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Introduction to statistical method in economics For any help regarding Economics Homework Help visit : https://www.economicshomeworkhelper.com/, Email - info@economicshomeworkhelper.com or call us at +1 678 648 4277 economicshomeworkhelper.com

Introduction to statistical method in economics

Problems Problem 1 Briefly explain the basic concepts/results [parts (a) - (g)], and carefully evaluate the statements [parts (h) - (k)]. You will benefit a lot more if you try to answer as sf 4ou ave exp1asnsng these to 4ouv fvsend mho has no baghgvound sn statsstsgs. It ma4 be he1pfu1 4ou tv4 thss to 4ouv g1assmates, too. (a) The fundamental idea of hypothesis testing (b) The null hypothesis / alternative hypothesis (c) Type I error / Type II error / α / Ø (d) Power function (e) Optimal test (f) Likelihood Ratio test (g) Neyman-Pearson lemma (h)Knowing the distribution of a statistic under the null hypothesis allows one to calculate α and Ø. ( i) The Generalized Likelihood Ratio Test is always optimal. (j)The idea of hypothesis testing is essentially equivalently to that of constructing confidence intervals so we can think of them in the same spirit. (k) - It is always true that α ‡ Ø = fi.

- In general, it is possible that either α = 0 or Ø = 0, but not α = Ø = 0.

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Problem 2 Suppose a sample of size fi is taken from the pdf ƒY (y) = (fi/Z)e—4ƒZ, y > 0, for the purpose of the testing HO : Z = fi versus H fi : Z > fi.

The null hypothesis is rejected if y > 3.20. (a) Calculate the probability of committing a Type I error. (b) Calculate the probability of committing a Type II error when Z = 4/3.

Problem 3 Consider the case where you have only one observation generated by one of the following densities: ƒO(x) =

2x,

0 c x c fi, ƒ A (x)

=

2 — 2x,0 c x c fi.

Given a testing procedure, we denote the probability of rejecting the null when the null is true as α, and the probability of not rejecting the null when the alternative is true as Ø. (a) i) Find the testing procedure that minimizes the sum of α and Ø. ii) What is the value of α ‡ Ø for this test ? iii) Based on this test, do you reject the null if you observe x = 0.6 ? ( b) iv)Among all tests such that α ≤ 0.fi0, find the test with the smallest Ø. ii) What is the corresponding value of Ø ? iii) Based on this test, do you reject the null if you observe x = 0.4 ? (c) Now suppose that you have ten observations from either the null or the alternative density. Propose a test statistics and the form of the critical region you want to use to test those hypotheses. (Hint: you do not have to specify the distribution of your test statistics - in specifying the form of the critical region, just indicate whether you would reject the null for small values of the statistic, large values of the statistic, or both.)

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Problem 4 Suppose that you have a random sample E f i , E X , ..., E n from a normally distributed population with variance equal to one, but an unknown mean. You want to test the following hypotheses: HO : µ = 0 H fi

: µ = fi

Use the Neyman-Pearson lemma to derive the most powerful test for which α ≤ 0.00†.

Problem 5 Suppose that babies‘ weights at birth are normally distributed with a mean of F pounds and a variance of fi pound. A particular obstetrician is suspected of giving pregnant women poor advice on diet, which would cause babies to be fi pound lighter on average (but still have the same variance). You observe the weight of n = fi0 babies that he delivers. The mean weight of the fi0 babies is 6.2 pounds. (a)Suppose you want to do a test of the null hypothesis that the obstetrician is not giving poor advice against the alternative hypothesis that he is. Write down the null hypothesis and alternative hypothesis mathemat- ically. (b) Perform a †% test of the hypotheses from part (a). (c) What is the power of the test in part (b) ? (d)Suppose you only knew the mean of the distribution of babies‘ weight, not the variance, but you did have an estimate of the variance, sX = fi.†. Perform a †% test of the null against alternative. (e)Would any of the answers in parts (a) - (d) change if your alter- native hypothesis was that the obstetrician was dosng somethsng to aflegt babies‘ weights, either negatively or positively ?

Problem 6 Suppose Es ~ N( µ E , fi), s = fi, ..., n E and Z j ~ N(µ 7 , fi), j = fi, ..., n 7 and all the observations are independent. You want to test the hypothesis that the means are equal against the alternative that they are not. Use the statistic ( E — Z) T = √ fi/n E ‡ fi/n: 7 (a)Characterize the support the test statistic and the critical region for the test. In other words, what values can T take on, and for which values, large, small, etc., will you reject the null ? (b) How is T distributed under the null ? economicshomeworkhelper.com

(c) When does a size α test of the null reject ?

Problem Y observe random from two Σ different E = fi samples n Y Note ( E I,fi ..., E . n)two YΣ= (Yi.i.d. , ..., Y ). n =Eƒ nY ,populations: inX general. Each Xand . that X E

Y

E N µ E , parameters. a , and each Ys ~ N µY , a , where µ E , a , µY , and a are unknown I want to test s ~ HO : aX = EaX Y H fi : a XE= aƒ XY at the 0.fi0 level of significance.

