1. CHAPTER 6 Statistical Inference & Hypothesis Testing
6.1 - One Sample
Mean μ, Variance σ 2, Proportion π
6.2 - Two Samples
Means, Variances, Proportions
μ1 vs. μ2 σ12 vs. σ π1 vs. π2
6.3 - Multiple Samples
Means, Variances, Proportions
μ1, …, μk σ12, …, σk π1, …, πk

2. CHAPTER 6 Statistical Inference & Hypothesis Testing
6.1 - One Sample
Mean μ, Variance σ 2, Proportion π
6.2 - Two Samples
Means, Variances, Proportions
μ1 vs. μ2 σ12 vs. σ π1 vs. π2
6.3 - Multiple Samples
Means, Variances, Proportions
μ1, …, μk σ12, …, σk π1, …, πk

3. CHAPTER 6 Statistical Inference & Hypothesis Testing
6.1 - One Sample
Mean μ, Variance σ 2, Proportion π
6.2 - Two Samples
Means, Variances, Proportions
μ1 vs. μ2 σ12 vs. σ π1 vs. π2
6.3 - Multiple Samples
Means, Variances, Proportions
μ1, …, μk σ12, …, σk π1, …, πk

4. “Parameter Estimation”
Women in U.S. who have given birth
Example: One Mean
POPULATION
“Random Variable”
X = Age (years)
Objective 1:
“Parameter Estimation”
Improve this point estimate of μ to an “interval estimate” of μ, via the…
Present: Assume that X follows a “normal distribution” in the population, with std dev σ = 1.5 yrs, but unknown mean μ = ?
That is, X ~ N(μ, 1.5).
Estimate the parameter value μ.
“Sampling Distribution of ”
standard deviation
σ = 1.5
This is referred to as a “point estimate” of μ from the sample.
mean μ = ???
Random Sample
size n = 400
mean
{x1, x2, x3, x4, … , x400}
FORMULA

5. Population Distribution of X
Sampling Distribution of
for any sample size n.
If X ~ N(μ, σ), then…
X = Age of women in U.S. who have given birth μ
“standard error” X μ
standard deviation
σ = 1.5 yrs

6. Sampling Distribution ofμ | μ
To achieve Objective 1 — obtain an “interval estimate” of μ — we first ask the following general question:
“standard error”
Suppose is any random sample mean.
Find a “margin of error” (d) so that there is a 95% probability that the interval contains μ.

7. standard normal distribution
d = (z.025)(s.e.) = (1.96)(.075 yrs) = yrs
Sampling Distribution of μ | μ
“standard error”
standard normal distribution
N(0, 1)
0.95
0.025
0.025 Z
-z.025
+z.025

8. IMPORTANT DEF’NS and FACTS standard normal distribution
d = (z.025)(s.e.) = (1.96)(.075 yrs) = yrs
d is called the “95% margin of error” and is equal to the product of the “.025 critical value” (i.e., z.025 = 1.96) times the “standard error” (i.e., ). | μ
The “confidence level” is 95%.
The “significance level” is 5%.
For any random sample mean the “95% confidence interval” is
It contains μ with probability 95%.
standard normal distribution
N(0, 1)
In this example, the 95% CI is
0.95
For instance, if a particular sample yields the 95% CI is
(25.6 – 0.147, ) =
(25.543, ) yrs.
It contains μ with 95% “confidence.”
0.025
0.025 Z
-z.025
+z.025

9. IMPORTANT DEF’NS and FACTS standard normal distribution
d = (z.025)(s.e.) = (1.96)(.075 yrs) = yrs
d = (zα/2)(s.e.)
d is called the “95% margin of error” and is equal to the product of the “.025 critical value” (i.e., z.025 = 1.96) times the “standard error” (i.e., ).
“100(1 – α)% margin of error” | μ
“α/2
zα/2)
1 – α
The “confidence level” is 95%.
1 – α.
The “significance level” is 5%. α.
For any random sample mean the “95% confidence interval” is
It contains μ with probability 95%.
“100(1 – α)% “confidence interval”
1 – α.
standard normal distribution
N(0, 1)
In this example, the 95% CI is
1 – α
0.95
For instance, if a particular sample yields the 95% CI is
(25.6 – 0.147, ) =
(25.543, ) yrs.
It contains μ with 95% “confidence.”
α/2
0.025
α/2
0.025 Z
-z.025
-zα/2
+z.025
+zα/2

