• 6.1 - One Sample

Mean μ, Variance σ 2, Proportion π

• 6.2 - Two Samples Means, Variances, Proportions μ1 vs. μ2 σ1

2 vs. σ22 π1 vs. π2

• 6.3 - Multiple Samples Means, Variances, Proportions μ1, …, μk σ1

2, …, σk2 π1, …, πk

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 σ1

2 vs. σ22 π1 vs. π2

• 6.3 - Multiple Samples Means, Variances, Proportions μ1, …, μk σ1

2, …, σk2 π1, …, πk

CHAPTER 6 Statistical Inference & Hypothesis Testing

1, Yes (Success)0, No (Failure)

I

“Do you like olives?”

I = 1 I = 0

POPULATIONTwo random binary variables I and J

TWO POPULATIONSRandom binary variable I

“Do you like Brussel sprouts?”

1

1, Yes (Success)0, No (Failure)

I

2

1, Yes (Success)0, No (Failure)

I

Alternative HypothesisHA: 1 ≠ 2 “There is a difference in liking Brussel sprouts bet two pops.”

= P(Yes to Brussel sprouts)

Null HypothesisH0: 1 = 2 “No difference in liking Brussel sprouts between two pops.”

Binary Response: P(Success) =

“Test of Homogeneity”

TWO POPULATIONSRandom binary variable I

“Do you like Brussel sprouts?”

1

1, Yes (Success)0, No (Failure)

I

2

1, Yes (Success)0, No (Failure)

I

Alternative HypothesisHA: 1 ≠ 2 “There is a difference in liking Brussel sprouts bet two pops.”

= P(Yes to Brussel sprouts)

Null HypothesisH0: 1 = 2 “No difference in liking Brussel sprouts between two pops.”

Binary Response: P(Success) =

“Test of Homogeneity”

1, Yes (Success)0, No (Failure)

J

“Do you like anchovies?”

J = 0

J = 1

POPULATIONTwo random binary variables I and J

TWO POPULATIONSRandom binary variable I

“Do you like Brussel sprouts?”

1

1, Yes (Success)0, No (Failure)

I

2

1, Yes (Success)0, No (Failure)

I

Alternative HypothesisHA: 1 ≠ 2 “There is a difference in liking Brussel sprouts bet two pops.”

= P(Yes to Brussel sprouts)

Null HypothesisH0: 1 = 2 “No difference in liking Brussel sprouts between two pops.”

Binary Response: P(Success) =

“Test of Homogeneity”

1, Yes (Success)0, No (Failure)

J

“Do you like anchovies?”

POPULATIONTwo random binary variables I and J

1, Yes (Success)0, No (Failure)

I

“Do you like olives?”

1 = P(Yes to olives) 2 = P(Yes to anchovies)

Alternative HypothesisHA: 1 ≠ 2 “An association exists between liking olives and anchovies.”

Null HypothesisH0: 1 = 2 “No association exists between liking olives and anchovies.”

“Test of Independence”

I = 1 I = 0

J = 0

J = 1

TWO POPULATIONSRandom binary variable I

“Do you like Brussel sprouts?”

1

1, Yes (Success)0, No (Failure)

I

2

1, Yes (Success)0, No (Failure)

I

= P(Yes to Brussel sprouts)

Binary Response: P(Success) =

“Test of Homogeneity”

1, Yes (Success)0, No (Failure)

J

“Do you like anchovies?”

POPULATIONTwo random binary variables I and J

1, Yes (Success)0, No (Failure)

I

“Do you like olives?”

“Test of Independence”

Sample, size n1

Sample, size n2

Sample, size n1

Sample, size n2

(Assume “large” sample sizes.)

I = 1 I = 0

J = 0

J = 1

1 = P(Yes to olives) 2 = P(Yes to anchovies)

If n 15 and n (1 – ) 15, then via the Normal Approximation to the Binomial… , (1 ) .X N n n

If n 15 and n (1 – ) 15, then via the Normal Approximation to the Binomial…

(1 ), .X Nn n

Sample 1, size n1

Sample 2, size n2

X1 = # Successes X2 = # Successes

11

1

ˆ Xn

22

2

ˆ Xn

Sampling Distribution of

Solution: Use

0

ˆFor CI, Else, :H

0

Problem: s.e. depends on !!

