advanced econometrics (i) chapter 9 - hypothesis...

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Advanced Econometrics (I)

Chapter 9 - Hypothesis Testing

Fall 2012

Wei Zhong

WISE & SOE

December 28, 2012

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Review: T Test

Steps for Hypothesis Testing (T-Test)

(1) Parameter of interest and the associated hypothesis testing.

H0 : µ = µ0, vs H1 : µ 6= µ0

(2) Test statistic and its distribution under H0 is true.

T =X̄n − µ0

Sn/√n∼ tn−1, under H0 is true

(3) Compute the observed test statistic T = x̄n−µ0sn/√n

(4) Apply Method 1 to construct critical region for a fixed sig-nificance level α; or apply Method 2 to compute p-value.

(5) Draw a conclusion (at the significance level α).

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Hypothesis Testing on Two Popula-tion Means

We will investigate the hypothesis testing on two normal pop-ulation means in the following 4 cases:

• Case 1. Two Independent Samples with KnownPopulation Variances.

• Case 2. Two Independent Samples with Un-known but Equal Population Variance.

• Case 3. Two Independent Samples with Un-known Unequal Population Variances.

• Case 4. Two Dependent Paired Samples withUnknown Variance.

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Case 1. Two Independent Sampleswith Known Population Variances.

• Example:

– The alkalinity, in milligrams per liter, of water in theupper reaches of rivers in a particular region is knownto be normally distributed with a standard deviationof 10 mg/l. Alkalinity readings in the lower reaches ofrivers in the same region are also known to be normallydistributed, but with a standard deviation of 25 mg/l.

– Ten alkalinity readings are made in the lower reaches ofa river in the region with the mean x̄1 = 99.0 and fifteenin the upper reaches of the same river x̄2 = 80.5.

– Investigate, at the 1% level of significance, the claim thatthe true mean alkalinity of water in the lower reaches ofthis river is greater than that in the upper reaches.

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• Solution: Let X1 and X2 denote the alkalinity readingsin the lower reaches and upper reaches, respectively. Then,we have

X1 ∼ N(µ1, σ21), X2 ∼ N(µ2, σ

22),

where σ21 = 252, σ2

2 = 102.

(1) Parameters of interest are µ1 and µ2 and the associatedhypothesis testing.

H0 : µ1 = µ2, vs H1 : µ1 > µ2

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• Solution:(2) Test statistic and its distribution under H0 is true.

Since X1 ∼ N(µ1, σ21), X2 ∼ N(µ2, σ

22), we have

X̄1 ∼ N(µ1, σ21/n1), X̄2 ∼ N(µ2, σ

22/n2),

and we know X̄1 and X̄2 are independent, thus,

X̄1 − X̄2 ∼ N

(µ1 − µ2,

σ21

n1

+σ2

2

n2

),

⇒ (X̄1 − X̄2)− (µ1 − µ2)√σ21n1

+ σ22n2

∼ N (0, 1)

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• Solution:(2) We propose the test statistic:

Z =X̄1 − X̄2√

σ21n1

+ σ22n2

∼ N (0, 1) , under H0 is true.

Intuitively, if Z is large enough, we will reject the null H0.

(3) Compute the observed test statistic

z0 = Z =99.0− 80.5√

252

15+ 102

10

= 2.57.

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• Solution:(4) Critical region {z > zα = 2.326}. Note this is one-sided

critical region for the one-tailed alternative hypothesis.

The observed test statistic value z0 does lie in the criticalregion so H0 is rejected.

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• Solution:(4) Or we may compute the p-value

p-value = P (Z > z0) = P (Z > 2.57)

= 0.0051 < α = 0.01,

then we reject H0, at the 1% level of significance.

(5) Conclusion: At the 1% level of significance, we reject H0.Thus, there is evidence, at the 1% level of significance, tosuggest that the true mean alkalinity of water in the lowerreaches of the river is greater than that in the upper reaches.

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Case 2. Two Independent Sampleswith Unknown but Equal Popula-tion Variance.

• Example:

– Mr. Brown is the owner of a small bakery in a largetown. He believes that the smell of fresh baking willencourage customers to purchase goods from his bakery.To investigate this belief, he records the daily sales for10 days when all the bakery’s windows are open, andthe daily sales for another 10 days when all windows areclosed.

– From the records, he get x̄1 = 202.18 and σ̂21 = 115.728,

x̄2 = 188.47 and σ̂22 = 156.653, where 1=open and

2=closed.

– Assuming that these data may be deemed to be ran-dom samples from normal populations with the samevariance, investigate the baker’s belief.

