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# STATISTICS MATH 203

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This 41 page Class Notes was uploaded by Sadie Schroeder on Wednesday September 30, 2015. The Class Notes belongs to MATH 203 at Western Kentucky University taught by David Neal in Fall. Since its upload, it has received 9 views. For similar materials see /class/216741/math-203-western-kentucky-university in Applied Mathematics at Western Kentucky University.

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Date Created: 09/30/15

MATH 203 Hypothesis Tests Synopsis Use the first six items from the STAT TESTS menu Tests about the Mean 1 A null hypothesis of H0 p M versus a oneesided alternative hypotheses of HazultM or HazugtM 1 When an assumed value of o is known then the test statistic for a ZiTest is This test statistic should be compared to the zescores on the N 0 1 curve in the following situations a When sampling a normally distributed measurement with any sample size In this case the test statistic follows an exact standard normal distribution b When sampling an arbitrary measurement with a large sample size In this case the test statistic follows an approximate standard normal distribution With large samples Y M you also can adjust the test stat to z S because S m 0 With very large n J For small samples with nonrnormal measurements the ZiTest is not appropriate 2 For a normally distributed measurement and 0 unknown the TiTest test statistic is This test statistic should be compared to the tescores on the tn 1 curve With normal measurements this test statistic is good for any sample size n 2 2 Notes For very large samples then the tescore on the tn 1 curve and the zescore will be about the same So the ZiTest and the TiTest should give the same conclusion For example with a sample of size n 301 the 90 tescore on the t300 curve is 165 while the zescore is 1645 So the oneesided rejection regions with x 005 will be about the same in either case That is with a large sample of a normal measurement a ZiTest can still be used x M with the test stat z S As noted above this test stat also can be used with none J normal measurements provided we have a very large sample because then S m o Tests about a Proportion p A null hypothesis of H0 p P versus a oneesided alternative hypotheses of HazpltP or HazpgtP The test statistic is for the 1 Pro ZTest is p P 7 p Z P1 P J which should be compared to the zescores on the N0 1 curve Note If P is the true proportion then the true standard deviation is o P1 P Example For the test H0 p 035 suppose a sample of size 900 gave 288 Yes Then 288 032 035 39 032 dth t t tat39 71887 p 900 an 6 es s 15 Z 035gtlt 065 J900 Tests about the Difference in Means A null hypothesis of H0 pl pa M versus a oneesided alternative hypotheses of Ha Ill H2 ltM 0r HaZHI Mz gtM 1 When values of 01 and 02 are given then the test statistic for the ZisampZTest is 39 39 M 2061 x2 2 2 0 0 12 n1 n2 This test statistic should be compared to the zescores on the N 0 1 curve in the following situations a When sampling normally distributed measurements with any sample sizes In this case the test statistic follows an exact standard normal distribution b When sampling arbitrary measurements with large sample sizes In this case the test statistic follows an approximate standard normal distribution For small samples with nonenormal measurements the ZisampZTest is not advisable For very large samples with nonenormal measurements the ZesampZTest can be used 93961 J3962 M S S 1 2 quot1 quot2 with the test stat because S1 m 01 and S2 m 02 with large ml and n2 Note When calculating this test on the T178384 add the value of M to E2 in the 27 SampZTest screen in order to test H0 m 2 M rather than H0 Ml p2 0 2 With normal measurments and 01 and 02 unknown then the test statistic is 761 762 M 2 2 S 1i quot1 quot2 t This test statistic should be performed with the NonePooled ZesampTTest in the following situation When sampling normally distributed measurements with any sample sizes 2 2 when the measurements on the two populations are not assumed to have the same standard deviation 3 Special Case When sampling normally distributed measurements that are assumed to have the same standard deviation the test statistic is tfi fz Mfi 7Cz M 2 2 l l Sp Sp n1 n2 JV I quot2 where Sp is the pooled deviation given by S n1 1S12n2 1S 7 n1n2 2 Now the test statistic by follows an exact redistribution with n1 n2 2 degrees of freedom The test can be computed with the Pooled ZesampTTest Notes When calculating either test on the T1783 84 add the value of M to 972 in the 27 SampTTest screen in order to test H0 in 2 M rather than H0 in v2 0 Tests about the Difference in Proportions 1 Test for equality ZiPropZTest A null hypothesis of H0 p1 p2 versus a oneesided alternative hypotheses of Ha p1 ltp2 or Ha p1 gtp2 x x Let p1 1 and p2 2 be the respective sample proportions and define the quotl quot2 pooled proportion by f7 x1 x2 the total Yes over the total sampled The test quot1 quot2 statistic for the ZiPropZTest is Z n p2 270 27 271 27 quot1 quot2 which should be compared to the zescores on the N 0 1 curve for large sample sizes 2 Test for the difference being something other than 0 This test is not builtein to the calculator A null hypothesis of H0 p1 7 p2 P versus a oneesided alternative hypotheses of Hazp17p2ltP or Hazp17p2gtP We no longer pool the proportions The test statistic is now I91 I72 P Z Pi1 Pi 17720 1772 quot1 quot2 which should be compared to the zescores on the N 0 1 curve for large sample sizes Test about the Average Difference such as BeforeAfter Measurements A null hypothesis of H0 pd M versus a oneesided alternative hypotheses of HazudltM or HazudgtM Subtract the measurements After 7 Before to obtain a new sample of the differences Then apply the ZeTest or TeTest as appropriate according to the guidelines of Tests about the Mean Dr Neal Spring 2009 MATH 203 Measurements Statistics is the study of numerical data Initially we will be concerned with a specific measurement X taken from a population Q For example we may want to measure the number of credit hours each fulletime student at WKU is taking this semester After obtaining such measurements from either the entire population or from a sample of the population we can begin