1 Constructing and Interpreting a Confidence Interval

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1 Itroductory Applied Ecoometrics EEP/IAS 118 Sprig 2014 WARM UP: Match the terms i the table with the correct formula: Adrew Crae-Droesch Sectio #6 5 March 2014 ˆ Let X be a radom variable with mea µ ad variace σ 2, ad we have a radom sample {x 1,..., x } with average x. ˆ Let Y be a biary radom variable that ca oly take the values 0 or 1, its mea is p ad we have a radom sample {y 1,..., y } with average ȳ = ˆp. Term 1 Observed Variace of x 2 Sample Variace of x 3 Variace of x 4 Sample Variace of x 5 Sample Variace of y 6 Sample Variace of ȳ (= ˆp) Formula s 2 = 1 1 (x i x) 2 ˆp(1 ˆp) i=1 1 (x i x) 2 σ 2 i=1 s 2 ˆp(1 ˆp) 1 Costructig ad Iterpretig a Cofidece Iterval Suppose I took a radom sample of 121 UCB studets heights i iches, ad foud that x = 65 ad s 2 = 4. Now, I d like to costruct a 95% cofidece iterval for the average height of UCB studets. Step 1. Determie the cofidece level. I the problem, it was give that we wat to be 95% cofidet that our iterval covers the true populatio parameter, so our cofidece level is Step 2. Compute your estimates of x ad s. Agai, we were give i this example that x = 65 ad s 2 = 4, so we do t eed to do the calculatios. Step 3. Fid c from the t-table. The value of c will deped o both the sample size () ad the cofidece level (always use 2-Tailed for cofidece itervals):

2 ˆ If our cofidece level is 80% with a sample size of 10: c 80 = ˆ If our cofidece level is 95%, with a sample size of 1000: c 95 =

3 I this problem: c 120 = 1.98 Step 4. Plug everythig ito the formula ad iterpret. The formula for a W% cofidece iterval is: CI W = [ ( ) ( )] s s x c W, x + c W Where c W is foud by lookig at the t-table for 1 degrees of freedom. So for our problem, we ca plug everythig i to get: CI 95% = [ ( ) ( )] , The 95% cofidece iterval is average height of the UCB studet populatio.. This iterval has a 95% chace of coverig the true Practice Use the Stata output below to costruct a 90% cofidece iterval for Michiga State Uiversity udergraduate GPA from a radom sample of the MSU studet body: Variable Obs Mea Std. Dev. Mi Max colgpa Cofidece level: 90% (Sigificace level is =10%) 2. x, s: x = ad s = Fid c 90 : c 90 = Compute & Iterpret iterval: [ ] , [ ] 2.922, Iterpretatio: We are 95% cofidet that this iterval covers the true MSU average GPA. 2 Hypothesis Testig Hypothesis testig is itimately tied to cofidece itervals. Whe we iterpret our cofidece iterval, we specify that there s a 95% chace that the iterval covers the true value. But the there s oly a 5% chace that it does t so after calculatig a 95% cofidece iterval, you would be skeptical to hear that µ is equal to somethig above or below your iterval. To see this, thik about the above practice problem, except suppose (just for the momet) that we kow the ukow true µ ad V ar( x); so we kow that the true MSU average GPA is 3.0 ad V ar( x) = Below is the distributio of x. 3

4 Is there a high probability that a radom sample would yield a x = 2.984? It certaily looks like it, give what we kow about the true distributio of x. Is there a high probability that a radom sample would yield a x = 3.1? Desity mu Hypothesis testig reverses this sceario, but uses the exact same ituitio. Suppose the dea of MSU, Lou Aa Simo, firmly believes the true average GPA of her studets is 3.1, ad she s coviced our radom sample is t a accurate reflectio of the caliber of MSU studets. Suppose we decide to etertai her beliefs for the momet that the true mea GPA is 3.1. Accordig to Lou Aa, the distributio of x looks like this: Desity mu Should we be skeptical of Lou Aa s claim? How to test a hypothesis: Now we ca formalize what we did above ito the 5 steps i hypothesis testig that we covered i lecture: 4

