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. σ2

2 π1 vs. π2

• 6.3 - Multiple Samples Means, Variances, Proportions

μ1, …, μk σ12, …, σk2 π1, …, πk

CHAPTER 6 Statistical Inference & Hypothesis Testing

Women in U.S. who have given birth

POPULATION“Random Variable”

X = Age (years)

That is, X ~ N(μ, 1.5).

Present: Assume that X follows a “normal distribution” in the population, with std dev σ = 1.5 yrs, but unknown mean μ = ?

mean x = 25.6FORMULA

mean μ = ???

{x1, x2, x3, x4, … , x400}

standard deviation σ = 1.5

This is referred to as a “point estimate” of μ from the sample.

Improve this point estimate of μ to an “interval estimate” of μ, via the…

“Sampling Distribution of ” X

size n = 400

Example: One Mean

Objective 1: “Parameter Estimation”

Estimate the parameter value μ.

X = Age of women in U.S. who have given birth

Sampling Distribution of XPopulation Distribution of X

X

μ

standard deviation σ = 1.5 yrs

If X ~ N(μ, σ), then… for any sample size n.,n

X ~ N ,

μ

X

“standard error”1.5 yrs

.075 yrs400n

n

XZ

X

Z

X d X d- +

Sampling Distribution of X

μ

X


.075 yrs400n

To achieve Objective 1 — obtain an “interval estimate” of μ — we first ask

the following general question:

Find a “margin of error” (d) so that there is a 95% probability that the interval contains μ. X d, X d- +

d

μP X - d X d< < + = 0.95

μ μP - d X d< < + = 0.95

X

. .

-d dP Z

+< < = 0.95

s.e s.e

XZ

s.e.

Suppose is any random sample mean.

X

X|μ

μP X - d X d< < + = 0.95

μ μP - d X d< < + = 0.95

. .

-d dP Z

+< < = 0.95

s.e s.e

XZ

s.e.

X|μX d X d- +


μ

X


.075 yrs400n

d

X

μP X - d X d< < + = 0.95

μ μP - d X d< < + = 0.95

. .

-d dP Z

+< < = 0.95

s.e s.e

XZ

s.e.

standard normal distributionN(0, 1)

Z

0.95

0.025 0.025

+z.025-z.025

d = (z.025)(s.e.) = (1.96)(.075 yrs) = 0.147 yrs

μP X - d X d< < + = 0.95

μ μP - d X d< < + = 0.95

. .

-d dP Z

+< < = 0.95

s.e s.e

XZ

s.e.

X|μX d X d- +

d

X

μP X - d X d< < + = 0.95

μ μP - d X d< < + = 0.95

. .

-d dP Z

+< < = 0.95

s.e s.e

XZ

s.e.


Z

0.95

0.025 0.025

+z.025-z.025

d = (z.025)(s.e.) = (1.96)(.075 yrs) = 0.147 yrs

The “confidence level” is 95%.

IMPORTANT DEF’NS and FACTS

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., ).

n

The “significance level” is 5%.

For any random sample mean the “95% confidence interval” is

It contains μ with probability 95%.

X,

- +X X ( margin of error, margin of error).

In this example, the 95% CI is- +X X ( 0.147, 0.147).

For instance, if a particular sample yields the 95% CI is (25.6 – 0.147, 25.6 + 0.147) = (25.543, 25.747) yrs. It contains μ with 95% “confidence.”

x = 25.6 yrs,

μP X - d X d< < + = 0.95

μ μP - d X d< < + = 0.95

. .

-d dP Z

+< < = 0.95

s.e s.e

XZ

s.e.

X|μX d X d- +

d

X

μP X - d X d< < + = 0.95

μ μP - d X d< < + = 0.95

. .

-d dP Z

+< < = 0.95

s.e s.e

XZ

s.e.