to analyze the data in some way The most elementary analysis involves computing the mean median mode and standard deviation Specifying the Population and the Measurement When studying a set of data we first must know the population Q under consideration and the specific measurement X that we are analyzing These terms should be properly defined before any study begins If possible we also should try to determine the actual population size N Example 1 Last Spring each student in a particular section MATH 132 were asked to give the number of credit hours they had earned so far in college Below are the responses Number of Credit Hours Earned 16 46 18 17 48 18 84 50 16 18 16 89 78 76 78 124 92 76 Sort the values in increasing order in a frequency chart and according to undergraduate classi cation Solution Although not explicitly asked for in the problem we first should state the population Q its size N and the measurement X before analyzing the data Here Q Students in this specific MATH 132 in Spring 2007 A specific person in Q is denoted by u the small Greek omega From the data we can determine that the population size is N 18 Then X total college credit hours earned before Spring 2007 If person 0 has earned 48 hours then we can write X 1 48 We now sort the data into increasing or decreasing order 16 17 18 50 78 89 16 18 46 76 78 92 16 18 48 76 84 124 Next we can make a frequency chart that displays all the possible measurements number of hours earned and the numbers of students frequencies having each possible measurement D Neal 5pm 2cm Frequency Chm We asr eah map the data an aatepanes that any aspay the numbers n perentapes a suaentshaynp eertanrahpes raneasurerhens path tns aase we Trseeanspeuaeneasurement 39 eameus Numbe em Percent x a 5r Un39ng the THEHA output Entzr the raw datafmm Brampxe 1mm yvux calculatax th n use the calmlatm tr 1 shrt the datamtamueaxmg amen 1 rnaRe a htstrprahn MLh range In my MLh hms ht length an m mrnpute the haste stausuts w Du the same by enteer the T nay chart data SnluLmn We hrstrnust enter the data mm a hst Press STAT Lhm press 1th hung up then STAT Edit srrern yn mdex tr clear any data that mpht he mLhz hsts press the Up anww tr hghhght u press CLEAR Lhm press ENTER If necessary hrphhpht mher hsts press CLEAR and press ENTER Then mnve the tursrr back under hst u n xzymymyzx List Editrr Haphhpht u press Enterdztz inw an cLEARthenpress umlylisl ENTER D Neal Spare 2cm Ta enter Lhz data mm hst H type ls press ENTER type 12 press ENTER mhuhue uhm a Lhehaurs earned are enteredmta n Then press ZdeulT m mum mLhe Hamsrxeen FramLhe sT D mean xemzve the mmman than mm nae cammand 50mm Thenwewthz saned dilemma sTAT Edit s een EnterSol39AUJ Luszuan Making Hiswgnm Ta make a hmagram m nae daca we mm adjust the wwnow aha sTAT PLOT setungs Here nae xahge afmeaxuremzntsxs Xuinta12 lexwAthms m lenth 2n 0M Alwaysmakz 39 butadjutYnnxasneoessarym see each bar Adjust the sTAT PLOT seturgsta a hnagam 3m type set me xnm m u wAthxequenues 1 and press GRAPH Press TEAcE m 22 hawmany mzamxements are m earhhm Adamum Sldlypeandphl Llwilh eqsl Theaanan SathSE sn Enter stuu u Fax Lhm u 3 me average humheh m rudxt hams earned u mm 2 as 33 The meuah number afhaurs earned was AM he uauaaee Wm he de ned and hveh phage haaauah later D m4 Spare 2cm um a szin a Pregnancy Clan Naw we Wm emu the mm maimemznts mm 1m L2 and enter the equmues mm 1th Adm the wwnow ashefaxe um adjust the 1m m the STAT PLOT setung m graph L2 mm L3 Vluts ff Enter frequency STAT PLOT I Aajmsums A Pm GRAPH dnndztzinw Sldlypeandphl L2 and L3 L2 Wm legs L3 Ta mmputedqe mums emu the mmmand xixquotSm L2Ls whzxe L2 represen39s themeaxurements and L3 representsthe frequenues STAT cALc I mm Hawta areLhehaskztha players atUCmr Dataxsgwen halvw Enter the Izquency than mm yaur mum and mnpute the ham mums fur the warnm men and a D Neal 5m 2m SnluLmn Let n A UCann haskztha players andlet X hught m mthes In all then are zaphym 3m 2 break up mm twa uhrpapulauans n wamen hasketha playzrswnh M Splayexs and n2 mzn haskztha players Wm M 14 players Entexdqemeasuremzmsmm u andLhesumesxwe x nmmxm L2 r aqua L3 and m Thenentexdqe mmmand LVas sum Litacamputedqe awn mm and enter LVas am LLLS and LVus am LLM m mmpute me mm m u wamen and u mznmdwmually Th mm may W 21 We may 5 my the m mag 11 M We may 5 5 3 and the mu mag 1mm 8 13 55 mm mm s 1 y a Dr Neal Spring 2009 MATH 203 Covariance and Correlation Suppose we have a set of paired measurements X Y Eventually we will be testing whether or not the measurements are independent ie whether or not one has absolutely no effect on the other For example let X be the sum of the last four digits in your social security number and let Y be the sum of the last four digits in your telephone number It would make sense that one sum39s value has no effect on the other sum39s value In other words there is no correlation between the two But when there is some correlation then we will measure the degree of linear dependence between the measurements For example for people buying their first life insurance policy let X be the age of the insured and let Y be the monthly premium In this case there probably is some correlation As the age increases then the monthly premium tends to increase also Covariance Recall that the variance of a measurement X is given by the average of the squares minus the square of the average When we have a census of measurements x1 x2 xN then we can write and then the true standard deviation is 0X 40 If we have two measurements X and Y then we can compute the covariance by the average product minus the product of the averages C0VX Y MXY MXHY Note that CovX X VarX We previously have noted that the average of a sum is the sum of the averages X X 11 Y but it is not always the case that the average of a product equals the product of the averages But we can say the following When measurements X and Y are independent then uXY uXuY hence CovX Y 0 Dr Neal Spring 2009 Example 1 Let X be an integer randomly chosen from 2 3 4 and let Y be an integer randomly chosen from 3 4 5 6 7 Compute uXY and X My Solution Because each of 2 3 4 are equally likely to be chosen 11X 3 Likewise Y 5 and thus 11X LY 15 lntuitively X and Y should be independent of each other so it should be the case that uXY 11X 11y 15 But we can compute the average product directly By