This does not have a normal distribution. It can be shown that this statistic, the t-statistic, has the t-distribution with n-1 degrees of freedom.

Transcription

1 Small sample tests Samplig problems discussed so far dealt with meas ad proportios. Evaluatio of their samplig errors was based o the ormal distributio. I the case of the mea, the samplig distributio was ormal because the variable was distributed ormally i the populatio or because the Cetral Limit Theorem esured ormality for large samples. the case of proportios, the ormal distributio was used as a approximatio for the uderlyig biomial distributio. I each case, we required a large sample ( 30). Whe samples are small, <30, whe the populatio is ormally distributed, ad whe the populatio variace has to be estimated from sample data, the distributio of the sample mea iso loger ormal. A small sample distributio, kow as the t-distributio, has to be used i this case. Whe samples are small ad the distributio of the variable i the populatio is ot ormal, there is o readily available samplig distributio. Whe dealig with proportios comig from small samples, it is ecessary to use the exact biomial distributio. 6. The t-distributio Assume that the variables is distributed ormally i the populatio with mea µ ad variace σ, i.e. X~N(µ,σ ). If σ is kow, the the sample mea is ormally distributed, ad we have o problem. However, i almost all cases we do ot i fact kow the populatio variace, σ, ad must estimate it. We have see that the estimator S ˆ = ( X X ) /( ) is a ubiased estimator of σ. We let i= ˆ i S ˆ = S. However, whe we replace σ with Ŝ i the usual formula for Z, we get: X µ t= Sˆ / This does ot have a ormal distributio. It ca be show that this statistic, the t-statistic, has the t-distributio with - degrees of freedom. For large, the t-distributio resembles the stadard ormal distributio, but we are iterest here i small samples. The formula for the t- distributio is quite complicated, ad depeds o the umber of degrees of freedom. However, it is symmetric about 0, so the same useful

2 shortcuts, such as P(t>-a)=P(t<a) ca be used as for the stadard ormal. It ca be show that E(t)=0, ad Var(t)=k/(k-), where k is the umber of degrees of freedom, so i this case, Var(t)=(-)/(-3). Tables of the cumulative t-distributio for differet umbers of degrees of freedom are available. There is also a t-distributio fuctio i Excel: For x>0, ad k degrees of freedom, the fuctio TDIST(x,,) will retur P(t>x), while the fuctio TDIST(x,,) will retur the -tailed test, P(t>x OR t<-x). There is also a fuctio TINV(p,) will retur the critical value X C for a -tailed t-distributio with degrees of freedom, such that P( t >X C )=p. The distributio of X i the populatio has to be ormal for the t-statistic to have the t-distributio. However, the t-distributio is quite robust, ad small deviatios from ormality i the populatio will ot ivalidate it. Tables of the t-distributio The t-distributio will deped o degrees of freedom. Typically, a table of the t-distributio will give the critical values correspodig to differet probability levels for a -tailed test. (For a -tailed test, you must halve the probability level, sice you are cosiderig that probability i each tail.) Part of a typical table by degrees of freedom (k) ad probability (α) is show below. k/α For example, P(t.5706) = 0.05 for the t-distributio with 5 degrees of freedom. We write.5706 = t α=.05,k=5, or t.05,5 = Uses of the t-distributio As the t-distributio is a samplig distributio, it ca be used to costruct cofidece itervals for the populatio mea µ ad to test hypotheses. Cofidece iterval

3 If a radom sample of size comes from a ormal populatio with mea µ ad variace σ (both µ ad σ beig ukow) we ca state X µ ˆ S / P tα /, < < tα /, =-α. Sice there is a probability α/ that the t-statistic will be higher tha the α/ critical value, ad aother α/ that it will be below mius that value. This expressio ca be re-arrage to give the (-α) cofidece iterval for µ, P[ X t Sˆ / < < X t Sˆ ] α /, µ + α /, / = -α. For example, if we wat a 95% cofidece iterval, the we choose α=.05. Compare this with the 95% cofidece iterval i the large sample case, ad whe we were assumig a kow σ. Here we had P[ X.96σ / < µ < X +.96σ / ] =0.95. Here,.96 is the critical value of the stadard ormal distributio, such that P(Z>.96) = (Sice this is a -tailed test). Thus, the α/ critical value of the stadard ormal distributio is replaced with the α/ critical value of the t-distributio. The populatio stadard deviatio σ is replaced by a ubiased estimate of the stadard deviatio, Ŝ. I each case, the cofidece iterval is measured i stadard errors of the sample mea; i the case of the kow S.D., SE( X )=σ/, i the case of the ukow SD, it is Ŝ /. Example A radom sample of 6 households is take from a large block of flats, ad shows that household expediture o food is 4 per week, with a stadard deviatio of 0. Assumig that household expediture o food is ormally distributed, fid the 95% cofidece iterval for the populatio mea. As S = ( X i X ) i=, we have

