STA 291 - Lecture 21 1STA 291Lecture 21All confidence intervals we learned here is of the formPoint estimator error boundInterchangeable wording: Error bound = margin of error±• If everything else held unchanged;increase confidence level à larger error bound • If everything else held unchanged;increase sample size n à smaller error boundSTA 291 - Lecture 21 2Planning on the sample size n• Usually we first fix a confidence level, e.g. 95%.• Then we would “trial and error” with different sample size n and see how small/large the error bound would be.STA 291 - Lecture 21 3Example• For 95% confidence intervals on a proportion p• If n = 1500 à error bound = 0.02530or 2.53%• If n = 1000 à error bound=0.03099or 3.10%STA 291 - Lecture 21 4• If n = 700 à error bound = 0.03704or 3.7%• If n = 500 à error bound = 0.0438or 4.38%Etc. etc. The formula I used is STA 291 - Lecture 21 5• Error bound =• If there is no reliable information on p, we can use the conservative value: p= 0.5 (the answer is not very sensitive to the change in the value of p )STA 291 - Lecture 21 60.5(10.5)1.96n−⋅Choice of sample size• Margin of error = error bound = BSTA 291 - Lecture 21 7Choice of sample size• In order to achieve a margin of error B, (with confidence level 95%), how large the sample size n must we get?• For the confidence interval of population mean, mu, the formula isSTA 291 - Lecture 21 8STA 291 - Lecture 21 9Choice of Sample SizeσXzXBnσ±⋅=±• So far, we have calculated confidence intervals starting with z, n and (plus, a possible t adjustment)• These three numbers determine the error bound B of the confidence interval • Now we reverse the equation: • We specify a desired error bound B• Given z and , we can find the minimal sample size n needed for achieve this.σSTA 291 - Lecture 21 10Choice of Sample SizezBnσ⋅=• From last page, we have• Mathematically, we need to solve the above equation for n• The result is• B must be in decimal form22 znBσ=⋅STA 291 - Lecture 21 11Example • About how large a sample would have been adequate if we merely needed to estimate the mean to within 0.5 unit, with 95% confidence?• (assume this may come from a pilot study)• B=0.5, z=1.96• Plug into the formula: 5σ=221.965 =384.160.5n=⋅• In reality, is usually replaced by s andWe need to replace z by t (with t-table).For example, if the number 5 is actually s, not thenSTA 291 - Lecture 21 12σσ221.9845 =393.620.5n=⋅• I want to stress that these are somewhat approximate calculations, as they rely on the pilot information about either or p, which may or may not be very reliable.• But it is much better than no planningSTA 291 - Lecture 21 13σChoice of sample size• The most lazy way to do it is to guess a sample size n and • Compute B, if B not small enough, then increase n;• If B too small, then you may decrease nSTA 291 - Lecture 21 14• For the confidence interval for p• Often, we need to put in a rough guess of p (called pilot value). Or, conservatively put p=0.5STA 291 - Lecture 21 15(1)ppzBn−⋅=• Suppose we want a 95%confidence error bound B=3% (margin of error + - 3%). Suppose we do not have a pilot p value, so use p = 0.5So, n = 0.5(1-0.5) [ 1.96/0.03]^2=1067.11STA 291 - Lecture 21 16Example 1(from last lecture):• Smokers try to quit smoking with Nicotine Patch or Zyban.• Placebo 160 subjects, 30 quit• Patch: 244 subjects, 52 quit • Zyban: 244 subjects, 85 quit• Zyban+patch: 245 subjects, 95 quit• Find the 95% confidence intervals for p: the success rate/proportionSTA 291 - Lecture 21 1795% confidence intervals for p• Placebo: [0.13, 0.25]• Patch: [0.16, 0.26]• Zyban: [0.29, 0.41]• Zyban+patch: [0.33, 0.44]STA 291 - Lecture 21 18Example 2• To test a new, high-tech swimming gear, a swimmer is asked to swim twice a day, one with the new gear, one with the old.• The difference in time is recorded: Time(new) – time(old) = -0.08, -0.1, 0.02, …. -0.004. There were a total of 21 such differences.Q: is there a difference?STA 291 - Lecture 21 19• First: we recognize this is a problem with mean mu.• And we compute the average X bar= -0.07• SD = 0.02• 90% confidence interval is:STA 291 - Lecture 21 20Plug-in the values into formulaSTA 291 - Lecture 21 210.020.02?? and ??2121XX−+0.020.020.07?? and 0.07??2121−−−+• What is the ?? Value.• It would be 1.645 if we knew sigma, the population SD. But we do not, we only know the sample SD. So we need T-adjustment.• Df= 21 -1 = 20• ??=1.725STA 291 - Lecture 21 22STA 291 - Lecture 21 23Example 3: Confidence Interval• Example: Find and interpret the 95% confidence interval for the population mean, if the sample mean is 70 and the pop. standard deviation is 12, based on a sample of size n = 100First we compute =12/10= 1.2 , 1.96x 1.2=2.352[ 70 – 2.352, 70 + 2.352 ] = [ 67.648, 72.352] nσSTA 291 - Lecture 21 24Example: Confidence Interval• Now suppose the pop. standard deviation is unknown (often the case). Based on a sample of size n = 100 , Suppose we also compute the s = 12.6 (in addition to sample mean = 70)First we compute =12.6/10= 1.26 , From t-table 1.984 x 1.26 = 2.4998[ 70 – 2.4998, 70 + 2.4998 ] = [ 67.5002, 72.4998] snSTA 291 - Lecture 21 25Error Probability• The error probability (a) is the probability that a confidence interval does not contain the population parameter -- (missing the target)• For a 95% confidence interval, the error probability a=0.05• a = 1 - confidence level orconfidence level = 1 – aSTA 291 - Lecture 21 26Different Confidence LevelsConfidence levelError a a/2 z90% 0.195% 0.05 0.025 1.9698% 0.02 0.01 2.3399% 2.57599.74% 386.64% 0.1336 0.0668 1.5STA 291 - Lecture 21 27Attendance Survey Question • On a 4”x6” index card–Please write down your name and section number–Today’s Question: Are you going to watch the NCAA Basketball game this weekend?a. All b. Somec. NoneSTA 291 - Lecture 21 28Facts About Confidence Intervals I• The width of a confidence interval – Increases as the confidence level increases–