Z

0.95

0.025 0.025

+z.025-z.025

d = (z.025)(s.e.) = (1.96)(.075 yrs) = 0.147 yrs




n




X,




x = 25.6 yrs,

1 – α

α/2 α/2

+zα/2-zα/2

1 – α

“α/2

“100(1 – α)% margin of error”

zα/2)

1 – α.

α.

“100(1 – α)% “confidence interval”

1 – α.

d = (zα/2)(s.e.)

μ μP - d X d α< < + = 1 -

. .

-d dP Z α

+< < = 1 -

s.e s.e

Example: α = .05, 1 – α = .95 Example: α = .10, 1 – α = .90 Example: α = .01, 1 – α = .99

+1.645-1.645

μP X - d X d< < + = 0.95

μ μP - d X d< < + = 0.95

. .

-d dP Z

+< < = 0.95

s.e s.e

XZ

s.e.

X|μX d X d- +

d

X

μP X - d X d< < + = 0.95

XZ

s.e.


Z

0.95

0.025 0.025

+z.025-z.025




n




X,


1 – α

α/2 α/2

+zα/2-zα/2

1 – α

“α/2

“100(1 – α)% margin of error”

zα/2)

1 – α.

α.

“100(1 – α)% “confidence interval”

1 – α.

d = (zα/2)(s.e.)

μ μP - d X d α< < + = 1 -

What happens if we change α?

-1.96 +1.96 +2.575-2.575

|0

Why not ask for α = 0, i.e., 1 – α = 1?Because then the critical values → ± ∞.

. .

-d dP Z α

+< < = 1 -

s.e s.e

μP X - X0.147 < < + 0.147 = 0.95

μ μP - X0.147 < < + 0.147 = 0.95

95% margin of error(z.025)(s.e.) = (1.96)(.075 yrs) = 0.147 yrs


X2

X3

0.147

X1

μ|

X4

X5

X6 … etc…

In principle, over the long run, the probability that a random interval contains μ will approach 95%.

standard normal distributionN(0, 1)0.95

0.025 0.025

+1.96-1.96

Z

?

BUT….



x = 25.6 yrs,

0.784

X1

μ|

In principle, over the long run, the probability that a random interval contains μ will approach 95%.

standard normal distributionN(0, 1)0.95

0.025 0.025

+1.96-1.96

Z

BUT….

In practice, only a single, fixed interval is generated from a single

random sample, so technically, “probability” does not apply.

NOW, let us introduce and test a specific hypothesis…

μP X - X0.147 < < + 0.147 = 0.95

μ μP - X0.147 < < + 0.147 = 0.95


IMPORTANT DEF’NS and FACTSIn this example, the 95% CI is

- +X X ( 0.147, 0.147).


x = 25.6 yrs,


μ > 25.4

POPULATION

“Random Variable” X = Age at first birth

mean μ = 25.4

H0:

“Null Hypothesis”

μ < 25.4 That is, X ~ N(25.4, 1.5).

μ < 25.4

standard deviation σ = 1.5 μ > 25.4

Year 2010: Suppose we know that X follows a “normal distribution” (a.k.a. “bell curve”) in the population.

Or, is the “alternative hypothesis” HA: μ ≠ 25.4 true?

mean x = 25.6

Statistical Inference and

Hypothesis Testing

{x1, x2, x3, x4, … , x400}

Study Question:Has “age at first birth” of women in the U.S. changed over time?

• public education, awareness programs• socioeconomic conditions, etc.

FORMULA

Does the sample statistic tend to support H0, or refute H0 in favor of HA?

x = 25.6

i.e., either or ? (2-sided)

Present: Is μ = 25.4 still true?

95% CONFIDENCE INTERVAL FOR µ

25.74725.543 = 25.4

BASED ON OUR SAMPLE DATA, the true value of μ today is between 25.543 and 25.747, with 95% “confidence.”

We have now seen:

x = 25.6

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.