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• Solution: Let X1 and X2 denote the daily sales when allthe windows are open and closed, respectively. Then, wehave

X1 ∼ N(µ1, σ2), X2 ∼ N(µ2, σ

2),

where the same variance σ2 is unknown.


H0 : µ1 = µ2, vs H1 : µ1 > µ2

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• Solution:(2) Test statistic and its distribution under H0 is true.

Since X1 ∼ N(µ1, σ2), X2 ∼ N(µ2, σ

2), we have

X̄1 ∼ N(µ1, σ2/n1), X̄2 ∼ N(µ2, σ

2/n2),


X̄1 − X̄2 ∼ N

(µ1 − µ2,

σ2

n1

+σ2

n2

),

⇒ (X̄1 − X̄2)− (µ1 − µ2)

σ√

1n1

+ 1n2

∼ N (0, 1)

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(2) However, the same variance σ2 is unknown, we use the sam-ple variance to replace it. Note that two independent sam-ples have the same variance, we can use the both samplesto obtain a pooled estimate of variance defined by

σ̂2p =

(n1 − 1)σ̂21 + (n2 − 1)σ̂2

2

n1 + n2 − 2.

Then, we propose the test statistic:

T =X̄1 − X̄2

σ̂p√

1n1

+ 1n2

∼ tn1+n2−2, under H0 is true.

Intuitively, if T is large enough, we will reject H0.

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• Solution:(3) Compute the observed test statistic

t0 =202.18− 188.47

11.67√

110

+ 110

= 2.63.

(4) Critical region {t > tα,n1+n2−2 = 1.734}. Note this is one-sided critical region for the one-tailed alternative hypothe-sis. The observed test statistic value t0 = 2.63 does lie inthe critical region so H0 is rejected.

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• Solution:

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• Solution:(4) Or we may compute the p-value

p-value = P (t18 > t0) = P (t18 > 2.63)

= 0.0085 < α = 0.05,


(5) Conclusion: At the 5% level of significance, we rejectH0. Thus, there is evidence, at the 5% level of significance,to suggest that the smell of fresh baking will encouragecustomers to purchase goods from Mr Brown’s Bakery.

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Case 3. Two Independent Sampleswith Unknown Unequal PopulationVariances.

• Example:

– A random sample of 10 hot drinks from Dispenser A hada mean volume of 203 ml and a standard deviation of 6ml. A random sample of 15 hot drinks from DispenserB gave corresponding values of 206 ml and 3 ml. Theamount dispensed by each machine may be assumed tobe normally distributed.

1. Test, at the 5% significance level, the hypothesis thatthere is no difference in the variability of the volumedispensed by the two machines.

2. Test, at the 5% significance level, the hypothesis thatthere is no difference in the mean volume dispensedby the two machines.

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• Solution: Let X1 and X2 denote the volume dispensed bymachine A and B, respectively. Then, we have

X1 ∼ N(µ1, σ21), X2 ∼ N(µ2, σ

22),

where variances σ21 and σ2

2 is unknown. Let n = 10 andm = 15 denote two sample sizes.

1. First, we use the F-Test (Chapter 6) to test the equality oftwo normal population variances.

(1) Parameters of interest are σ21, σ

22 and the associated hy-

pothesis testing.

H0 : σ21 = σ2

2, vs H1 : σ21 = σ2

2.

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• Solution (Cont’d):

(2) Test statistic and its distribution under H0 is true.A F test statistic is based on the sample variance ratio

F =S2

1

S22

.

Intuitively, if H0 : σ21 = σ2

2 is true, the F test statistic

should be close to 1, because S21

p→ σ21 and S2

2

p→ σ22 as

n→∞. On the other hand, if the F test statistic is fardifferent from 1, then we may reject H0.

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(2) Test statistic and its distribution under H0 is true.Under H0 : σ2

1 = σ22, we have

F =S2

1

S22

=S2

1/σ21

S22/σ

22

=

(n−1)S21/σ

21

n−1

(m−1)S22/σ

22

m−1

∼ χ2n−1/(n− 1)

χ2m−1/(m− 1)

∼ Fn−1,m−1.


F0 = s21/s

22 = 62/32 = 4.

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(4-1) Construct critical region for a fixed significance level α:Reject the hypothesis H0 : σ2

1 = σ22 at the significance

level α = 0.05 if

F > CFn−1,m−1,α2, or F < CFn−1,m−1,1−α

2

where CFn−1,m−1,α2

is the upper-tailed critical value of theF distribution at level α

2, determined by P (Fn−1,m−1 >

CFn−1,m−1,α2) = α

2.

Here, CF9,14,0.025 = 3.21 and CF9,14,0.975 = 0.26. There-fore, the critical region is {F0 > 3.21 or F0 < 0.26}.Because F0 = 4 does lie in the critical region, we rejectH0.

