, T-score > 2.
… the “T-score" increases (from ≈ 2 to a max of for a 95% confidence level) as n decreases  larger margin of error  less power to reject, even if a genuine statistically significant difference exists!
If n is large, T-score ≈ 2.

19. Given:
X = Age at first birth ~ N(μ , σ )
H0: μ = 25.4 yrs Null Hypothesis
HA: μ ≠ 25.4 yrs Alternative Hypothesis
Previously…
σ = 1.5 yrs, n = 400,
statistically significant at  = .05
Now suppose that σ is unknown, and n < 30.
Example: n = 16, s = 1.22 yrs
standard error (estimate) =
.025 critical value =
t15, .025

20. Lecture Notes Appendix…

21. Given:
X = Age at first birth ~ N(μ , σ )
H0: μ = 25.4 yrs Null Hypothesis
HA: μ ≠ 25.4 yrs Alternative Hypothesis
Previously…
σ = 1.5 yrs, n = 400,
statistically significant at  = .05
Now suppose that σ is unknown, and n < 30.
Example: n = 16, s = 1.22 yrs
95% margin of error
= (2.131)(0.305 yrs)
= yrs
standard error (estimate) =
.025 critical value =
t15, .025 =
95% Confidence Interval =
(25.9 – 0.65, ) =
(25.25, ) yrs
p-value =
Test Statistic:

22. Lecture Notes Appendix…

23. Given:
X = Age at first birth ~ N(μ , σ )
H0: μ = 25.4 yrs Null Hypothesis
HA: μ ≠ 25.4 yrs Alternative Hypothesis
Previously…
σ = 1.5 yrs, n = 400,
statistically significant at  = .05
Now suppose that σ is unknown, and n < 30.
Example: n = 16, s = 1.22 yrs
95% margin of error
= (2.131)(0.305 yrs)
= yrs
standard error (estimate) =
.025 critical value =
t15, .025 =
95% Confidence Interval =
CONCLUSIONS:
(25.9 – 0.65, ) =
The 95% CI does contain the null value μ = 25.4.
(25.25, ) yrs
The p-value is between .10 and .20, i.e., > .05.
(Note: The R command
2 * pt(1.639, 15, lower.tail = F)
gives the exact p-value as .122.)
p-value =
= 2 (between .05 and .10)
Not statistically significant; small n gives low power!
= between .10 and .20.

24. … with sample mean = 25.9 and sample sd = 1.22.
Edited R code:
y = rnorm(16, 0, 1) z = (y - mean(y)) / sd(y) x = *z
sort(round(x, 1))
Generates a normally-distributed random sample of 16 age values…
c(mean(x), sd(x))
[1]
… with sample mean = 25.9 and sample sd = 1.22.
t.test(x, mu = 25.4)
One Sample t-test
data: x
t = , df = 15, p-value =
alternative hypothesis: true mean is not equal to 25.4
95 percent confidence interval:
sample estimates:
mean of x
25.9

27. See…