Confidence Intervals for the Mean of Non-normal Data Class 23, 18.05, Spring 2014 Jeremy Orloff and Jonathan Bloom

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1 Cofidece Itervals for the Mea of No-ormal Data Class 23, 8.05, Sprig 204 Jeremy Orloff ad Joatha Bloom Learig Goals. Be able to derive the formula for coservative ormal cofidece itervals for the proportio p i Beroulli data. 2. Be able to compute rule-of-thumb 95% cofidece itervals for the proportio p of a Beroulli distributio. 3. Be able to compute large sample cofidece itervals for the mea of a geeral distributio. 2 Itroductio So far, we have focused o costructig cofidece itervals for data draw from a ormal distributio. We ll ow will switch gears ad lear about cofidece itervals for the mea whe the data is ot ecessarily ormal. We will first look carefully at estimatig the probability p of success whe the data is draw from a Beroulli(p) distributio recall that p is the mea i this case. The we will cosider the case of a a large sample from a ukow distributio; i this case we ca appeal to the cetral limit theorem to justify the use z-cofidece itervals. 3 Beroulli data ad pollig Oe commo use of cofidece itervals is for estimatig the proportio p i a Beroulli(p) distributio. For example, suppose we wat to use a political poll to estimate the proportio of the populatio that supports cadidate A, or equivalet the probability p that a radom perso supports cadidate A. I this case we have a simple rule-of-thumb that allows us to quickly compute a cofidece iterval. 3. Coservative ormal cofidece itervals Suppose we have i.i.d. data x,x 2,...,x all draw from a Beroulli(p) distributio. the a coservative ormal ( α) cofidece iterval for p is give by x ± z α/2 2. () V The proof give below uses the cetral limit theorem ad the observatio that σ = p( p) /2.

2 8.05 class 23, Cofidece Itervals for the Mea of No-ormal Data, Sprig You ll also see i the derivatio below that this formula is coservative, providig a at least ( α) cofidece iterval. Example. A pollster asks 96 people if they prefer cadidate A to cadidate B ad fids that 20 prefer A ad 76 prefer B. Fid the 95% rule-of-thumb cofidece iterval for p, the proportio of the populatio that prefers A. aswer: We have x = 20/96 =.62, α =.05 ad z.025 =.96. The formula says a 95% cofidece iterval is.96 I.62 ± =.62 ± Proof of Formula The proof of Formula will rely o the followig fact. Fact. The stadard deviatio of a Beroulli(p) distributioisatmost.5. Proof: Let s deote this stadard deviatio by σ p to emphasize its depedece o p. The variace is the σ2 p = p( p). It s easy to see usig calculus or by graphig this parabola that the maximum occurs whe p = /2. Therefore V the maximum variace is /4, which implies that the stadard deviatio σ p is less the /4 =/2. Proofofformula (). The proof relies o the cetral limit theorem which says that (for large ) the distributio of x is approximately ormal with mea p ad stadard deviatio σ p /. For ormal data we have the ( α) z-cofidece iterval σ p x ± z α/2 The trick ow is to replace σ p by.sice σ p the resultig iterval aroud x 2 2 x ± z α/2 2 always at least as wide as the iterval usig ± σ p /. A wideritervalismorelikelyto cotai the true value of p so we have a coservative ( α) cofidece iterval for p. Agai, we call this coservative because 2 overestimates the stadard deviatio of x, resultig i a wider iterval tha is ecessary to achieve a ( α) cofidece level. 3.3 How political polls are reported Political polls are ofte reported as a value with a margi-of-error. For example you might hear 52% favor cadidate A with a margi-of-error of ±5%. The actual precise meaig of this is if p is the proportio of the populatio that supports A the the poit estimate for p is 52% ad the 95% cofidece iterval is 52% ± 5%. Notice that reporters of polls i the ews do ot metio the 95% cofidece. You just have to kow that that s what pollsters do.