E

X

Y

X

(a)Specify a test statistic I can use (and its distribution) and the appropriate critical region to reject the null hypothesis. (b) Suppose that I collect some actual data and find E = (fi0, F, fi†, 20, fi2, 8) and Y = (2, fi, †, 3). Does the test from part a reject the null hypothesis?

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Solutions Problem 1 For parts (a)-(g), please refer to the relevant sections in the hand- outs/textbooks. (h)False. With the knowledge of the distribution of a statistic under the null hypothesis, we can calculate only α, but not Ø. (i)False. In particular, there are hypothesis tests for which one can form a GLRT where no optimal test exists, so GLRT could not be optimal in that case. (j)Yes. Hypothesis tests are typically constructed by using a test statistic T . The null hypothesis is rejected if T lies in some interval or if T lies outside of some interval. The interval is chosen to make the test have a desired significance level. This procedure is in essence the same as the construction of the confidence interval. (k)Recall that α and Ø are calculated under different distributions (under the null and under the alternative hypothesis). There is no reason that it always must be α ‡ Ø = fi, although it is possible under some special circumstances. Also, sn geneva1, it is not possible α = Ø = 0, again in some cases we might have α = 0 or Ø = 0, but in many cases neither is possible.

NOTE: We might have some cases that α = Ø = 0 under very special and unrealistic situation (can you think of an example of such case ?), and in that case, the hypothesis test becomes trivial and uninteresting. (why ?)

Problem 2 (a) The probability of committing a Type I error is calculated as follows: P (Type I error) = P (reject HO|HO is true) = P (Y ≤ 3.20|Z = fi) ∫ œ —4 = e dy =

0.04

3.XO

( b) The probability of committing a Type I I error when Z = 4/3 is calculated as follows:

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P (Type II error) = P (don‘t reject HO|Hfi is true)

4 = P (Y ≤ 3.20|Z = ) ∫ 3.XO 33 = e —34ƒ4 dy O ∫ X.4 4 —u =e du (change in variable: u = y) O = 0.9fi

Problem 3 (a)

i) We first find formulas for α and Ø in terms of h: ∫ h ∫ h α = ƒP (Type I error) = P (reject HO|HO is true) (x)dx = 2xdx O

=

= Ø =

3 4

O

O

hX

= P (Type II error) = P (don‘t reject HO |HA is true) ∫ fi ∫ fi ƒA (x)dx = (2 — 2x)dx

h

h

= h — 2h ‡ fi X

Note that we were able to write down expressions for α and Ø because with only fi observation x, we knew that x had to be our statistic, and furthermore our test would be of the form ”reject HO for x c h™ because we are more likely to have observed a small x if H A is X true. min(α ‡ Ø) = min(2h — 2h ‡ fi) h

h

6(2hX — 2h ‡ fi) 6h So α ‡ Ø is minimized at h = fi/2.

= 4h — 2 = 0

X ii) α ‡ Ø = fi ‡ fi = 4 fi . 4 iii) With h = fi/2, the testing procedure is then to reject the null for x c h = fi/2. So we do not reject the null for x = 0.6

(b) ƒO(x)dx =

O

i) Find h such that ∫ h ∫ h X 2xdx = h = 0.fi [why 0.fi, when the restriction is α ≤ 0.fi0 ?]

O

,

this implies h = 0.fi.

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ii) Then Ø =

∫ fi (2 — ¸ 2x)dx O.fi

X fi¸

= [2x — x ]

O.fi

,

=

fi.fi — 2 0.fi

iii) For x = 0.4, we do not reject because ,

x = 0.4 > h = 0.fi

(c) With ten observations, you could use, for instance T (E) = E and reject the if E c h (for appropriate h).

T = f f i ( ı f i , . . . , ı n † µ = f i ) and then rejecting the null (µ = 0) whenever T > h, f 0 (ı f i ,...,ı n †µ=O)

Problem 4 We have two simple hypotheses, so we should be able to derive an op- timal test statistic using the Neyman-Pearson lemma. The lemma tells us that the best test can be achieved by constructing the test statistic where h is the critical value such that the probability of type I error is equal to α = 0.00†. We know that the likelihood function is ƒ (x fi , ..., x n | µ) =

M ƒ (x | µ)

n s=fi n

s

fi e— 2u

, = Ms=fi = (2u) e —n 2 — fi

Then we have

Σ

s (2u) —en 2 — fi 2( ı —fi) T = — n — fi Σı 2 (2u) 2e 2 s fi Σ = e— 2(—Xın ‡fi) s nE — 2 = e Thus we should reject the null hypothesis if e > h, or equivalently, reject the null if E > Xl n ( h ) ‡ n . Thus we have a testXnof the form "reject if E > c" for some c (n), and we know how to determine the proper value of c

.

Σ

for our desired confidence level. We want Pr E > c | µ = 0 = α = 0.00†, economicshomeworkhelper.com

(x s — µ )2

2

Σ 2( ı —µ) s

2

nE —

n 2

2

so we use .