“point estimate” for μ

Objective 2: Hypothesis Testing… via Confidence Interval

NOTE THAT THE CONFIDENCE INTERVAL ONLY DEPENDS ON THE SAMPLE, NOT A SPECIFIC NULL HYPOTHESIS!!!



We have now seen:

= 25.4

What if…? 95% CONFIDENCE INTERVAL FOR µ

25.74725.543

FORMAL CONCLUSIONS:

The 95% confidence interval corresponding to our sample mean does not contain the “null value” of the population mean, μ = 25.4.



x = 25.6


Objective 2: Hypothesis Testing… via Confidence Interval

x = 25.2


95% CONFIDENCE INTERVAL FOR µ

25.34725.053

younger than in 2010, rather than older.

? NOTE THAT THE CONFIDENCE INTERVAL ONLY DEPENDS ON THE SAMPLE, NOT A SPECIFIC NULL HYPOTHESIS!!!

μ μP - X0.147 < < + 0.147 = 0.95

μP X - X0.147 < < + 0.147 = 0.95

P - X0.147 < < + 0.147 = 0.9525.4 25.4 P X< < = 0.9525.253 25.547

Objective 2: Hypothesis Testing… via Confidence IntervalObjective 2: Hypothesis Testing… via Acceptance Region

X|

μ = 25.4


IF the null hypothesis H0: μ = 25.4 is indeed true, then…

μ

X


.075 yrs400n


25.4

“Null” Distribution of X

25.253 25.547

95% ACCEPTANCE REGION FOR H0

… we would expect a random sample mean to lie in here, with

95% probability… X

… and out here…

…with 5% probability.

0.025 0.025

0.95

= 25.4 x = 25.6

IF H0 is true, then we would expect a random sample mean to lie between 25.253 and 25.547, with 95% probability.

x

25.54725.253


Objective 2: Hypothesis Testing… via Acceptance Region

FORMAL CONCLUSIONS:

The 95% acceptance region for the null hypothesis does not contain the sample mean of



We have now seen:

x = 25.6.

Our data value lies in the

5% REJECTION REGION.

NOTE THAT THE ACCEPTANCE REGION ONLY DEPENDS ON THE NULL HYPOTHESIS, NOT ON THE SAMPLE!!!

what is the probability of obtaining a random sample mean that is as, or more, extreme than the one actually obtained?

= 0.0077

X

Zs.e.


Objective 2: Hypothesis Testing… via Acceptance RegionObjective 2: Hypothesis Testing… via “p-value”

|

μ = 25.425.253 25.547


0.025 0.025

0.95

x = 25.6

i.e., 0.2 yrs OR MORE away from μ = 25.4, ON EITHER SIDE (since the alternative hypothesis is 2-sided)?

P X X= 25.2 U 25.6

P X P X= 25.2 25.6

P X= 2 25.6 P Z= 2 2.667= 2(0.00383)

0.003830.00383

< .05statistically significant

25.2

Z

25.6 25.4.075

> 1.96

- measures the strength of the rejection

what is the probability of obtaining a random sample mean that is as, or more, extreme than the one actually obtained?

= 0.0077

X

Zs.e.


Objective 2: Hypothesis Testing… via Acceptance RegionObjective 2: Hypothesis Testing… via “p-value”

|

μ = 25.425.253 25.547


0.025 0.025

0.95

x = 25.6

i.e., 0.2 yrs OR MORE away from μ = 25.4, ON EITHER SIDE (since the alternative hypothesis is 2-sided)?

P X X= 25.2 U 25.6

P X P X= 25.2 25.6

P X= 2 25.6 P Z= 2 2.667= 2(0.00383)

0.003830.00383

< .05statistically significant

25.2

Z

25.6 25.4.075

> 1.96

α

1 – α

α / 2 α / 2 100(1 – α)% ACCEPTANCE REGION FOR H0

-zα/2 +zα/2

- measures the strength of the rejection

= .05

If p-value < , then reject H0; significance!

... But interpret it correctly!