Recall…

11

1

ˆ Xn

If n11 15 and n1 (1 – 1 ) 15, then via Normal Approximation to the Binomial

1 1 11 1

1 1

(1 )ˆ , .X Nn n

Sample 1, size n1

Sample 2, size n2

X1 = # Successes X2 = # Successes

22

2

ˆ Xn

ˆ ˆ 1 2 ???π πSampling Distribution of

If n22 15 and n2 (1 – 2 ) 15, then via Normal Approximation to the Binomial

2 2 22 2

2 2

(1 )ˆ , .X Nn n

Mean(X – Y) = Mean(X) – Mean(Y)Recall from section 4.1 (Discrete Models):

and if X and Y are independent…

Var(X – Y) = Var(X) + Var(Y)

1 2

1 1 2 21 2

1 2

ˆ ˆ

(1 ) (1 ),Nn n

0

ˆ ˆ1 2π πSampling Distribution of

Sample 1, size n1

Sample 2, size n2

X1 = # Successes X2 = # Successes

1 2

1 1 2 21 2

1 2

ˆ ˆ

(1 ) (1 ),Nn n

Similar problem as “one

proportion” inference s.e.!

For confidence interval, replace 1 and 2 respectively, by

standard error

1 2ˆ ˆ . and

For critical region and p-value, replace 1 and 2 respectively, by….. ????

Null Hypothesis H0: 1 = 2

…so replace their common value by a “pooled” estimate.

1 2

1 2

ˆp

X Xn n

1 2

ˆ ˆ ˆ ˆ(1 ) (1 )p p p p

n n

standard error estimate

1 2

1 1ˆ ˆ(1 )p p n n

= 0 under H0

“Null Distribution”

• Study Question: “Is there an association between liking Bruce Willis movies and gender, or not?”

Example: Two Proportions (of “Success”)

Test of Homogeneity or Independence?

Example: Two Proportions (of “Success”)

• Design: Randomly select two large samples of males and females, and record their binary responses (Yes = 1, No = 0) to the question “Do you like Bruce Willis movies?”

Let the discrete random variable X = “# Successes” (i.e., “Yes” responses) in each gender of the samples, and use these data to test…

• Data: Sample 1) n1 = 60 males, X1 = 42 Sample 2) n2 = 40 females, X2 = 16

1ˆ 42 / 60 0.7 2

ˆ 16 / 40 0.4

• Analysis via Z-test: Point estimates ˆ /X n

1 2ˆ ˆ 0.3

• Null Hypothesis H0: P(“Yes” among Males) = P(“Yes” among Females), i.e.,

H0: π1 = π2 where π = P(Success) in each gender population. “No association exists.”π1 – π2 = 0,

NOTE: This is > 0.

pooled42 16 58ˆ60 40 100

0.58 Therefore, 0 pooled pooled

1 2

1 1ˆ ˆs.e. (1 )n n

01 1s.e. (0.58)(0.42)60 40

0.10075

1 2ˆ ˆ- 2 ( 0.3)P valuep

0.3 0 2.97750.10075

Z

2 ( 2.9775)P Z REJECT H0

Conclusion: A significant association exists at the .05 level between “liking Bruce Willis movies” and gender, with males showing a 30% preference over females, on average.

.05 .0029

Test of Homogeneity (between two populations)• Study Question: “Is there an association between liking Bruce Willis

movies and gender, or not?”

TWO POPULATIONSRandom binary variable I

“Do you like Bruce Willis movies?”

1

1, Yes (Success)0, No (Failure)

I

2

1, Yes (Success)0, No (Failure)

I

Alternative HypothesisHA: 1 ≠ 2 “There is a difference in liking Bruce Willis bet two pops.”

= P(Yes to Bruce Willis movies)

Null HypothesisH0: 1 = 2 “No difference in liking Bruce Willis between two pops.”

Binary Response: P(Success) =

“Test of Homogeneity”

1, Yes (Success)0, No (Failure)

J

“Do you like anchovies?”

POPULATIONTwo random binary variables I and J

1, Yes (Success)0, No (Failure)

I

“Do you like olives?”

Alternative HypothesisHA: 1 ≠ 2 “An association exists between liking olives and anchovies.”

Null HypothesisH0: 1 = 2 “No association exists between liking olives and anchovies.”

“Test of Independence”

I = 1 I = 0

J = 0

J = 1

Males Females

1 = P(Yes to olives) 2 = P(Yes to anchovies)

Conclusion: A significant association exists at the .05 level; “liking Bruce Willis movies” and gender are dependent, with males showing a 30% preference over females, on average.