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(4-2) Compute p-value.

p-value = 2P (F9,14 > F0) = 2P (F9,14 > 4) = 0.0206.

Since the p-value < α = 0.05, we reject H0.

(5) Conclusion: At the 5% significance level, we may re-jectH0. That is, there is evidence, at the 5% significancelevel, of a difference in the variability of the volume dis-pensed by the two machines.

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• Solution:2. Next, we will test the hypothesis of the equality of two

normal means with unknown unequal variances. Here, wedenote n1 = 10 and n2 = 15.


H0 : µ1 = µ2, vs H1 : µ1 6= µ2

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Since X1 ∼ N(µ1, σ21), X2 ∼ N(µ2, σ

22), we have

X̄1 ∼ N(µ1, σ21/n1), X̄2 ∼ N(µ2, σ

22/n2),


X̄1 − X̄2 ∼ N

(µ1 − µ2,

σ21

n1

+σ2

2

n2

),

⇒ (X̄1 − X̄2)− (µ1 − µ2)√σ21n1

+ σ22n2

∼ N (0, 1)

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(2) Since both variances are unknown and unequal, naturally,we may use their sample variances to replace the corre-sponding variances. Then, we can propose the test statistic:

T =X̄1 − X̄2√

S21

n1+ S2

2

n2

∼ tν, under H0 is true,

where ν is the estimated degree of freedom defined by

ν =(S2

1/n1 + S22/n2)

2

(S21/n1)2/(n1 − 1) + (S2

2/n2)2/(n2 − 1).

Intuitively, if T is large enough, we will reject H0. ThisT-test is also called the Satterthwaite’s T-Test.

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t0 =203− 206√

62

10+ 32

15

= −1.464.

(4) To construct the critical region or compute the p-value, weneed compute the estimated d.f.

ν =(s2

1/n1 + s22/n2)

2

(s21/n1)2/(n1 − 1) + (s2

2/n2)2/(n2 − 1)= 12.

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(4) Critical region {|t| > tα/2,ν = t0.025,12 = 2.18}. Note this istwo-sided critical region for the two-tailed alternative hy-pothesis. The observed test statistic value t0 = −1.464doesn’t lie in the critical region so H0 can not be rejected.

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• Solution:(4) Or we may compute the p-value (two-sided)

p-value = 2P (t12 > |t0|) = 2P (t12 > 1.464)

= 2× 0.084 = 0.168 > α = 0.05,

then we can not reject H0, at the 5% level of significance.

(5) Conclusion: At the 5% level of significance, we can notreject H0. Thus, there is no evidence, at the 5% level ofsignificance, to suggest that there is some difference in themean volume dispensed by the two machines.

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Case 4. Two Dependent PairedSamples with Unknown Variance.

• Example: A convenience food, known as ‘QuicKnosh’,was introduced into the British market in January 1992.After a poor year for sales the manufacturers initiated anintensive advertising campaign during January 1993. Thetable below records the sales, in thousands of pounds, for aone-month period before and a one-month period after theadvertising campaign, for each of eleven regions.

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• Example: Determine, at the 5% significance level,whether an increase in mean sales has occurred by usingthe T-Test for paired values.

• Solution: Two samples are dependent, but they arepaired. Therefore, we consider the difference of each sub-ject before and after the campaign. Denote X1 and X2

be the the monthly sales before and after the advertisingcampaign. Then, we know X1 and X2 are dependent andpaired. Next, we denote

D = X1 −X2,

and assume that the difference is normally distributed withmean µd = µ1 − µ2 and variance σ2

d. That is,

D ∼ N(µd, σ2d).

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H0 : µ1 = µ2, vs H1 : µ1 > µ2,

which is equivalent to

H0 : D = 0, vs H1 : D > 0.

Then, the problem is simplified to one-sample T-Test.


T =D̄n

σ̂d/√n∼ tn−1, under H0 is true.

Intuitively, if T is large enough, we will reject the null H0.

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t0 =D̄n

σ̂d/√n

=0.855

0.907/√

11= 3.13.

(4) Critical region {t > tα,11−1 = 1.81}. Note this is one-sidedcritical region for the one-tailed alternative hypothesis. Theobserved test statistic value t0 = 3.13 does lie in the criticalregion so H0 is rejected.

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(4) Or we may compute the p-value

p-value = P (t11−1 > t0) = P (t10 > 3.13)

= 0.0053 < α = 0.05,


(5) Conclusion: At the 5% level of significance, we reject H0.Thus, there is evidence, at the 5% level of significance, tosuggest that an increase in mean sales has occurred afterthe advertising campaign.

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