multiplying each possible X value by each possible Y value we obtain all 15 possible products 6 8 10 12 14 9 12 15 18 21 12 16 20 24 28 Thus 681028225 437 15 15 Hence CovX Y uXY Xuy 15 715 0 Example 2 Let X be an integer randomly chosen from 0 1 2 3 4 and let Y be an integer randomly chosen from 0 to 4 7 X Compute CovX Y Solution Clearly 11X 2 But computing y is a little more involved Because each of the five X values are equally likely the overall Y average is 2151050 1 MY 5 By multiplying each possible X value by each possible Y value we again obtain 15 possible products 0 0 000 0 1 2 30 2 4 0 30 Thus 123243 15 qu 15 15 1 Hence CovX Y uXY 1X Ly 1 7 2 71 In this case the possible values of Y are clearly dependent on the value of X Example 3 Compute CovX Y if Y is a direct function ofX given by Y 4X2 3 and X has the following distribution X 72 71 0 1 2 Prob 01 02 005 025 040 Dr Neal Spring 2009 Solution First consider the possible values of X Y and X Y To find the averages we weight the values according to their assigned probabilities 1X 2 gtlt 01 1gtlt020 gtlt0051gtlt 0252 X 04 065 ILY 19 gtlt 017 gtlt023 gtlt 0057gtlt 02519 gtlt04 128 Mg 38gtlt 01 7 gtlt020 gtlt0057gtlt 02538gtlt 04 1175 Hence CovX Y ILXY Xuy 1175 7 065gtlt 128 343 Paired Measurements When we have paired measurements of the form xl yl x2 y2 191 yn where each single measurement is equally likely then the covariance is given by 1 1 1 C0VXYMXY MXMY39inJi in Zy1 71 7 n 7 n 7 11 11 11 Example 4 Compute CovX Y if Y is a direct function of X given by Y 4X2 3 with X and Y having the following equally likely values I X 2 I 1 0 I 1 Z I Y I 19 I 7 0 I 7 19 Solution The possible products are 38 7 0 7 38 hence the covariance is 1 1 1 CovXY xlyi xl gtlt yi n n n 11 11 11 2 1012 Xw707w 5 38 70738 5 5 0 0 X 104 0 Note that Y is clearly dependent on X even though the covariance is 0 Dr Neal Spring 2009 Logical Summary If X and Y are independent then the covariance is 0 1f the covariance is not 0 then X and Y must be dependent 1f the covariance is 0 then X and Y may or may not be independent Correlation Suppose we have paired measurements of the form xl yl x2 y2 xn yn Then the true correlation p between the measurements is defined by MXY XVY 7 C0VX Y p 0X 0r T 0X or where X and y are the respective true averages of the populations 0X and oY are the true standard deviations and uXY is the average product of the measurements The correlation measures the strength and direction of the linear relationship between the measurements X and Y Properties of Correlation 1 It is always the case that 1 S p S 1 If p is near 1 then there is a strong positive linear relation in the sense that as the values of one measurement increases then the values of the other measurement tend to increase If p is near 1 then there is a strong negative linear relation so that as one set of values increases the other set of values tend to decrease Moreover p i1 if and only if there is an exact linear relationship 2 1f the measurements are independent then the correlation will equal 0 Hence there is no linear relationship between X and Y 3 1f the correlation does not equal 0 then the measurements must be dependent 4 The converse is not necessarily true If the correlation equals 0 then it may or may not be the case that the measurements are independent There is no linear dependence but there may be some other type of dependence 5 When each pair of measurements is equally likely and not assigned varying weights then the correlation is computed by 1 quot 1 quot 1 quot 2x1yi sz39 X 2 p MXY MXMY quot11 quot11 quot11 0X 0Y n n 2 n n 2 1 2 1 1 2 1 sz39 299 X Zyz39 2sz n n 7 7 11 11 11 11 D Neat 5pm 2cm Bangle 5 Fmd the mnelauan hetween tar and mmtne fur the raunwrng m brands as ngarettes T my Mm quotw Ap1ne Bensan st Hedges Bu Durham Gnlden Lrghts Kent Knnl LampM Snluunn Frrst we enter the datamta hsts sayhsts u and L2 and then mmpute the haste stausttes wALh the ZeVertzts mmmand tenetqu 1644321ne1397xn991N 41an 6525496xn40357 m Bemuse p rs near 1 we have a stranghnear retauanshrp hetween tar and mcmne As the tar levehmreases themmtrne leveltendsta rmrease rn ahnear manner The LinReyzx b mmmand mm the STAT GALC menu wtn earnpute the mnelaum arrerny Enter the mmmand LinReyzx b LLLZ m use whatever hsts bald the parred data we as r Nate nthe value m r dues nat appear then hung up and enter the Wms con mmmand rramthe CATALOG Thenxeeentex the Lian mmde The LinReg mmmand arsptays the cmxelauan as the symhat r whmh rs used ta denme the mp ennehtnn 1c the parrea data n yr x2 2 represents a passrhte measurements then we use p m the true can use u ar the true avnage Buul39we anly have a sample then 7 Hawzvex1u1 as 7 mm a gt elau n Just as we m rs mmputed the am way as u the sample mnelauan t rs pute e sameway as p Dr Neal Spring 2009 Sample Correlation Formally the sample correlation r on sample data xl yl x2 y2 cn yn is defined by 7 670007 7 quot11 F 2 JT 2 n n 39 39 1 1 x x y Jin2x2XJZyi2y2 quot11 quot11 This value of r is used to estimate the true correlation p Of course it is easily calculated with the LinReg command after entering the paired data into lists Dr Neal Spring 2009 MATH 203 Estimate for Difference in Means Consider a measurement X on two distinct populations 91 where X has mean m and variance 012 and 92 where X has mean 12 and variance 0 Our goal is to estimate the difference of averages ul 12 To do so we obtain a random sample of size ml from 21 and let f1 be its sample mean and S1 be its sample deviation Then we obtain an independent random sample of size n2 from population 92 and let E2 be its sample mean and S2 be its sample deviation Ultimately we want to conclude with a certain level of confidence whether one average is significantly higher than the other average or whether there is no significant difference in the averages With large populations and nonetrivial measurements it is almost certainly the case that the averages m and 2 will be different However they may be close enough to each other so that a confidence interval for the difference ul v2 contains the number 0 and has a small margin of error In this case we might say that there is no statistical difference in the means Average and Variance of a Difference Let X and Y denote random measurements on a population and let W X Y denote all random differences in