• Study Question: “Is there an association between liking Bruce Willis movies and gender, or not?”

Example: Two Proportions (of “Success”)

• Design: Randomly select two large samples of males and females, and record their binary responses (Yes = 1, No = 0) to the question “Do you like Bruce Willis movies?”

Let the discrete random variable X = “# Successes” (i.e., “Yes” responses) in each gender of the samples, and use these data to test…

• Data: Sample 1) n1 = 60 males, X1 = 42 Sample 2) n2 = 40 females, X2 = 16

1ˆ 42 / 60 0.7 2

ˆ 16 / 40 0.4

• Analysis via Z-test: Point estimates ˆ /X n

1 2ˆ ˆ 0.3

• Null Hypothesis H0: P(“Yes” among Males) = P(“Yes” among Females), i.e.,

H0: π1 = π2 where π = P(Success) in each gender population. “No association exists.”π1 – π2 = 0,

NOTE: This is > 0.

pooled42 16 58ˆ60 40 100

0.58 Therefore, 0

1 1s.e. (0.58)(0.42)60 40

0.10075

1 2ˆ ˆ- 2 ( 0.3)P valuep

0.3 0 2.97750.10075

Z

2 ( 2.9775)P Z REJECT H0 .05 .0029

Test of Homogeneity or Independence

“Do you like olives?”

TWO POPULATIONSRandom binary variable I

“Do you like Bruce Willis movies?”

1

1, Yes (Success)0, No (Failure)

I

2

1, Yes (Success)0, No (Failure)

I

Alternative HypothesisHA: 1 ≠ 2 “There is a difference in liking Bruce Willis bet two pops.”

= P(Yes to Bruce Willis movies)

Null HypothesisH0: 1 = 2 “No difference in liking Bruce Willis between two pops.”

Binary Response: P(Success) =

“Test of Homogeneity”

1, Yes (Success)0, No (Failure)

J

“Do you like anchovies?”

POPULATIONTwo random binary variables I and J

1, Yes (Success)0, No (Failure)

I

1 = P(Yes to Bruce) 2 = P(Yes to Male)

Alternative HypothesisHA: 1 ≠ 2 “An association exists between liking Bruce and Male.”

Null HypothesisH0: 1 = 2 “No association exists between liking Bruce and Male.”

“Test of Independence”

I = 1 I = 0

J = 0

J = 1

Males Females

“Gender: Male?”“Do you like Bruce Willis?”

“Do you like olives?”

TWO POPULATIONSRandom binary variable I

“Do you like Bruce Willis movies?”

1

1, Yes (Success)0, No (Failure)

I

2

1, Yes (Success)0, No (Failure)

I

Alternative HypothesisHA: 1 ≠ 2 “There is a difference in liking Bruce Willis bet two pops.”

= P(Yes to Bruce Willis movies)

Null HypothesisH0: 1 = 2 “No difference in liking Bruce Willis between two pops.”

Binary Response: P(Success) =

“Test of Homogeneity”

1, Yes (Success)0, No (Failure)

J

“Do you like anchovies?”

POPULATIONTwo random binary variables I and J

1, Yes (Success)0, No (Failure)

I

1 = P(Yes to Bruce) 2 = P(Yes to Male)

Alternative HypothesisHA: 1 ≠ 2 “Liking Bruce” and “Gender” are statistically dependent.

Null HypothesisH0: 1 = 2 “Liking Bruce” and “Gender” are statistically independent.

“Test of Independence”

I = 1 I = 0

J = 0

J = 1

Males Females

“Gender: Male?”“Do you like Bruce Willis?”

• Study Question: “Is there an association between liking Bruce Willis movies and gender, or not?”

Example: Two Proportions (of “Success”)

• Data: Sample 1) n1 = 60 males, X1 = 42 Sample 2) n2 = 40 females, X2 = 16

H0: π1 = π2 where π = P(Success) in each gender population. “No association exists.”π1 – π2 = 0,

• Null Hypothesis H0: P(“Yes” among Males) = P(“Yes” among Females), i.e.,

• Design: Randomly select two large samples of males and females, and record their binary responses (Yes = 1, No = 0) to the question “Do you like Bruce Willis movies?”

Let the discrete random variable X = “# Successes” (i.e., “Yes” responses) in each gender of the samples, and use these data to test…

~ ALTERNATE METHOD ~

I = 1 I = 0

J = 0

J = 1

• Study Question: “Is there an association between liking Bruce Willis movies and gender, or not?”