5 Step 1: Defie hypotheses: H 0 ad H 1 Here we write dow the ull ad alterative hypotheses i precise mathematical laguage. H 0 : H 1 : Sice we do t have a good reaso to thik the average GPA should be higher or lower tha 3.1, our alterative hypothesis is: H 1 : µ 3.1 Step 2: Compute test statistic: t = x µ SE( x) = x µ s This step is pretty mechaical. To compute the t-statistic (ofte referred to as the t-stat), we eed the sample mea x, the sample variace s 2, ad the sample size : x = s 2 = = 101 t = = You ca thik of the t-stat as: the sample mea coverted to stadard uits, assumig the ull hypothesis is true. Equivaletly, if the ull hypothesis is true, the we drew the sample mea from a ormal distributio with mea 3.1 ad usig this distributio, we ca re-write x i stadard uits. Step 3: Choose sigificace level (α) ad fid the critical value: c α from table The sigificace level is: Which is the same as the probability of Type 1 Error: Sice Type I error is obviously bad, we wat the probability that we make this type of error to be small. We ca set this probability to be low by choosig a low sigificace level. Oce we ve picked a sigificace level, we ca fid the critical value from a statistical table. Let s choose a 5% sigificace level for our example, so that the probability of Type I Error is just 5%. Check the t-table for the critical value whe our sigificace level for a 2-tailed test is α = 0.05 ad we have 100 degrees of freedom: c α = c.05 = Step 4: Reject the ull hypothesis or fail to reject it: Reject the ull if t > c α If t > c α, the we reject the ull hypothesis. We established above that if the ull hypothesis were true ad the true mea is 3.1, the the probability of observig a sample mea with a t-stat of t > c.05 is equal to the sigificace level, or 5%. So if we do observe a sample mea with a t-stat of t > c.05 the we say we reject the ull hypothesis. Just as before, we ca thik of a picture to help with the ituitio here: 5

6 If t < c α, the we fail to reject the ull hypothesis. Note that we ever really accept the alterative or ull hypothesis, we simply do or do ot fid evidece agaist the ull. I our example: t = 3.13 = 3.13 > = c.05 Reject the ull Step 5: Iterpret If we reject the ull hypothesis: There is statistical evidece at the 5% level that the average GPA at MSU is differet from 3.1. It s ulikely that Lou Aa s assertio is correct. If we fail to reject the ull hypothesis: There is o statistical evidece at the 5% level that the average GPA at MSU is differet from 3.1. It s possible that Lou Aa s assertio is correct. 3 Two Variatios: Proportios ad Differece Betwee Two Meas There are two variatios o hypothesis testig that we ve covered so far: 1. Whe x is actually a proportio, i.e. the average of a biary variable. Hypothesis tests for biary variables vary from the steps outlied above i two ways: (a) Uder the ull hypothesis, we kow the true proportio, p. Recall the formula give for the mea ad variace of a biary variable kowig the mea implies that you kow the variace. Thus, whe we use the ull hypothesis to compute the test statistic, we do t use the sample variace. Istead we use the variace specified by the ull hypothesis. E[x] = p V ar(x) = p(1 p) (b) Remember that if we kow the true mea ad the true variace, we use a ormal distributio istead of a t-distributio. So to fid the critical value, we use the ormal table istead of the t-table. 2. Whe we re iterested i how two sample meas differ. Hypothesis tests for a differece i meas vary from the steps outlied above i oe way: (a) To compute the test statistic, we use the stadard deviatio of the differece. Example 1. Suppose that a military dictator i a uamed coutry holds a plebiscite (a yes/o vote of cofidece) ad claims that he was supported by 65% of the voters. Call X the biary votig variable. A NGO suspects that the dictator is lyig ad cotracts you, a skilled ecoometricia, to ivestigate this claim. You have a small budget, so you ca oly collect a radom sample of 200 voters i the coutry. From your sample of 200, you fid that 115 supported the dictator, so the sample proportio ˆp = Step 1. H 0 : p = 0.65 H 1 : p <