these measurements The important facts that we need are l The average of the difference equals the difference of the averages That is HW MX Mr H When measurements X and Y are obtained independently then the variance of the difference is the sum of the variances That is 05V o2o Ill The previous results apply to all possible differences in sample means f1 972 from independent samples on 21 and 22 We know that all possible values of f1 have mean 2 0 0 m and standard dev1ation T1 so their variance is 1 L1kew1se all posmble values of quotl quotl 2 U x2 have mean 3922 and variance 2 Now let W x1 x2 be all posmble differences in quot2 sample means Then 2 2 2 2 01 c 2 o1 o2 uwM1 M2 0W 0W quot1 quot2 quot1 quot2 Moreover for large sample sizes ml and n2 W is approximately normally distributed Dr Neal Spring 2009 IV Because W m NpLW 0W with probability r the values of W 971 E2 will be within P W i ZuZ 0 W where zaZ is the appropriate ziscore That is with probability r 1 a the values of Tel 9E2 will be within 2 2 01 02 M1 P 2J r ZaZ quot1 quot2 But then with probability r 1 x the values of 111 112 will be within 2 2 01 02 quot1 x2iZa2 quot1 quot2 Confidence Interval for Difference in Means From the previous result we obtain the confidence interval formula for the difference in means 2 2 0 0 H1 M2 quot1 x2iZa2 1 2 quot1 quot2 ZisampZInt on Calc Arbitrary Measurements Need large samples Normal Measurements Any sample sizes work With large samples we may replace 01 and 02 with estimates or upper bounds 2 2 2 2 S1 32 U1 U2 H1 M2 quot1 x2iZa2 0F H1 M2 quot1 x2iZa2 quot1 quot2 quot1 quot2 0 If we disregard the 22 statistics then this formula reduces to the confidence interval 2 o z a fora single mean 111 It becomes 111 w x1 izuZ 1 x i i 711 J D Neal 5pm zum Ta aemunun walk papuJauans ht sues M and M we muld ahtam a miller margm ht em hyuseng urw 7r721ampn There are the same umalpxahlem wuh hsfaxnmla as wuh the gznexal mn dmce mtexvalfax the mean Namzly 1 Fax aahmaay measurnnents we need large sample sues a ablam amuaey and small mannns ht nm u a the zesmxe names mm an appmxjmalextandard harmed msmhuaan apphed m a passehxe d 39fexmce m samplemeam W i e i an The uue standard dewaums are a en unlmawn As wuh the cun denoe Interval m the mean we ean mpre the amuaey and Wexmmz these pxahlem when samphngrmm mma ymmmutedmeamemzms unmer The ensue suengh m mums pee sauaae nun s hung neasuaeu an M maeaenamanuaaeauaes a a Lheuc ber Fae a sample a 35 mans havmg 15 mum nesanue saaasuesweae s e 9 mm s e 35 Fae anameaemena samue a 3n aheshavna 35 em ahe saausaesweae s 1 1wALh 2 e 3 a Fmd a 95 mn dmce Interval fur the aurexense m average tensile suengm amnng a such manufavlures a 15 mttan hexs and 25 mum mans Explain the Interval m wards 352 342 e i 7 ex mnem 35 m Thaus the average tensile mmgvh anhe 159a mttan husxsfmml as use hm a 12x pnhlgharthanthat ufd123596 cmtan hexs wuh Ohm Interval u e M muld equal n sathe Vang term suerghs cauld he equal me Ohm dam aJane Lhzre xsnm a staasueany nyn cant aurexense m the Vang smile suengah as ewdenczd by the dasenessm 71 Xand72 1n1 SnluLmn lByhand New script eeuztmznenxsurwsm e a mn denoe Interval aJsa can he camputed wuh the zeSannzxm feature the STATS TESTS menu Set the lnpt a sun than entzr the vanahxes 1e dewaums m a and 412 and calmlate Thas yp Item a ham useng the samu D Neal Spare 2cm Bangle 2 We wrsh tn see r there rs any apparent durerence rn hrgh schddx grade pmnt age between guts and buys m the state Wha chadse tn gn tn mnege The rdudwrng da arsaran am cnuededndrhrghschndm s mm a m Kentucky h schddx gaduatesm thnr rustyear gr mllege Fm hetwseen average female and average male PA gmup gh d a was mn denoe Interval fur the durerence gade pmnt average Explain th e rntm m Snluunn Becausethe data sets are d the meme we can mmpute the statesucs dnadth samples at ancewrth the zevrsnts mmmand Enter the data mm hsts u funds and L2 zrnase m the m7 Edit screen Then enter the mmmand zevr Stats L1 L2 A zr cgrnpuurg the staustecs we see that the average female GPA rs 7 3353 wALhasample devraudndrs z n 3414 The averagemale GPA 72 31 5557wALh 2 n n 51254 In each case there are an rneasurernents Note u the data sets are m d1 me srzes then the War Stats cdnrnand wrn ndt wmk Lnthrs case srrnpty execute the LVzr Stats mmmand an each hst and ndte the Values m hr each sample Ndwtdrrnd a was cun denoe Interval hr the true durerence rn average GPA 2 2 New essssestnsssn 11545 A 57254 nussurwsndxs 25122 n 221242 at Thus we can assert that arnnng a mlleg ha d students In the state the min female GPArsrrdrn n as grade paints hrgher n ma grade paints hrgher than the avztage male GPA Dr Neal Spring 2009 Difference in Proportions We also can apply the formula to the special case of a difference in proportions p1 p2 Recall that for a proportion p the population variance is 02 p1 p which is estimated by S2 f71 f7 So for large populations a confidence interval for the difference in proportions is f 1 1 pl 7 pzm p1 7 p2 i zaZ Ww ZiPropZInt on Calc l 2 025 025 oruse p17p2mp17p2i zaZ quot1 quot2 0 When one or more population sizes is small then we could include its small population correction factor to decrease the margin of error For two small populations we have l 711 71N1 quot1J P21 P2N2 quot2J 7 m 7 i p1 p2 p1 p2 zaZJ 711 N1 1 n2 N2 1 025 N n 025 N 0r PI P2 P1 P2iZa2 52 3112 Example 3 A poll commissioned by the Center on Addiction and Substance Abuse at Columbia University found that 1340 of 2000 adults and 304 out of 400 youths interviewed believe that popular culture encourages drug use Find a 95 confidence interval for the true difference in proportions between adults and youths who have this belief Explain the interval in words Solution By hand Using the confidence interval for two large populations we have 1340 304 067 f dlt 076f h39ld dth p1 2000 ora u 5 p2 400 orc 1 ren an en 067 033 076 024 pl pzm 067 076i196 7009i004665 2000 400 025 025 O 7 m 067 076 i 196 m 7009i00537 r p1 p2 2000 400 Using the first interval 7009 i004665 we have 7 01366 S p1 p2 S 7004335 or equivalently 004335 S p2 p1 S 01366 So the proportion of all children who believe that popular culture encourages drug use is from about 43 percentage points higher to about 137 percentage points higher than that of adults D 1191 5m 2m Note Du hm eay ThaT The pmpamm 1e mm A 2 heghex Ta 