3 8.05 class 23, Cofidece Itervals for the Mea of No-ormal Data, Sprig The 95% rule-of-thumb cofidece iterval. Recall that the ( α) coservative ormal cofidece iterval is x ± z α/2. 2 If we use the stadard approximatio z.025 = 2 (istead of.96) we get the rule-of thumb 95% cofidece iterval for p: x ±. Example 2. Pollig. Suppose there will soo be a local electio betwee cadidate A ad cadidate B. Suppose that the fractio of the votig populatio that supports A is θ (we ve switched from p to θ just to provide practice with aother stadard otatio). Two pollig orgaizatios ask voters who they prefer.. The firm of Fast ad First polls 40 radom voters ad fids 22 support A. 2. The firm of Quick but Cautious polls 400 radom voters ad fids 90 support A. Fid the poit estimates ad 95% rule-of-thumb cofidece itervals for each poll. Explai how the statistics reflect the ituitio that the poll of 400 voters is more accurate. aswer: For poll we have Poit estimate: x =22/40 = 0.55 Cofidece iterval: x ± =0.55 ± =0.55 ± 0.6 = 55% ± 6%. 40 For poll 2 we have Poit estimate: x = 90/400 = Cofidece iterval: x ± =0.475 ± =0.475 ± 0.05 = 47.5% ± 5%. 400 The greater accuracy of the poll of 400 voters is reflected i the smaller margi of error, i.e. 5% for the poll of 400 voters vs. 6% for the poll of 40 voters. Other biomial proportio cofidece itervals There are may methods of producig cofidece itervals the proportio p of a biomial(, p) distributio. For a umber of other commo approaches, see: 4 Large sample cofidece itervals Oe typical goal i statistics is to estimate the mea of a distributio. Whe the data follows a ormal distributio we could use cofidece itervals based o stadardized statistics to estimate the mea. But suppose the data x,x 2,...,x is draw from a distributio with pmf or pdf f(x) that may ot be ormal or eve parametric. If the distributio has fiite mea ad variace ad if is sufficietly large, the the followig versio of the cetral limit theorem shows we ca still use a stadardized statistic.

4 8.05 class 23, Cofidece Itervals for the Mea of No-ormal Data, Sprig Cetral Limit Theorem: For large, the samplig distributio of the studetized mea is approximately stadard ormal: x μ N(0, ) s/ So for large the ( α) cofidece iterval for μ is approximately [ ] s s x z α/2, x + z α/2 where z α/2 is the α/2 critical value for N(0, ). This is called the large sample cofidece iterval. Example3. How largemust be? Recall that a type CI error occurs whe the cofidece iterval does ot cotai the true value of the parameter, i this case the mea. Let s call the value ( α) the omial cofidece level. We say omial because uless is large we should t expect the true type CI error rate to be α. We ca ru umerical simulatios to approximate of the true cofidece level. We expect that as gets larger the omial cofidece level of the large sample cofidece iterval will coverge to the true value. We ra such simulatios for x draw from the expoetial distributio exp() (which is far from ormal). For several values of ad omial cofidece level c we ra 00,000 trials. Each trial cosisted of the followig steps:. draw samples from exp(). 2. compute the sample mea x ad sample stadard deviatio s. s 3. costruct the large sample c cofidece iterval: x ± z α/2. 4. check for a type CI error, i.e. see if the true mea μ = is ot i the iterval. With 00,000 trials, the empirical cofidece level should closely approximate the true level. For compariso we ra the same tests o data draw from a stadard ormal distributio. Here are the results. omial cof. omial cof. α simulated cof. α simulated cof

5 8.05 class 23, Cofidece Itervals for the Mea of No-ormal Data, Sprig Simulatios for exp() Simulatios for N(0, ). For the exp() distributio we see that for = 20 the simulated cofidece of the large sample cofidece iterval is less tha the omial cofidece α. But for = 00 the simulated cofidece ad omial cofidece are quite close. So for exp(), somewhere betwee 50 ad 00 is large eough for most purposes. Thik: Why whe = 20 the simulated cofidece for the N(0, ) distributio is smaller tha the omial cofidece? This is because we used z α/2 istead of t α/2. For large these are quite close, but for = 20 there is a oticable differece, e.g. z.025 =.96 ad t.025 =2.09.

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