Example: Two Proportions (of “Success”)

• Data: Sample 1) n1 = 60 males, X1 = 42 Sample 2) n2 = 40 females, X2 = 16

H0: π1 = π2 where π = P(Success) in each gender population. “No association exists.”π1 – π2 = 0,

Males FemalesYes 42 16

No60 40

Males Females

Yes E11 = ? E12 = ? 58

No E21 = ? E22 = ? 42

60 40 100

Observed

Expected(under H0)

Males FemalesYes 42 16 58

No 18 24 42

60 40 100

• Null Hypothesis H0: P(“Yes” among Males) = P(“Yes” among Females), i.e.,

• Design: Randomly select two large samples of males and females, and record their binary responses (Yes = 1, No = 0) to the question “Do you like Bruce Willis movies?”

Let the discrete random variable X = “# Successes” (i.e., “Yes” responses) in each gender of the samples, and use these data to test…

Recall Probability Tables from Chapter 3….

Under the null hypothesis, the binary variable I is statistically independent of the binary variable J, i.e., P(I ∩ J) = P(I) P(J).

J = 1 J = 2

I = 1 π11 π12 π11 + π12

I = 2 π21 π22 π21 + π22

π11 + π21 π12 + π22 1

Recall Probability Tables from Chapter 3….

Contingency Table

Under the null hypothesis, the binary variable I is statistically independent of the binary variable J, e.g., P(“I = 1” ∩ “J = 1”) = P(“I = 1”) P(“J = 1”).

J = 1 J = 2

I = 1 π11 π12 π11 + π12

I = 2 π21 π22 π21 + π22

π11 + π21 π12 + π22 1

J = 1 J = 2

I = 1 E11 E12 R1

I = 2 E21 E22 R2

C1 C2 n

J = 1 J = 2

I = 1 E11/n E12/n R1/n

I = 2 E21/n E22/n R2/n

C1/n C2/n 1

Probability TableTherefore…

11 1 1E R Cn n n

1 111

R CEn

, etc.

H0: π1 = π2 where π = P(Success) in each gender population. “No association exists.”

• Null Hypothesis H0: P(“Yes” among Males) = P(“Yes” among Females), i.e.,

Check: Is the null hypothesis true?

1 2

34.8 23.2 5860 40 100

• Study Question: “Is there an association between liking Bruce Willis movies and gender, or not?”

Example: Two Proportions (of “Success”)

• Data: Sample 1) n1 = 60 males, X1 = 42 Sample 2) n2 = 40 females, X2 = 16 Males Females

Yes 42 16

No60 40

Males Females

Yes E11 = ? E12 = ? 58

No E21 = ? E22 = ? 42

60 40 100

Observed

Expected(under H0)

Males FemalesYes 42 16 58

No 18 24 42

60 40 100

11(58)(60)

100E 34.8

34.8

12(58)(40)

100E 23.2

23.2

21(42)(60)

100E 25.2

25.2

22(42)(40)

100E 16.8

16.8

= 1, 2,..., # rowsIn general,

= 1, 2,..., # cols.i j

i j

R C iE

jn

“Chi-squared” Test Statistic2

2 ( )

all cells

Observed ExpectedExpected

2df~

where “degrees of freedom” df = (# rows – 1)(# cols – 1), = 1 for a 2 2 table.

π1 – π2 = 0,

• Design: Randomly select two large samples of males and females, and record their binary responses (Yes = 1, No = 0) to the question “Do you like Bruce Willis movies?”

Let the discrete random variable X = “# Successes” (i.e., “Yes” responses) in each gender of the samples, and use these data to test…

21

• Study Question: “Is there an association between liking Bruce Willis movies and gender, or not?”

Example: Two Proportions (of “Success”)

Males FemalesYes 34.8 23.2 58

No 25.2 16.8 42

60 40 100

Observed Expected(under H0)Males Females

Yes 42 16 58

No 18 24 42

60 40 100

“Chi-squared” Test Statistic2

2 ( )

all cells

Observed ExpectedExpected

2 2 2 2(42 34.8) (16 23.2) (18 25.2) (24 16.8)34.8 23.2 25.2 16.8

= 8.867 on 1 df p = ?????

• Design: Randomly select two large samples of males and females, and record their binary responses (Yes = 1, No = 0) to the question “Do you like Bruce Willis movies?”