131 heghex The mrexehee mpmpamahe a1waye shauld he expressedmtztms ht pxcznmge pnlnb The hm Type ht mn denoe Interval fur a du39fexencz eh naege papula an pmpamme aka eah he faund wALh The zepeepzhu reamee eTem 11 mm The STAT T mean mm a g T M m crLevel95I Calculate Patrice Emlciszs 1 A bank 1e arrenhg Twa ememe rudn eam p1ahe Sample gmupe are hemg murmured m eee 1f Thexe 1e a 1ngth dx 39exmce eh The avnage amnunts The gmupe rhargepex quarter The my eTaTaeTaeem ahe quarter are helww FlanA 152 3191151 329251 FlanB 111 sznss as 11 12 Fmd a 95 cun denoe ehmvaJ fax The mrexehee eh The average ammehTe ehaeged Thee quarter ammg an peaple eh The Twa gmupe Exp1aeh The Interval Th wards h v he1ThaT zsn Hut ex 1111 peaple under age an regularly e1 Dad maeuemg peemepeex eemeee 13m 11th 1x Hut ex 975 peaple aT1eaeT an yeaee ald regularly da ea Fmd a 91 mn dmce ehTewaJ at The mrexehee eh The Tm pwpmuahe arThaee wha dawnlaad mueeeeh Theee age gaups mphquot The ehTewaJ Th wards Axswus 1 The LMZINQWSJSH s s New s219n1 The average amwunt ehaeged per quarter by Plan A memmm 1e mmSlS sn hm Ta 32191 we Than The avnage amnuntrhargedhy FlanB mstamexs 2 The zepeepzwgvee 1 13279 g pry s 121121 The pmpamm ufpeapleundex an Whn dawnlaad mueeeuemg peenepeex 1e 1mm 13 21 p11 mm 111th Ta 2112peT pmntshngr Than The pmpm uan ht pewple an ax hue wha da ea D Neal 5pm 2cm MATH 203 scones Let 2 m 1 he the man naxma mmhuuan havmg mean u and man dwnuanl The zeemxermpxahmmy y Isthevalue 1 such um Pp s z s Thaus Lhexuspxahahdny y between the Fame 1 and 4 Huh y We denate the mm mm Hutu pxahahxhty w the ymbal a whzxe y 1 Thenthexe I a 2 pxahahxhty at eauhtad hum 93820112 lemme are many denaud w M where 412 xemm the ngmeuu pxahahxhty Hzre are the man mmmanly used lemme M WWW m ms ms 225 1 ms 2m Note Any athex meme can he mm MLh the hudan inva mind mm the DISTR menu Fax example 1 we want the m lemme men than n m pxahahdny m the mddh xegmn Sn Lhzxe I a 2 e was at eaLh ml The Lpanuve meme men came mm mdxrg the Inverse uf was mmulauve pmhahdny a the standard mxma dmnhuuan andxsappmmmatdy W 12x me do innNam annn1zx Fm Inner pmhahdny Fax agennd nmma dmnbuuan X Mum the haunds um cantam a are Qven by u it D ea spare m Bangle Hnghts h hamwmheh he hmhheuy dmnhuted M011 5 mehh a s5 5 mLhes mi 5 standard mmh a 515m 3 Inches Us heme m salve the mth 5 What hexghts he uLh mm a 511 When he between these heeghm h Nhathnghts he meh um 55 a 511 When he between these heeghm a What heehhh he meh um 15 a 511 When he between these heeghm shhmhh a The haunds 11m mhmh an at naxmally dmnhuted measurnnents he quot1 545e whmh here hes551m5xs Thus an a wamzn he mm an 555 h 7 435ilmh smhught h The hmmdsdm mm 95 hnhe mzaxuremmts are 11955 whmh hue are 5551mm Thus95afwamznaxefmm55SZhHSXimhzsmhught e Fem weheemhe 15 heme Ether m 75mner pxahahllny than than he 12 5 teachtml Heme the panuve heme casz mm ndmg the Inverse ht x15 mmulauvepmhahxhty hnhe mmquot mm mmhuuhh 5mm ahautl 15 sh the haundsdqat mm 15 wt themeaxuremmts are u 1 15a whmh hue are 5551115 Thus75afwamznaxefmm62 5hSXESimhzsmhught mm IQ some are gehenny mm m be nmma y dmnhuted wALh 5 mean ht mu and 5 mam dewauanan Us lrsmreslafmdlhehwundslhalmnlmn a 95 ufallscme h mam scan a mahmxmes h 94m a scmes Axswus a uti6 gtm tl 6xl mmshlzad hu12326agtmU12326x15mrS llhlsdxa e umsm mnntnsmnsh 52 134 m1 xss UseinvNannUJJJJ du11885gtmU1188x15mr713571232 Usemvmmmm Dr Neal Spring 2009 MATH 203 Survey Proposal Survey Question 1 On the Mean of a Measurement 1 Think of a measurement X that you would like to analyze For example the number of hours that a typical student sleeps per weeknight or the number ofjobs that a person has had 2 Specify the population Q that you wish to target for your survey for example Undergraduates at WKU or Adults living in Bowling Green 3 Break your population down into two disjoint subepopulations 21 and 22 for example 91 female and 22 male or 21 married and 22 single Survey Question 2 Population Proportion Using the same population and subepopulations as in Survey Question 1 4 Think of a YesNo question that has only two responses For example Do you favor capital punishment or Are you a smoker or nonesmoker State your question and which response you want to measure This will be called a favorable response When you collect your responses you should have at least 30 from each subpopulation Ideally the number asked in each subepopulation should be in the same proportion to the actual number of people in the subpopulation Data for your responses can be easily sorted 9 Person 2 Person 3 15 A Yes A39 Not Yes total in 91 total in 92 Total Yes Total Not Yes Total Dr Neal Spring 2009 MATH 203 Least Squares Regression As well as finding the correlation of paired data xl yl x2 y2 xn yn we also can plot the data with a scatterplot and find the Least Squares Line of Best Fit through the data This line having equation y a x b provides an approximate linear functional relationship between the values of x1 and yi Of course it is only a good fit if the correlation is near i1 which means that there is a strong linear dependence between the measurements X and Y The slope at is given by After the slope at is calculated then the intercept b is given by a xy rcW ZC2 by ax 1n 1n where xy Zgyi and x2 le2 n 7 11 11 ThenY aXb This least squares line is the line that minimizes the sum of squared errors between the actual yl values and the linear approximations axl b That is the sum n 2axl b yZ2 is minimized with these choices of a and b iil To compute both the correlation and the leastesquares line enter paired data into lists L1 and L2 or some other pair of lists press STAT scroll to CALC press 4 for LinRegaXb then enter the command LinRegaXb L1 L2 Some Other Quick Facts 1 The point E i is always on the leastesquares regression line NX r0 2 The slope of the leastrsquares regression line also is given by a N where r is 0r the sample correlation 6X l x2 93902 and 6y Jyz 72 3 The value r2 is the coef cient of determination It measures the proportion of the observed values accounted for by the regression fit Because 0 S r2 S 1 an r2 near 1 means that there is a strong fit and an r2 near 0 means that there is virtually no fit of the data D Neal Slaw 2cm Banglzs u Make asutterplaL u Campute y and expluh whatltmeans lu pm the equaum anhe leauesquueuegesuuhhhe and gaphmhmugh