7 (The reaso why we re usig a 1-tailed test here is because we suspect the proportio supportig the dictator is below 65%.) Step 2. Remember that the variace of X uder the ull hypothesis is: V ar(x) = p(1 p) V ar(ˆp) = p(1 p) This meas that our test statistic, z = The for large samples, we kow that: z = =.65(1.65) 200 = SD( ˆp) = ˆp p SD(ˆp) N(0, 1): ˆp p = SD(ˆp) = Step 3. Here we ll choose a 5% sigificace level agai, so α =.05 c.05 = 2.57 Step 4. Reject ull hypothesis if z > c.05, so we: Reject Fail to reject Step 5. Iterpret: We fail to reject the ull hypothesis (at the 5% sigificace level) that the proportio of people who support the dictator is 65%. I this sample, there is o evidece that the dictator is fixig the vote results. Example 2. Now let s cosider a example from actual data for a poverty alleviatio program i Mexico. I 1997, 24,059 households i rural Mexico were radomly allocated betwee treatmet ad cotrol groups for a coditioal cash trasfer program called Oportuidades to keep kids i school. Whe aalyzig the results of a radomized experimet, the first step is to verify that the cotrol group is, o average, very much like the treatmet group i terms of characteristics that we observe ad have data for. For example, data was collected o household assets. Your data reveals that while 14.47% of the 14,846 treatmet households have a refrigerator, ad 16.53% of the 9,213 cotrol households have oe. I order to cofirm that about the same proportio of households i each group have a refrigerator, we eed to perform a hypothesis test. Call the sample proportio of households with a refrigerator i the treatmet group ˆp t, the true treatmet proportio p t, the sample proportio of households with a refrigerator i the cotrol group ˆp c, ad the true cotrol proportio p c. Also, call the whole sample proportio of households (i either treatmet or cotrol) with a refrigerator ˆp. Step 1. Defie the hypotheses H 0 : p t p c = D = 0 H 1 : p t p c = D 0 Step 2. Compute a test statistic (t-stat) How do we compute this test statistic? We kow that the ull hypothesis specifies E[p t p c ] = 0, so what s left is the stadard deviatio. Wheever we re testig a differece of meas, remember the formula: V ar( x ȳ) = V ar( x) + V ar(ȳ). So applyig the formula, we have that: V ar(ˆp t ˆp c ) = V ar(ˆp t ) + V ar(ˆp c ) V ar(ˆp t ) = ˆp(1 ˆp) t V ar(ˆp c ) = ˆp(1 ˆp) c ˆp(1 ˆp) Which meas SD(ˆp t ˆp c ) = t + ˆp(1 ˆp) c The trickiest part here is keepig track of what your ull hypothesis is! Now we re ready to calculate our z-statistic: 7

8 ˆD = ˆp t ˆp c =.0206 ˆp = (.1447) + (.1653) = SD( ˆD).1526(1.1526).1526(1.1526) = + = z = = Step 3. By the ull hypotheses we chose, we re doig a two-sided test. Let s choose the 5% sigificace level as this is the most commo test that ecoomists evaluate. Check the ormal table to fid that c = Step 4. Reject Fail to reject Step 5. Iterpret:At the 1% sigificace level, there is statistical evidece that the proportio of households with a refrigerator i the cotrol group is ot the same as the proportio of households with a refrigerator i the treatmet group. What does this mea for the study? Probably ot much. I radomized experimets such as this oe, may household characteristics are checked for balace across treatmet ad cotrol. Statistically, we expect that some of our hypothesis tests will reject the ull simply because a 5% sigificace level idicates that 5% of the time we will reject the ull eve though it s true. 8

1 Constructing and Interpreting a Confidence Interval

1 Constructing and Interpreting a Confidence Interval Itroductory Applied Ecoometrics EEP/IAS 118 Sprig 2014 WARM UP: Match the terms i the table with the correct formula: Adrew Crae-Droesch Sectio #6 5 March 2014 ˆ Let X be a radom variable with mea µ ad

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