ECO220Y1 Lecture Notes - Lecture 14: Statistical Hypothesis Testing, Central Limit Theorem, Confidence Interval

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3 Jul 2018
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Department
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ECO220
Lecture 14
July 3, 2018
1
Chapter 11: Confidence Intervals for Population Proportion
Point estimator for p is
Confidence interval for p is  
with n being large np 10, nq 10
Two populations:
To compare p1 to p2, we estimate p1 - p2
Point estimator for p1 - p2 is

Confidence interval for p1 - p2: Point estimator ±
 


±


To find SE

; from the Law of Expectation
Var(aX + bY) = V(aX) + V(bY) + 2abCov(X, Y)
If X, Y are independent, Cov(X, Y) = 0; and
V(aX + bY) = V(aX) + V(bY) = a2V(X) + b2V(Y)
We select two independent samples form the two populations. Central Limit Theorem says

~ N(p2,
) with n2 large
Hence Var(

  

= 

=
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ECO220
Lecture 14
July 3, 2018
2
SE(

Confidence interval for p1 - p2 is
±
We use 
to estimate p1 and
to estimate p2. That is, the estimated standard error of 
is
Estimated SE(

Confidence Interval for p1 - p2 is


±
Assuming n1 and n2 are large. That is n1p1 10, n2p2 10
And n1q1 10, n2q2 10
Example 1: Compare the % of casino players in Ontario with Manitoba
(a) Point estimation for p1 - p2 is


0.665 - 0.752 = -0.087
In percentages, Ontario has 8.7% less people who go to the casino compared to Manitoba
(b) A 95% confidence interval for p1 - p2 is
Point estimator ±
 


±
= 0.665 - 0.752 ± 1.96
 

= -0.087 ± 0.0453
(-0.132, - 0.042)
95% of the times, p1 - p2 is between -0.132 and -0.042
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ECO220Y1 Full Course Notes
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Document Summary

Confidence interval for p is (cid:1868) (cid:1852)(cid:3118) (cid:3043) (cid:3044) with n being large. To compare p1 to p2, we estimate p1 - p2. Point estimator for p1 - p2 is (cid:1868)(cid:883) (cid:1868)(cid:884) . Confidence interval for p1 - p2: point estimator (cid:1852)(cid:3118) (cid:4666)(cid:1867)(cid:1866)(cid:1872) (cid:1871)(cid:1872)(cid:1865)(cid:1853)(cid:1872)(cid:1867)(cid:1870)(cid:4667) To find se((cid:1868)(cid:883) (cid:1868)(cid:884) ); from the law of expectation. Var(ax + by) = v(ax) + v(by) + 2abcov(x, y) If x, y are independent, cov(x, y) = 0; and. V(ax + by) = v(ax) + v(by) = a2v(x) + b2v(y) We select two independent samples form the two populations. Central limit theorem says (cid:1868)(cid:884) ~ n(p2, (cid:3043)(cid:3118)(cid:3044)(cid:3118)(cid:3118) ) with n2 large. Confidence interval for p1 - p2 is (cid:1868)(cid:883) (cid:1868)(cid:884) (cid:1852)(cid:3118) (cid:3043)(cid:3117)(cid:3044)(cid:3117)(cid:3117) +(cid:3043)(cid:3118)(cid:3044)(cid:3118)(cid:3118) We use (cid:1868)(cid:883) to estimate p1 and (cid:1868)(cid:884) to estimate p2. That is, the estimated standard error of (cid:1868)(cid:883) (cid:1868)(cid:884) is. Estimated se((cid:1868)(cid:883) (cid:1868)(cid:884) (cid:4667)= (cid:3043)(cid:3117)(cid:3044)(cid:3117)(cid:3117) +(cid:3043)(cid:3118)(cid:3044)(cid:3118)(cid:3118) (cid:1868)(cid:883) (cid:1868)(cid:884) (cid:1852)(cid:3118) (cid:3043)(cid:3117)(cid:3044)(cid:3117)(cid:3117) +(cid:3043)(cid:3118)(cid:3044)(cid:3118)(cid:3118)