10. IMPORTANT DEF’NS and FACTS standard normal distribution
d = (zα/2)(s.e.)
d is called the “95% margin of error” and is equal to the product of the “.025 critical value” (i.e., z.025 = 1.96) times the “standard error” (i.e., ).
“100(1 – α)% margin of error” | μ
“α/2
zα/2)
1 – α
The “confidence level” is 95%.
1 – α.
The “significance level” is 5%. α.
For any random sample mean the “95% confidence interval” is
It contains μ with probability 95%.
“100(1 – α)% “confidence interval”
1 – α.
standard normal distribution
N(0, 1)
What happens if we change α?
Example: α = .05, 1 – α = .95
Example: α = .10, 1 – α = .90
Example: α = .01, 1 – α = .99
1 – α
0.95
+2.575
-2.575
-1.96
+1.96
+1.645
-1.645 |
α/2
0.025
α/2
0.025 Z
Why not ask for α = 0, i.e., 1 – α = 1?
Because then the critical values → ± ∞.
-z.025
-zα/2
+z.025
+zα/2

11. IMPORTANT DEF’NS and FACTS
95% margin of error
(z.025)(s.e.) = (1.96)(.075 yrs) = yrs
In this example, the 95% CI is
For instance, if a particular sample yields the 95% CI is
(25.6 – 0.147, ) =
(25.543, ) yrs.
It contains μ with 95% “confidence.” ? μ |
standard normal distribution
N(0, 1)
0.95
0.025
+1.96
-1.96 Z
In principle, over the long run, the probability that a random interval contains μ will approach 95%.
… etc…
BUT….

12. IMPORTANT DEF’NS and FACTS
95% margin of error
(z.025)(s.e.) = (1.96)(.075 yrs) = yrs
In this example, the 95% CI is
For instance, if a particular sample yields the 95% CI is
(25.6 – 0.147, ) =
(25.543, ) yrs.
It contains μ with 95% “confidence.” μ |
standard normal distribution
N(0, 1)
0.95
0.025
+1.96
-1.96 Z
In practice, only a single, fixed interval is generated from a single random sample, so technically, “probability” does not apply.
In principle, over the long run, the probability that a random interval contains μ will approach 95%.
NOW, let us introduce and test a specific hypothesis…
BUT….

13. Random Sample Statistical Inference and Hypothesis Testing
Women in U.S. who have given birth
Study Question:
Has “age at first birth” of women in the U.S. changed over time?
Statistical Inference and
Hypothesis Testing
POPULATION
“Random Variable”
X = Age at first birth
“Null Hypothesis”
Year 2010: Suppose we know that X follows a “normal distribution” (a.k.a. “bell curve”) in the population.
Present: Is μ = still true?
H0:
public education, awareness programs
socioeconomic conditions, etc.
Or, is the “alternative hypothesis” HA: μ ≠ true?
That is, X ~ N(25.4, 1.5).
i.e., either or ?
(2-sided)
μ < 25.4
μ > 25.4
standard deviation
σ = 1.5
μ < 25.4
μ > 25.4
Does the sample statistic tend to support H0, or refute H0 in favor of HA?
mean μ = 25.4
Random Sample
{x1, x2, x3, x4, … , x400}
FORMULA
mean

14. Objective 2: Hypothesis Testing… via Confidence Interval
We have now seen:
95% CONFIDENCE INTERVAL FOR µ
25.543
25.747
“point estimate” for μ
BASED ON OUR SAMPLE DATA, the true value of μ today is between and , with 95% “confidence.”
FORMAL CONCLUSIONS:
The 95% confidence interval corresponding to our sample mean does not contain the “null value” of the population mean, μ = 25.4.
Based on our sample data, we may reject the null hypothesis H0: μ = 25.4 in favor of the two-sided alternative hypothesis HA: μ ≠ 25.4, at the α = .05 significance level.
INTERPRETATION: According to the results of this study, there exists a statistically significant difference between the mean ages at first birth in 2010 (25.4 yrs) and today, at the 5% significance level. Moreover, the evidence from the sample data suggests that the population mean age today is older than in 2010, rather than younger.
NOTE THAT THE CONFIDENCE INTERVAL ONLY DEPENDS ON THE SAMPLE, NOT A SPECIFIC NULL HYPOTHESIS!!!

15. Objective 2: Hypothesis Testing… via Confidence Interval
What if…?
We have now seen:
95% CONFIDENCE INTERVAL FOR µ
25.347
25.053
95% CONFIDENCE INTERVAL FOR µ
25.747
25.543
“point estimate” for μ
“point estimate” for μ
BASED ON OUR SAMPLE DATA, the true value of μ today is between and , with 95% “confidence.”
BASED ON OUR SAMPLE DATA, the true value of μ today is between and , with 95% “confidence.” ?
FORMAL CONCLUSIONS:
The 95% confidence interval corresponding to our sample mean does not contain the “null value” of the population mean, μ = 25.4.
Based on our sample data, we may reject the null hypothesis H0: μ = 25.4 in favor of the two-sided alternative hypothesis HA: μ ≠ 25.4, at the α = .05 significance level.
INTERPRETATION: According to the results of this study, there exists a statistically significant difference between the mean ages at first birth in 2010 (25.4 yrs) and today, at the 5% significance level. Moreover, the evidence from the sample data suggests that the population mean age today is older than in 2010, rather than younger.
NOTE THAT THE CONFIDENCE INTERVAL ONLY DEPENDS ON THE SAMPLE, NOT A SPECIFIC NULL HYPOTHESIS!!!
younger than in 2010, rather than older.