the scattnplat l I Lhzre a relaumshp between the m and mcmne levels lh ugueuea Eland Aplne Bensm amp Hedgzs Bu Durham STAT EDlT STAT PLOT A mASenings zoom We see the gehenl uehel A the m level mmeases than the mmnne level mueases The leauesquueuegresuuhhhe u an by yaxb e DD61U47SXD138167 Due mhe elase m Lhmhneal luhmuh mlle he used in pmlm ammune level y at a antar level x Fannstanchf x 2nmgunulhehy e l ssahguthlmuhe tr e n Flam VARS Flam lc STATISTICS EQ znd nmci Press ENTER VIII2x2 Bemuse e um hemuhe level lg 97 detnmmed by m level when uuhg y n mums x EI138167 as aprednlm and 2 5 detexmnedhy mhex avian D Neal Spare 2cm 2 1r apexsnn hashxgh hady denmy men uney shauld havelesshady m The rnunvnng datahxtsmzaxuremmts a hady denseues and peruntagzs uf hady m mm a Iandam s pewple 1s une relauunshp ahsn mm STAT ED STAT PLOT A mASenings zoom We see me suanguend Ashndy denmy mmeases unen hady m decreases we as Bemuse y 188D dnse quot memhemsa irmgngaave hnear relauanshlp between hady denmy and hady STAT cALc Enerde y mam rat The Teastesqunesxegessmn hne 1s Wen by y axb wAsaszAAsxsT whrhmhsmsegvesmdmnxtperfemhmarm u x 1 x uzzsv xouzxssvzz a v usxxxsuzzsv xsz oxoux597zz vZasene szvzez Sux oo5 57uo rvsgvl tr an VARS STATISTICS EQ ness GRAPH ness ENTER Bemuse H was we can say that hady m Ts 99 99 determmed by anes hady denmy whenuseng y 74045927 x 445 2575 as aprednlax D Neal Spare 2cm 3 as that a rdauarashap between dnvmg speed and MPG fur yvux gasguzuarrg sum STAT EDIT A mASemugs zoom arly s eraas ah same sun ufnlauanshp pnhaps duadraae The mpg There de e araareases as speed anueases m a senaara pamL but naera as speed anueases ruruaer the mpg drugs m f There as rad mrreaauaraa The mnelauan measures are maglh aaad duemun a are aaraear assad a hetweenthevanahle Iunhemusedq earreaauaraeduaasnaadaesraaarraearauaamaere as raa reaauarashap In uaas sase uaere amply as nut a pnmanent aaraear reaauarashap hetwem speed and mag humaere cznamJyxs areaauarashap Here are aeasaesduaresregressaara aarae as earasaaraa and as Ewen by a nauaas sase nae aeastesduares regressaara hn xsnmagaad m amae data Hawevex n as nae he ear appruuraaaunra m nae data whmh reany daesraa gaadherehesause daere asaad pnmanmt aaraear reaauarashap between d and mpg as a Vzr ea 5927362 Euxooo asvsuo a r7927 v v Press GRAPH Bemuse r2 n nae aarae daes nut m nae data at an Usarag nae lane 4 195 as a predadar daera nae rrapgasrad at an detexmnedhyns speed D Neal 5pm 2cm A Isthzre axdauanshphetwemhught and 0pm The man data a cullemun a mzamxements ham a xandam sample m WKU students Elude Heigh nehes GPA s 5 s s 1 STAT EDIT Adamaeme zoom There daesnm appear Ta he any relauanshlp heTween hnght and grade pnnT avztage HTng and law andmddh cm are named framsmden39s a a hexghm The mnelauanxs marly STAT CALC When Thexe 1 na relauunshp whatsaevex heTween The variables Then we say ThaT They are Jndepmdml When variables are mdzpmdmL Then The True cmnlauan wm equal SD The emehuan caef uentfmm anndam sample afmeamxemmts shde he very alas Ta u when The Twa vanahles have nu assauaum heTween Thenn Hzre The leanesquareuegeenan lmexsgwenhy y an EIEI7754X 3 3663446 B e mnela s nearly n Lhex 1 na he aunnshp heTween heTQhT and cm In TaeT Then 1 na Idauanshp aT a because cm I he mdem a hzxgqtl Sn gun The leaneaquarea regresnm hn 1s aT agnadm aTThe data Press GRAPH Bemuse T a n was less Than 1 m a pnsan39s GPA Ts deTnmened by hexght when unngthelmey r 7754x33663446 asapxedmax D Neal 5pm 2cm 5 Sixteen hinge drinkers at Ohm State Umvnnty had a beer party Thmy mmutes hm campuspahse measured Lhzu mm almhnl mntmt BAG Hue ue1he am 1 z s A s s 1 x Eeels 5 2 9 x s 1 s 5 EAC um nu ma n12 MA was nn1 nus a m n 12 12 M 15 Is Eeels s 5 A s 5 1 1 A EAC nnz nus um nm was any um nus kthen a mnelauanhetweenBAC and the numhzr afhezrs drunk by the stude Nhatwwuld yau predmuhe average BAcm hem peaplehawrg 5 beers7 zoom The BAC can vary mm pnsanta pexsm Fax ms39ance the BAC anhe studen39 hawngShezrswzu Z A mdnm Butagznenlu39endemsk Asthenumhex a heexsmueases sh daesdqe mm u m mntmt The hgh mnelauan shwwsthauhexe es a xehuvexy sun paseuve lmear relauunshlp yn m the relanamhp es hm punser lmear perhaps due in the varyxrg items m almhnl an dx 39exmtpeaple Here the leastesquues xegessmh hhe es gweh by y z n ma 1 e n D127whmh can he mtnpxeted asgwlrg themx eBACfathevanausamvuntsafheu Evaluaung thehnefm x shexs we ahtam ah Vang BAC m haul n n95 whuhxswe have the legal 11mm sh dam dunk and dnve me GALC zmrmcm value wuh y n1he mm almhal mutant was detexmnedhy the number m been whehusuw ma 1 e n 1121 as aprednlax The BAC es 2m determined by mhex femurs such shady wughL male an39emale etc Dr Neal Spring 2009 MATH 203 Population Proportion A special case of a population mean is the proportion p of those having a certain designation For example we may ask what proportion of the population approves of the President s performance Because p is a proportion it is always the case that 0 S p S 1 however p is often stated as a percentage If p 053 then we usually say that p is 53 However we always should work with p in decimal form When determining a proportion questions may be asked in YesNo form so that the responses are not numerical but instead are categorical Responses may also be of the form 1 Strongly Approve 2 Approve 3 Indifferent 4 Disapprove 5 Strongly Disapprove But in order to analyze the approval rate mathematically we must assign a numerical value to the categories that we are trying to measure In this case the responses Strongly Approve and Approve could be counted as one category Yes and the other responses are Not Yes We then assign the values 1 for Yes and 0 for Not Yes so that the responses actually become numerical The True Mean and True Standard Deviation 1 if Yes There are only two possible measurements x1 0 f If we average all possible 139 not 1 N Yes measurements over the population of size N then we obtain u F299 T p 139 l which is the true proportion that we are trying to measure Thus The variance is the average of the squares minus the square of the average But N l Y because x12 9 still 1 or 0 we have 02 32x12 7 2 Te 7 2 p pz il p1 p By taking the square root we obtain the true standard deviation Thus 0Jp1p A random measurement X from the population is then either 1 or 0 Such a measurement used to describe a population