Because 8.867 is much greater than the α =.05 critical value of 3.841, it follows that p << .05.More precisely, 7.879 < 8.867 < 9.141; hence .0025 < p < .005.

The actual p-value = .0029, the same as that found using the Z-test!

Yes = c(42, 16)No = c(18, 24)Bruce = rbind(Yes, No)chisq.test(Bruce, correct = F)

Pearson's Chi-squared test

data: Bruce X-squared = 8.867, df = 1, p-value = 0.002904

α =.05

• Study Question: “Is there an association between liking Bruce Willis movies and gender, or not?”

Example: Two Proportions (of “Success”)

Males FemalesYes 34.8 23.2 58

No 25.2 16.8 42

60 40 100

Observed Expected(under H0)Males Females

Yes 42 16 58

No 18 24 42

60 40 100

21“Chi-squared” Test Statistic

22 ( )

all cells

Observed ExpectedExpected

2 2 2 2(42 34.8) (16 23.2) (18 25.2) (24 16.8)34.8 23.2 25.2 16.8

= 8.867 on 1 df p = .0029

The α =.05 critical value is 3.841.

Recall…

• Design: Randomly select two large samples of males and females, and record their binary responses (Yes = 1, No = 0) to the question “Do you like Bruce Willis movies?”

H0: π1 = π2 where π = P(Success) in each gender population. “No association exists.”

• Study Question: “Is there an association between liking Bruce Willis movies and gender, or not?”

Example: Two Proportions (of “Success”)

• Data: Sample 1) n1 = 60 males, X1 = 42 Sample 2) n2 = 40 females, X2 = 16

1ˆ 42 / 60 0.7 2

ˆ 16 / 40 0.4

• Analysis via Z-test: Point estimates ˆ /X n

1 2ˆ ˆ 0.3

• Null Hypothesis H0: P(“Yes” in Male population) = P(“Yes” in Female population), i.e.,

π1 – π2 = 0,

NOTE: This is > 0.

pooled42 16 58ˆ60 40 100

0.58 Therefore, 0

1 1s.e. (0.58)(0.42)60 40

0.10075

1 2ˆ ˆ- 2 ( 0.3)P valuep

0.3 0 2.97750.10075

Z

2 ( 2.9775)P Z REJECT H0

Conclusion: A significant association exists at the .05 level; “liking Bruce Willis movies” and gender are dependent, with males showing a 30% preference over females, on average.

.05 .0029

• Design: Randomly select two large samples of males and females, and record their binary responses (Yes = 1, No = 0) to the question “Do you like Bruce Willis movies?”

Let the discrete random variable X = “# Successes” (i.e., “Yes” responses) in each gender of the samples, and use these data to test…

• Study Question: “Is there an association between liking Bruce Willis movies and gender, or not?”

Example: Two Proportions (of “Success”)

Males FemalesYes 34.8 23.2 58

No 25.2 16.8 42

60 40 100

Observed Expected(under H0)Males Females

Yes 42 16 58

No 18 24 42

60 40 100

21“Chi-squared” Test Statistic

22 ( )

all cells

Observed ExpectedExpected

2 2 2 2(42 34.8) (16 23.2) (18 25.2) (24 16.8)34.8 23.2 25.2 16.8

p = .0029

The α =.05 critical value is 3.841.

NOTE: (Z-score)2 = (2.9775)2 Connection between Z-test and Chi-squared test !

= 8.867 on 1 df NOTE: (Z-score)2 = (2.9775)2 Connection between Z-test and Chi-squared test !

= 8.867 on 1 df

• Design: Randomly select two large samples of males and females, and record their binary responses (Yes = 1, No = 0) to the question “Do you like Bruce Willis movies?”

• Categorical data – contingency table with any number of rows and columns

• See notes for other details, comments, including “Goodness-of-Fit” Test.

• 2 2 Chi-squared Test is only valid if:

Null Hypothesis H0: 1 – 2 = 0. One-sided or nonzero null value Z-test!

Expected Values 5, in order to avoid “spurious significance” due to a possibly inflated Chi-squared value.

• Paired version of 2 2 Chi-squared Test = McNemar Test

Formal Null Hypothesis difficult to write mathematically in terms of 1, 2,… “Test of Independence” “Test of Homogeneity”

Informal H0: “No association exists between rows and columns.”

80% of Expected Values 5