16. IF the null hypothesis H0: μ = 25.4 is indeed true, then…
Objective 2: Hypothesis Testing… via Acceptance Region
Objective 2: Hypothesis Testing… via Confidence Interval
IF the null hypothesis H0: μ = 25.4 is indeed true, then…
95% margin of error
(z.025)(s.e.) = (1.96)(.075 yrs) = yrs
Sampling Distribution of
“Null” Distribution of μ
“standard error”
… and out here…
…with 5% probability.
0.95
… we would expect a random sample mean to lie in here, with 95% probability…
0.025
0.025
95% ACCEPTANCE REGION FOR H0 |
μ = 25.4
25.253
25.547
25.4

17. Objective 2: Hypothesis Testing… via Acceptance Region
Our data value lies in the
5% REJECTION REGION.
We have now seen:
95% ACCEPTANCE REGION FOR H0
25.253
25.547
IF H0 is true, then we would expect a random sample mean to lie between and , with 95% probability.
FORMAL CONCLUSIONS:
The 95% acceptance region for the null hypothesis does not contain the sample mean of
Based on our sample data, we may reject the null hypothesis H0: μ = 25.4 in favor of the two-sided alternative hypothesis HA: μ ≠ 25.4, at the α = .05 significance level.
INTERPRETATION: According to the results of this study, there exists a statistically significant difference between the mean ages at first birth in 2010 (25.4 yrs) and today, at the 5% significance level. Moreover, the evidence from the sample data suggests that the population mean age today is older than in 2010, rather than younger.
NOTE THAT THE ACCEPTANCE REGION ONLY DEPENDS ON THE NULL HYPOTHESIS, NOT ON THE SAMPLE!!!

18. IF the null hypothesis H0: μ = 25.4 is indeed true, then…
Objective 2: Hypothesis Testing… via Acceptance Region
Objective 2: Hypothesis Testing… via “p-value”
- measures the strength of the rejection
IF the null hypothesis H0: μ = 25.4 is indeed true, then…
> 1.96
what is the probability of obtaining a random sample mean that is as, or more, extreme than the one actually obtained?
i.e., 0.2 yrs OR MORE away from μ = 25.4, ON EITHER SIDE (since the alternative hypothesis is 2-sided)?
< .05
statistically significant |
μ = 25.4
25.253
25.547
95% ACCEPTANCE REGION FOR H0
0.025
0.95

19. IF the null hypothesis H0: μ = 25.4 is indeed true, then…
Objective 2: Hypothesis Testing… via Acceptance Region
Objective 2: Hypothesis Testing… via “p-value”
- measures the strength of the rejection
IF the null hypothesis H0: μ = 25.4 is indeed true, then…
> 1.96
what is the probability of obtaining a random sample mean that is as, or more, extreme than the one actually obtained?
i.e., 0.2 yrs OR MORE away from μ = 25.4, ON EITHER SIDE (since the alternative hypothesis is 2-sided)?
< .05 α
statistically significant |
μ = 25.4
25.253
25.547
95% ACCEPTANCE REGION FOR H0
0.025
0.95
1 – α
α / 2
100(1 – α)% ACCEPTANCE REGION FOR H0
-zα/2
+zα/2

20. p-value If p-value < , then reject H0; significance!
 = .05
If p-value < , then reject H0; significance!
... But interpret it correctly!
p-value

21. ~ Summary of Hypothesis Testing for One Mean ~
Assume the population random variable is normally distributed, i.e., X  N(μ, σ).
NULL HYPOTHESIS H0: μ = μ0 (“null value”)
Test null hypothesis at significance level α.
ALTERNATIVE HYPOTHESIS HA: μ  μ0
i.e., either μ < μ0 or μ > μ0 (“two-sided”)
CONFIDENCE INTERVAL
Compute the sample mean
Compute the 100(1 – α)% “margin of error” = (critical value)(standard error)
Then the 100(1 – α)% CI =
Formal Conclusion:
Reject null hypothesis at level α, Statistical significance! Otherwise, retain it.
zα/2
if CI does not contain μ0.