proportion is called a Bernoulli trial denoted by X b1 p mammam Wm 5Lhmmaxn 5 p n n 5 w methatthefunmma pip2 Isunnlan39m p s1 Whm p 1whexe mm uf the pvpulauan n um um every measureman m 1 and mmrm a hemusedme mm dewauun thn p u when mm a the pvpuhum n my than every measureman a gum mm mm dewauan SD n n The mmnmm mmquot dewauan amswhen p 12whmh Ewes a value a a s as Sampwpamammmwaysme caselhata s 0 ins Con dence men11f quot0 QO Weemmatedqevalue uf 1 Wu smplevapmum my 7 Rnponsas n whzxe n nth sample we and m nth number affavmahluexpmses Because p I unlmawn um 41 W 7p I alsaunknawn Butwe can 2mm by s W 7 m hynsuppex haund u u n5 qu mn dmce mtztvals m u I a Re mmmhm Mfmtmwm PW s Why mby U 5wehave plaung a by 1 L 750 m p 1 fm 153 pummm PN 7 cva 1 I raupupmmrsm N The mn d m Interval m a lag papulauan pmpamun can he fwund wnh the muun LPNlenl emu Item A mm the STATTESTS mznu n alwaysuses a z W 7 D Nest 5pm 2cm x thvesugttms asked zsn undergaduate students at a large uhwexsety haul p uh reasanahly sure that the pwpmuams ttxetsttwsthemg 195 u st E 1x Assuth large papmtuahhtstudmtswehtve tthns gurus shjhthh am i g emf W m 721nn5135 ZUWtUU Z Thusmstssspsusms WMSPSUW WthLhe mes mgh he p s tttetstn 113 thus we em dearly say that tths srhanl mhethth 2 uthe suaehtspuy Unng the budLm LPmpZ 1m feature mmthe STAT TESTS menu 3 quot 25h Level 95 catehtste Note yhthe semhd shhmah ahvve we used the waist case stamina ht a 5 a ahthh n 72 n ma Repmtexs arteh 9v thesmrmmuah as 73 wt students pray 39 hsd on s survey ast students the has s quothth warm ofmuxuf z pemuee hams Bangle z A Iandam survey a a sehaars ssnn Oink students fwund that 2H5 But a sun had ghhe awaxshp shunh temple ete ththe hst mahth Fmd a 93 mn dmce mtervalfar the true pxapaman ut the sehaars Greek wha had been a wmsh h the last mahth 1s thexe mhduswe evidence that the pwpmuams lessthan45967 2n5 shjhthh Fust p 5m an Thm ushg the Inn pvpulauun sue a N36EIEI we have uaJPO IU New 2326f 41x 59 3mm 7 1 7 7 I mm m 3599 IAIJIUMAX Thtus EI36252 SpSDAS748WhEhmeansLhatp sun mghtexseednts 2 326 x n 5 ann Note ththes case the mammum margm ht hm ts 31mm 3599 nmz1 Dr Neal Spring 2009 Choosing the Sample Size As with confidence intervals for the mean we may like to know in advance what sample size would provide a certain maximum margin of error e with a certain level of confidence r By using U 05 as a bound for o the required sample size n satisfies 05 2 202 X for large populations e 2 n2 NX za2gtlt05 e 052 for a population of size N z x N 1 e where zaZ is the appropriate ziscore depending on the level of confidence Note We always round up to the nearest integer Example 3 What sample size will guarantee a maximum margin of error of 0035 for any 99 confidence interval of a proportion What sample size would guarantee the result from a population of size 1200 Solution For a large population the required sample size must satisfy 05 2 2576 05 2 71 4M 135424 6 0035 thus n must be at least 1355 From a population of size 1200 the sample size must satisfy 2 z 2 X05 N 0 X e 7 1200gtlt 135424 63648 N 1za2 msjz 1199135424 e 712 thus n must be at least 637 Dr Neal Spring 2009 Practice Exercises 1 In a nationwide survey only 378 of 900 adults surveyed approved of the President s performance a Find a 95 confidence interval for the true proportion of adults who approved of the President s performance at the time of the survey b What is the maximum margin of error for a 95 confidence interval based on a survey of 900 people c Based on the results in Part b can you be relatively sure that less than a majority approve d If you wanted to estimate the true proportion within 0025 with 95 confidence then how many people must be surveyed Z A pollster wants to survey the 535 members of Congress on a completely none partisan issue The question will be Did you watch the Super Bowl The true proportion p will be based on a random sample of 60 members of Congress a Suppose 42 out of 60 said that they did watch the Super Bowl Find a 98 confidence interval with the largest margin of error for the true proportion of members of Congress who watched the Super Bowl Can you say with certainty whether or not at least twoithirds of Congress watched the Super Bowl b If you wanted to estimate the true proportion p within 003 with 98 confidence then how many members of Congress would you need to survey Dr Neal Spring 2009 Answers 378 z 1 196 V042 058 1 a Note that p 042 Then pm 243M 042 i 900 J J900 042 i 0032246 That is 0387754 S p S 0452246 which also can be found easily with the liPropZInt screen zuZ gtlt05 7196 gtlt05 b T W 00326 c Using the maximum possible margin of error we have p S 042 00326 lt 050 so we can be certain that less than a majority approved at the time of the survey 05 2 2 d 7 Z Za2 X J 196 X 0 5 153664 so sample 1537 8 0025 42 Z a f7 a 07 and then using N 535 and n 60 we have 2 gtlt05 N 2326 05 475 pupi m 07r X 07i01416 J71 N l J60 534 That is 05584 S p S 08416 which means that p might be below 23 2 z X 05 2 NX 112 J 535X2326X05 e 003 535x15028544 N1M2 5342326 X 052 53415028544 e 003 b n 2 39474 so sample 395 members of Congress nuasnnm MATH 203 The erieriltmlions rhetearaaaaansunaeasean as nnweaaanneanayasnneneann than y he human A tantrum s am an a amen new as he aegnes n aeeaan a when a s an huge heater than a eaaa n 1 Sun a and when unwary r e at name a tantrum a man by hung a a wane anesn w when r and s an the We mean and sanau aenaaanann madam sanue n a area nan a nnuyaenuta neasuenent Thrsusage was rst drsedyered by w s Gassetwhue ennp1dy1ng statusuua1 rnethgds as ahrewer at curnnessBrewery 1n Duhhn1re1and Arter studyrng under 1ltar11gtearsgn 1n Landau 1n 19m Gusset s 1n 1911 under the seuddnyrn 1 gaduated 1mm oarnhrrdge Unryersrty He spent the next 13 years deye1dgrng th rnathernatrea1 themy and rurther statusuea1 agpheatugns g1 Students Ldmnbuum Aswrth the standard nmma drstrrhuuan a Ldmnbuum deates a he11eshaped yeunatrssyrnrnetrreahdutunegmnt n Mmevvu as the degrees ht 1reeddrn n 1ndease the demmuums mnvexge m the standard nmma drstrrhutrdn z wrth the tar1s nadenrng and the rnrdd1e herght 1ndeasrng Theuneareuaueuryersgryentay x 9 m 1ara11x x2 2 7T where 3 1s a mnstant that depends an n Recall Wrtha standard narrna1 dmnbuuan 2 Mn 1 we have zrsmres M that are the hdundsthatgrye anrnner pmhahdny uf t 11 Fur examp1e 5 