4 Ŝ =S */(-) = (6/5)*0 = 06.67, so Ŝ =0.33. From tables, t.05,5 =.34. Thus the cofidece iterval is µ=4 ±.34(0.33/ 6) = 4±5.50 or 36.5<µ<47.5. The cofidece iterval is quite wide, sice is small ad Ŝ is quite large. Test of hypothesis The procedure for testig a hypothesis is similar to that used for large samples, i.e. based o the ormal distributio, but istead of usig the z- statistic, we ow use the t-statistic. Procedure: Set up the ull hypothesis, H 0 :µ=µ 0 (say) ad the alterative hypothesis, H : µ µ 0. Choose the sigificace level α at which H 0 is to be X µ tested. The test statistic is t= 0. The critical value of t is t Sˆ α/,- as / this is a -tailed test, ad is foud from tables. The decisio rule is to reject H 0 if t > t α/,-, ad accept H 0 otherwise. If the alterative hypothesis were H :µ>µ 0 or H :µ<µ 0, the we would use a -tailed test, with the critical value beig t α,-. Note agai that our decisio rule is based o measurig how may stadard errors the sample mea is from the hypothesised populatio mea. Differece betwee two sample meas We may If two small radom samples are take from two ormal populatios with the same variace, it ca be show that the statistic: [ X X ) ( )]/ Sˆ (/ ) (/ ) t = ( µ µ p + has the t-distributio with ( + -) degrees of freedom where ˆ S + S S p = is a pooled estimate of the commo populatio + variace, ad ad are the sample sizes. Whe the variables X ad

5 X are ot ormally distributed, or whe the populatio variaces are ot equal, the test (sometimes called the studet t-test) is ot strictly valid. However the t-distributio is quite robust, so that small deviatios from ormality or small differeces i the variaces ca be igored i practice. Very ofte, our ull hypothesis will be that the two populatio meas are equal, so that µ -µ i the above formula will be equal to 0. Example Cotiuig the last example, suppose that a radom sample of households take from aother large block of flats showed a average household food expediture of 36 per week with a stadard deviatio of 9 per week. Assumig that household expediture o food is ormally distributed i each block, ad that the populatio variaces are equal, test the hypothesis that the two populatio meas are the same. H 0 :µ -µ =0 H :µ µ. Assume α=0.05. We first calculate the estimated populatio variace, [ (9 ) + 6(0 )]/( + 6 ) ˆ p = S = 98.9, hece Ŝ p =9.95 (4 36) 0 t= 9.95 (/6) + (/) =.58. The critical value of t obtaied from tables is t.05,6 = (There are 6+- = 6 d.f.). As.58<.055, H 0 caot be rejected at the 5% level of sigificace. The t-statistic is also crucial i regressio aalysis, as the differece betwee a estimated regressio parameter ad the populatio parameter, divided by its stadard error, has the t-distributio. We therefore use t- statistics to test hypotheses about regressio parameters, for example the hypothesis that the parameter is equal to zero (i.e. o relatioship betwee the variables). 6. The χ distributio

6 The χ distributio has may applicatios; it ca be used to test hypotheses about populatio variaces, ad about the distributio of two or more populatios amogst differet categories. (For example, are the distributios of differet ethic groups amogst differet classes of job the same?) It also appears i may cotexts i regressio aalysis. We itroduce it iitially i terms of populatio variaces. Whe a radom sample of size is take from a populatio i which a variable X follows the ormal distributio, it ca be show that the statistic i= ( X i X ) χ =S /σ = where σ is the populatio variace has the χ σ distributio with (-) degrees of freedom. The distributio depeds o the umber of degrees of freedom, it has a complicated formula ad is positively skewed. The variable χ lies betwee zero ad, E(χ )=(-) ad Var(χ )=(-). As icreases, the distributio slowly approaches the ormal distributio. Whe 00, the approximatio is quite close. A typical χ distributio is show below: f(χ ) - χ Tables show the area (α) uder the χ curve to the right of a particular value of χ for a give umber of degrees of freedom, k. For example, the etry i the table for k=4 ad α=0.95 is This meas that