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.

x.

if CI does not contain μ0.

x x ( - margin of error, + margin of error).

~ Summary of Hypothesis Testing for One Mean ~

NULL HYPOTHESIS H0: μ = μ0 (“null value”)

ALTERNATIVE HYPOTHESIS HA: μ μ0 i.e., either μ < μ0 or μ > μ0 (“two-sided”)

Test null hypothesis at significance level α.

zα/2 σ n

Assume the population random variable is normally distributed, i.e., X N(μ, σ).

ACCEPTANCE REGION


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.

μ μ0 0 ( - margin of error, + margin of error).





zα/2 σ n


x.if AR does not contain

x.

p-value


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.”





x - μσ n

0


x. (shown)

or

Z ~ N(0, 1)

z-score

if p < α.

The alternative hypothesis

usually reflects the

investigator’s belief!

|

μ = 25.4


0.95

25.253 25.547

0.025 0.025.00383.00383

x = 25.6

= .0077

Objective 2: Hypothesis Testing… 1-sided tests

P X X= 25.2 U 25.6

P X P X= 25.2 25.6

P X= 2 25.6 P Z= 2 2.667= 2(.00383) < .05

statistically significant

2-sided test H0: μ = 25.4HA: μ 25.4

1-sided tests“Right-tailed”

H0: μ 25.4HA: μ > 25.4

Here, all of = .05 is in the right tail.

In this case, = .05 is split evenly between the two tails, left and right.

p-value




25.2

= .0077.00383

25.2

.00383

25.54725.253

P Z= 2 2.667

0.0250.025



P X X= 25.2 U 25.6

P X P X= 25.2 25.6

P X= 2 25.6= 2(.00383) < .05




p-value

|

μ = 25.4

0.95

.00383


H0: μ 25.4HA: μ > 25.4



x = 25.6

0.05

?

= 2(.00383) P Z= 2 2.667 P Z 2.667

(.99617)

P X X= 25.2 U 25.6

P X P X= 25.2 25.6

P X= 2 25.6


< .05




p-value

|

μ = 25.4

0.95


H0: μ 25.4HA: μ > 25.4



0.05

?

25.2

25.2

25.2

>> .05

strong support of null hypothesis

x = 25.2

“Left-tailed”H0: μ 25.4HA: μ < 25.4

Here, all of = .05 is in the left tail.




x = 25.6

= .00383.99617

STATBOT 301Subject: basic calculation of p-values for

z-test

sign of z-score?

1 – table entrytable entry

HA: μ ≠ μ0?HA: μ < μ0 HA: μ > μ0

2 × table entry 2 × (1 – table entry)

– +

CALCULATE… from H0

Test Statistic

“z-score” =

0x

n

CALCULATE… from H0

Test Statistic

“z-score” =

sign of z-score?

1 – table entrytable entry

HA: μ ≠ μ0?HA: μ < μ0 HA: μ > μ0

2 × table entry 2 × (1 – table entry)

– +

0x

n


POPULATION“Random Variable”

X = Age (years)

Present: Assume that X follows a “normal distribution” in the population, with std dev σ = 1.5 yrs, but unknown mean μ = ?

mean x = 25.6FORMULA

mean μ = ???

{x1, x2, x3, x4, … , x400}

Estimate the parameter value μ.

standard deviation σ = 1.5

This is referred to as a “point estimate” of μ from the sample.

Improve this point estimate of μ to an “interval estimate” of μ, via the…

“Sampling Distribution of “ X

size n = 400

Example: One Mean

Objective 1: “Parameter Estimation”

But how do we know that the variance is

the same as in 2010?

That is, X N(μ, 1.5).

(1.5)2 = 2.25 in our example

… which leads us to…

All have postively-skewed tails.

HOWEVER……

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

CHAPTER 6 Statistical Inference & Hypothesis Testing

Documents

Transcript of CHAPTER 6 Statistical Inference & Hypothesis Testing