Hi the standard narrna1 drryerswrthrn 1195 Far X Mm 95 uf the measurements are wrthrn u 119641 nngenera1 erwrthrn u r a wAthxahahLmy t 1 Hawevzr wrth the demmuums we have a set ht tesmres 1hr my degree a 1reedgrn These trsmns arerurther nut than the resmres but they s1dw1y derrease a the zrsmxes asthe degrees uffnedammuease Gusset and Frsher sgent rnany years ea1dr1atung aedrrate tah1es1nr vanaus degees uf 39eedam and vanaus 1eye1s g1 mn dmce At rst these tah1es were used exdusrye1y at Gassefshxewny hut ndw they are standard 1n statusues tex hr the can he faund usrng a sgeda1 ea1eu1atgr pragram aw the tesmres are wrde1y used 1hr undrng thehdunds g1 mn denoe mtuvakfax the mean whm sarn 1ng anarma y dmnbuud rneasurernent Mdstrrngdrtanuy the rnethgdagphestd srnau sarnp1es Dr Neal Spring 2009 t SCOI ES 090 095 098 099 deg of fr 1 6314 12076 31821 63657 2 2920 4303 6965 9925 3 2353 3182 4541 5841 4 2132 2776 3747 4604 5 2015 2571 3365 4032 6 1943 2447 3143 3707 7 1895 2365 2998 3499 8 1860 2306 2896 3355 9 1833 2262 2821 3250 10 1812 2228 2764 3169 11 1796 2201 2718 3106 12 1782 2179 2681 3055 13 1771 2160 2650 3012 14 1761 2145 2624 2997 15 1753 2131 2602 2947 16 1746 2120 2583 2921 17 1740 2110 2567 2898 18 1734 2101 2552 2878 19 1729 2093 2539 2861 20 1725 2086 2528 2845 21 1721 2080 2518 2831 22 1717 2074 2508 2819 23 1714 2069 2500 2807 24 1711 2064 2492 2797 25 1708 2060 2485 2787 26 1706 2056 2479 2779 27 1706 2052 2473 2771 28 1701 2048 2467 2763 29 1699 2045 2462 2756 30 1697 2042 2457 2750 on 1645 1960 2326 2576 Dr Neal Spring 2009 The Sampling Distribution The redistributions can be applied to obtain a more accurate confidence interval for the mean u of a normally distributed measurement X when the true standard deviation 0 is C unknown Gosset discovered that the values E J follow a tn 1 distribution where n n is the sample size Most of the time with probability r l a the values Elfquot n will be within the interval tu2 to tuZ That is 4 95 l aZS l aZS l uZS l uZS t S St or S S or S S aZ S aZ J H x J x J H x J J Confidence Interval for 11 When sampling a normally distributed measurement we now can say that ltot2 S J2 Normal measurements any sample size n2 2 unit where n is the sample size and tel2 is the tescore from the tn 1 distribution This formula is analogous to the confidence interval for the mean of an arbitrary Za 2 o n measurement which was given by u m x i But now the tescore replaces the zescore and S replaces a o The result requires that we actually use S In other words we are not merely substituting S as an estimate for the unknown 0 We are supposed to use S in the formula 0 When sampling a normally distributed measurement the new result applies for all sample sizes n 2 2 not just for large sample sizes We now can work with small samples of a normally distributed measurement when a is unknown o The terms Eff are exactly tn 1 Thus when sampling a normally distributed n measurement the resulting confidence intervals are mathematically accurate Among all 95 confidence intervals there is exactly 095 probability that the true mean will be in an interval 0 When 0 is known go back and use the zescores p m x i z o a because in this n case 7c is exactly N p o J71 if sampling a normally distributed measurement D Neal 5pm 2cm name A pmducex as deemed we I emrg the fame xequued m pun sphsed whes apan Ideauy eaeh shvuld wAthtand aneast 2n paunds a pu fume and the xequned rams have a1ways hem Hun a he narma y dmnhuted Bexnw are a mndam sample m the xequued pun fumes m paunds Fmd a 99 cun denoe mtzrva r Lh e average requxredpu fame Is the average least 2n 1hs7 Snluunn The desemx mn deme can he faund useng the TInkerval feature Item x m the STATTESTS menu Enter the data mm a 1m than exenne the Tinker sneen after adjumrg the semngs 1 LB FE reeae vmwlsu Calculate Ta wmk the prahlnn by hand we hm must mmpme he sample mean 7 and he sampledmauan Upan anngsa we ahtam 7 25415 and s a 24925 wALh Lhm sample m sue m n 2n Fm a 99 Interval the zesmxe a the 209 shsmhuuan 1s 2351thus mzzsmtlsus lmy 2sus In gt zsxxsusnm Thus the avnage requued pu fame 1 mm 22 xx lbs in 21 M as and 1s oenamJy abavez lhs Note u we are gwen the summary stausaes n 7 and than we can use the TInkerval sneen semngthe hapm Statsndqexdqan Data Bangle 2 Axandamgaup afZ7hea1Lhy maJeshad amean sysmhemm pressure a 7 m 9 wALh a sample dewauan a s 93 ON a 95 mn dencz mtzrva fax the mean ynnhchlaadpxesxure annngan semuannaJes Snluunn wnh the TInkerval sneen we ablam a 95 mn dmce Interval wt 11122 m 531 sq heanhy males have a mean sysmhe mm pressure between 11122 and m 5x Dr Neal Spring 2009 Exercise Born on Dating A solution s pH level is the measurement of its acidity The concept of pH was first proposed by the Danish scientist SPL Sorenson in 1909 to measure the acidity of water in brewing beer A Brewmaster is studying the freshness of a beer by testing the acidity levels from a sample of bottles that have been sitting on a shelf for two months The acidity levels are known to be normally distributed and the ideal taste should have a pH level of 36 A level of 34 or lower is undrinkable bitter beer Here are the sample pH levels 335 346 349 338 337 340 354 350 296 340 339 334 a Find a 95 confidence interval for the true mean pH level of this brew after two months b Using the highest bound of your interval for the mean p and using S for the standard deviation what percentage of bottles still become bitter beer after two months That is compute PX S 34 for X Np S The pH levels for a sample of bottles that remained refrigerated for two months are as follows 350 353 352 359 351 349 354 352 349 352 355 350 345 355 c Find a 98 confidence interval for the true mean pH level of this brew after two months of refrigeration d Now use the worst case scenario with the lowest bound of your interval for the mean u Use S from this data set for the standard deviation Find the percentage of refrigerated bottles that become bitter beer after two months e What advice should the Brewmaster give to marketers regarding the shipping and storing of his brew Answer a 32881 S p S 34752 S 01472 b Using X N34752 01472 then PX S 34 m 03047 or about 3047 are bitter c 34947S p S 35424 S 00337 d Using X N34947 00337 then PX S 34 m 00025 or less than 1 are bitter

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