22. ~ Summary of Hypothesis Testing for One Mean ~
Assume the population random variable is normally distributed, i.e., X  N(μ, σ).
NULL HYPOTHESIS H0: μ = μ0 (“null value”)
Test null hypothesis at significance level α.
ALTERNATIVE HYPOTHESIS HA: μ  μ0
i.e., either μ < μ0 or μ > μ0 (“two-sided”)
ACCEPTANCE REGION
Compute the sample mean
Compute the 100(1 – α)% “margin of error” = (critical value)(standard error)
Then the 100(1 – α)% AR =
Formal Conclusion:
Reject null hypothesis at level α, Statistical significance! Otherwise, retain it.
zα/2
if AR does not contain

23. ~ Summary of Hypothesis Testing for One Mean ~
Assume the population random variable is normally distributed, i.e., X  N(μ, σ).
NULL HYPOTHESIS H0: μ = μ0 (“null value”)
Test null hypothesis at significance level α.
ALTERNATIVE HYPOTHESIS HA: μ  μ0
i.e., either μ < μ0 or μ > μ0 (“two-sided”)
p-value
Compute the sample mean
Compute the z-score
If +, then the p-value = 2 P(Z ≥ z-score ).
If –, then the p-value = 2 P(Z ≤ z-score ).
Formal Conclusion:
Reject null hypothesis Statistical significance! Otherwise, retain it.
Remember: “The smaller the p-value, the stronger the rejection, and the more statistically significant the result.”
Z ~ N(0, 1)
z-score
if p < α.

24. statistically significant
Objective 2: Hypothesis Testing… 1-sided tests
The alternative hypothesis usually reflects the investigator’s belief!
2-sided test
H0: μ = 25.4
HA: μ  25.4
p-value
In this case,  = .05 is split evenly between the two tails, left and right.
1-sided tests
“Right-tailed”
H0: μ  25.4
HA: μ > 25.4
Here, all of  = .05 is in the right tail.
< .05
The alternative hypothesis usually reflects the investigator’s belief!
statistically significant |
μ = 25.4
95% ACCEPTANCE REGION FOR H0
0.95
25.253
25.547
0.025
.00383

25. ? < .05 statistically significant Use 1-sided tests sparingly!
Objective 2: Hypothesis Testing… 1-sided tests
2-sided test
H0: μ = 25.4
HA: μ  25.4
p-value
In this case,  = .05 is split evenly between the two tails, left and right.
1-sided tests
“Right-tailed”
H0: μ  25.4
HA: μ > 25.4
Here, all of  = .05 is in the right tail.
< .05
statistically significant
Use 1-sided tests sparingly!
0.95
0.05
0.025
0.025
.00383
.00383
95% ACCEPTANCE REGION FOR H0
95% ACCEPTANCE REGION FOR H0 |
μ = 25.4
25.253
25.547 ?

26. ? < .05 >> .05 statistically significant p-value 25.2 25.2
Objective 2: Hypothesis Testing… 1-sided tests
2-sided test
H0: μ = 25.4
HA: μ  25.4
p-value
25.2
25.2
In this case,  = .05 is split evenly between the two tails, left and right.
25.2
1-sided tests
“Right-tailed”
H0: μ  25.4
HA: μ > 25.4
Here, all of  = .05 is in the right tail.
“Left-tailed”
H0: μ  25.4
HA: μ < 25.4
Here, all of  = .05 is in the left tail.
< .05
>> .05
The alternative hypothesis usually reflects the investigator’s belief!
strong support of null hypothesis
statistically significant
0.95
0.05
95% ACCEPTANCE REGION FOR H0 |
μ = 25.4 ?

27. Subject: basic calculation of p-values for z-test
STATBOT 301
Subject: basic calculation of p-values for z-test
Calculate… from H0
Test Statistic
“z-score” =
sign of z-score?
1 – table entry
table entry
HA: μ ≠ μ0?
HA: μ < μ0
HA: μ > μ0
2 × table entry
2 × (1 – table entry) – +
Calculate… from H0
Test Statistic
“z-score” =
HA: μ ≠ μ0?
HA: μ < μ0
HA: μ > μ0
1 – table entry
table entry
sign of z-score?
2 × table entry
2 × (1 – table entry) – +

28. “Parameter Estimation”
Women in U.S. who have given birth
Example: One Mean
POPULATION
“Random Variable”
X = Age (years)
Objective 1:
“Parameter Estimation”
Improve this point estimate of μ to an “interval estimate” of μ, via the…
Present: Assume that X follows a “normal distribution” in the population, with std dev σ = 1.5 yrs, but unknown mean μ = ?
That is, X  N(μ, 1.5).
Estimate the parameter value μ.
“Sampling Distribution of “
But how do we know that the variance is the same as in 2010?
standard deviation
σ = 1.5
This is referred to as a “point estimate” of μ from the sample.
mean μ = ???
Random Sample
size n = 400
{x1, x2, x3, x4, … , x400}
FORMULA
mean

29. … which leads us to…
(1.5)2 = 2.25 in our example

30. All have postively-skewed tails.
HOWEVER……

31. In practice,  2 is almost never known, so the sample variance s 2 is used as a substitute in all calculations!