7 P(χ 4)>0.707=0.95. The etry for k=4 ad α=0.05 is 9.488, so P(χ 4)>9.488=0.05. Cofidece iterval for the populatio variace (σ ) As the statistic χ =S /σ has the χ distributio with - d.f., we ca write P[χ.975,-<(S /σ )<χ.05,-]=0.95. Rearragig, we get a 95% cofidece iterval: P[S /χ.975,-<σ <S /χ.05,-]=0.95. Similarly, we may test hypotheses about σ usig the χ statistic. 6.3 The F-distributio The F-distributio ca be used to test equality of two populatio variaces. It also occurs frequetly i regressio aalysis. It is used to test whether a set of regressio results as a whole is sigificat, ad it ca be used to test whether a more complicated model is to be preferred to a simpler model. We itroduce it i terms of populatio variaces. If samples of size ad respectively are take from two ormal populatios with variaces σ ad σ, it ca be show that the statistic ˆ ( S / σ ) F= has the F-distributio with k ( ˆ = - ad k = - d.f., S / σ ) where Ŝ ad Ŝ are the ubiased estimates of the populatio variaces, that is Ŝ = i= ( X i X ) ad similarly for Ŝ. The F-distributio has a complicated formula, ad depeds o two degrees of freedom, k ad k. It is positively skewed, takig values betwee 0 ad. It ca be show that E(F)=k /(k -) for k >.

8 Tables The F-tables show critical values of F correspodig to differet values of α (tail probabilities) ad differet combiatios of degrees of freedom, k = - i the umerator ad k = - i the deomiator. The table etry will show F k,k,α s.t. P(F >F k,k,α)=α. A extract from a table is show below: k /k 3 4 α=.05 α=.05 α=.05 α=.05 3 α=.05 α=.05 For example, P(F>F.05,4, =9.5)=.05, P(F>F.05,4, = 39.5) = That is, if our F distributio has 4 d.f. i the umerator, ad i the deomiator, the the 95% critical value is 9.5, ad the 97.5% critical value is Note that oly the upper-tailed values of the F-distributio are tabulated. This is because it is always possible to place the larger value of S ˆ / σ i the umerator of the F ratio, so that the observed values of F will always fall i the right-had tail. Test of hypothesis Set up hypotheses, H 0 :σ =σ H :σ σ. Select level of α=0.05 (say, to get a -tailed test for sigificace level of 5%). The test statistic is ( Sˆ / σ ) F = = ( Sˆ / σ ) Sˆ Sˆ sice uder H 0, σ =σ. Covetio: The larger estimate of the commo populatio variace is placed i the umerator of the F-ratio, so if Ŝ > Ŝ, we let F= Ŝ / Ŝ i order to esure that F falls i the upper tail of the F-distributio. The critical value of F is obtaied by lookig at the F-table with k d.f. o the top of the table (horizotal) ad k d.f. o the left had side of the

9 table (vertical). Look up the appropriate box, ad select the value of F for the appropriate value of α i that table box. Decisio rule: For a oe-tail test, if F>F α, k,k, H 0 ca be rejected at the α level of sigificace. For a -tailed test (usually the case), H 0 ca oly be rejected at the α level of sigificace, e.g. if we wat a 5% level of sigificace we must take α=.05. Example We wat to test whether male ad female studets have differet variaces i their test scores o a certai course. The 5 male studets have a sample variace of σ m = 5, ad the 3 female studets have a sample variace of σ f =. Test the hypotheses that the variaces are equal at the 5% level of sigificace, usig a -tailed test. Our ull hypothesis is H 0 : σ m = σ f. First of all, we must calculate Ŝ ad Ŝ. We have that S =5, so Ŝ = S * /( -)=5*5/4=34.4, ad Ŝ =S * /( -) = *30/9=5.. So the test statistic is F=34.4/5.=.87. There are 5-=4 d.f. i the umerator ad 30- =9 d.f. i the deomiator. We use the FINV fuctio i Excel, where FINV(p,k,k) gives the value of F* s.t. P(F>F*)=p, where the F- distributio has k ad k d.f. Hece we wat FINV(0.05,4,9)=.54. (Sice we wat a -tailed test). Sice.87<.54, we caot reject H 0, so we do ot have sufficiet evidece to coclude that male ad female studets have differet variaces. (NB: it seems here that we have take as our sample the whole class, so what is the differece betwee the sample variace ad the populatio variace? I this case, we would be takig our populatio to be male ad female studets i geeral, or hypothetical future studets o the course. Of course, we would eed to cosider carefully whether it is legitimate to extrapolate from our sample, this year s class, to the geeral case. This is a commo problem i statistical ad regressio aalysis; we might have quite a limited sample, ad the questio of whether we ca extrapolate to future cases, or to say, differet coutries or differet circumstaces, is ofte quite ucertai.)

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