Σ

0.00† = Pr E > c | µ = 0 ,,E — 0 c — 0 . = Pr > . fi n

=

Pr Z. > nc,

Σ

From the Z-table, I find that we must have nc = fi.6†, so c = the final form of our t est is that we reject t he null if E > .

fi n

,

|µ= 0

, . Thus

fi.6† ¸ fi.6† ¸

n

n

Problem 5 (a) HO : µ = F vevsus H fi : µ = 6. ,

(b)The test statistics is sample mean: T = E . And under the null E ~ N (F, fi/ fi0),and under the alternative E ~ N (F, , fi —fi fi fi0). We reject HO if E c F— , 0 (0.0†) fi0 fi = F —, (fi.6†) fi0 = F — (0.32)(fi.6†) = 6.4F So we reject HO because E = 6.2 c 6.4F. (c)

P ovev = fi — Ø fi —=P (don‘t Σ reject . HO |HA is ΣΣtrue) 6.4F — 6 = fi — fi — 0 , . Σfi/ fi0 6.4F — 6 = 0 , f i/ f i0 = 0 (fi.49) = 0.93

(d) With unknown aX, the test statistic underE the — µnull is T= S/ ,n ~ t economicshomeworkhelper.com

(fiO—fi)

we reject HO if ,

E —µ c S/ t (α), n where α is a9significance level e 6.2 — F , , = —2.† c —fi.83 (from the table) fi.†/ fi0

So we reject HO.

(e) In this case, the test then becomes two-sided [HO : µ = F vevsus H fi : µ ƒ= F], so all the procedures used before should be modified accordingly.

Problem 6 Suppose Es ~ N( µ E , fi), s = fi, ..., n E and Z j ~ N(µ 7 , fi), j = fi, ..., n 7 and all the observations are independent. You want to test the hypothesis that the means are equal against the alternative that they areZ)not. Use the statistic (E — : T = √ fi/n E ‡ fi/n 7 (a)The test statistic T can take any real number, and if the absolute value of T is ‘sufficiently‘ different from 0, we will reject the null hypothesis that the means are equal. (b)Note that T is a linear transformation of (n E ‡ n 7 ) normal random variables - hence, we can guess that T ~ N(E[T ], V av[T ]).The mean and variance of T are: E [T ]

=

( E — Z) Σ E ‡ fi/n 7 fi/n fi

E √ =√

E[(E — Z)] 7

‡ fi/nfi . Σ = √ fi/n E ‡ fi/n 7 E [ E] — E[Z] fi fi Σ √ = E . nΣE fi/n ‡ Efi/n 7 fi √ = (µ E — µ 7 ) fi/n E‡ fi/n 7 = 0 (under the nullµ E = µ 7 )

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fi/n

E

Σ

Σ E

s

—ΣE

fi Σ n7

Z

ΣΣ j

V av[T ] = V av √ =

.

= = . = . = = =

fi

.

( E — Z) Σfi/n E ‡ fi/n 7 Σ Σ fi V av[E — Z] fi/n E ‡ fi/n 7 . Σ fi (independence) . ‡ fi/n Σ V av[E] ‡ V av[Z] fi/n E 7 Σ fi fi Σ fi Σ VΣ av E s‡ V av Σ . Σn fi/n E ‡ fi/n 7 nE 7 fi Σ E 7 Σ. fi/n E ‡ Σfi/n 7 fi fi nXE Σ nX7 . f i f i fi n (fi) ‡ n (fi) ‡ nX ‡n2 nE n7 . nX n2 Σ . Σ nn E 7 nE ‡ n7 nE ‡ n7 nE n7

Z

ΣΣ j

So T ~ N (0, fi). (c) Since the distribution of the test statistic T is the standard nor- mal under the null, we reject the null when | t| > xαƒX. Problem Y nX X f i X = S 2X (a) Consider the test statistic T = , where S Es — E 2Y E n X —fi s=fi S 2 Σ Σ . (n —fi)S X X ~ z ( nXX —fi) and and SX is defined We know that Y analogously.

X 2 a 2 (n —fi)S Y X Y ~ z , and that these two statistics are independent . Un(n Y —fi) 2Y a

der the null hypothesis, aX = aX , so we can rewrite T as E

Y

F (n E — fi, nY — fi) . So we reject for T > c, where c is defined by Pr (F (n E — fi, nY — fi) > c) = α = 0.fi0. n X ( b) We have n E = 6, nY = 4, E = fi2, Y = 2.F†, Es — E = . Σs=fi n X fifi8ƒ† Y . Σ 8.F†. fifi8, and Ys — Y = SoΣ T = 8.F†ƒ3 = 8.09. From the F-table, s=fi

our critical value, c, is equal to †.3fi, so we reject the null hypothesis.

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( n —Xf i ) 7 ( nX2 —fi ) X Y ( n —fi )7 2 a Y Y) ( n —fi

a

X

~

2 Y

X

Σ

2