Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION 10-1 Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION As has been mentioned repeatedly, we use sample information to estimate the unknown information about the population we are interested in. EXAMPLES 1. Rats raised in an enriched environment appear to have larger cortexes than those raised in unenriched environments. What is the average cortex weight, µ, for the population of cortex weights for every rat which could be raised in an enriched environment? We’ll grow a sample of rats in the enriched environment and use our sample data to estimate µ. 2. Fiddler crabs are known as such because they have asymmetrically sized pincers. Like human handedness, it is estimated that approximately 10% of fiddler crabs are “left-pincer”, i.e. the left pincer is larger than the right. What is the true proportion, π, of left-pincered crabs on a remote island in the South Pacific? We’ll fly to the S. Pacific, take a random sample of crabs and use the sample proportion p to estimate π. Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION 10-2 Several ways we could report the results: We could report a POINT ESTIMATE, which is a single number from the sample. It has little context or meaning unless additional information is provided. For example, when reporting the sample mean it usually is better to also report something about its sampling variability, such as including the sample size n and the sample standard deviation s. So, other possibilities for reporting: 1. Report x and s. Not useful since s is not the standard deviation of x 2. Report x and σ x =σ n . Usually can’t because we don’t know the value of σ 3. Report x and SEM = s n . • SEM is STANDARD ERROR OF THE MEAN • SEM is an unbiased estimate of σ x (under random sampling) Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION 10-3 • one common report is to list x ± SEM but it’s not the best choice since it isn’t directly interpretable as an interval estimate We would like to report a range or interval of plausible values for the population parameter we’re estimating. Such a report is called an INTERVAL ESTIMATE of the population parameter. When done right, the interval estimator includes our point estimate and an estimate of the accuracy of the point estimator and some measure of our “comfort” with the estimate we are providing. That is, we would like to be somewhat confident that the interval we report actually covers the true value of the parameter we are trying to estimate. Defn: A CONFIDENCE INTERVAL for a population characteristic is an interval of plausible values for that parameter. It is constructed so that the true value of the parameter is captured inside the interval with a chosen, specified level of confidence. EXAMPLE Political polls are almost always reported as follows: “The proportion of voters who would vote for Gore today is 48% ± 4% with 95% confidence”. What that Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION 10-4 means is that based on their sample, the true proportion π of voters who would choose Gore is somewhere inside the interval 44% and 52%. The 95% confidence refers to the probability that the interval captures π. Defn: The CONFIDENCE LEVEL associated with a confidence interval is the probability that the interval estimate covers the true value of the parameter. One way of thinking of it is as the success rate of the method we are using to do the estimation! The method includes the choice of point estimator, the assumptions about the distribution of that estimator, and the sampling design. The confidence level is chosen by the researcher doing the reporting. Common levels in the life sciences are 90%, 95% and 99%. In social sciences you sometimes see 85% as well. When intervals are constructed in a way that uses knowledge of the sampling distributions of the point estimators, we can assign these probabilities and not just guess what they might be. Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION • 10-5 π • • • • e.g. 68% of all sample proportions are within the interval above. If we put a confidence interval of the same width around each estimate shown above, what percent of the new intervals would include π ? Important Point: the confidence level is not a probability about a particular interval including the true value. The interval either does or does not cover the true value (but you’ll never know that). The confidence level is the probability that the method you used will give an interval that includes the true value!!! Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION 10-6 1) INTERVAL ESTIMATION OF THE POPULATION PROPORTION π How does the Gallup polling company calculate the 95% confidence interval they report? They take a sample of size n and calculate the sample proportion of successes p (in this case people who indicate they would vote right now for Gore). Then they use: A LARGE SAMPLE 95% CONFIDENCE INTERVAL FOR THE POPULATION PROPORTION π is p ± 1.96 i.e. p(1 − p ) n ⎛ p(1 − p ) ⎜⎜ p − 1.96 , n ⎝ p(1 − p ) ⎞ ⎟⎟ p + 1.96 n ⎠ • Large-sample refers to a sample size that is sufficiently large to invoke the Central Limit Theorem. • p is the point estimator of π so is at the center of the interval Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION • 10-7 p(1 − p ) is the estimator of σ p - we use this n because it is the estimate of how much sampling variability there is in p • 1.96 is the z-score, z*, that makes the following statement true: 0.95 = Pr(- z* < Z < + z*). We use this because we are using the CLT which states that sample proportions are normally distributed for large samples The formula is easily adapted for other confidence levels. Simply replace 1.96 with the appropriate number from the table below. The z critical values for common confidence levels are: Confidence Level 80% 90% 95% 98% 99% 99.9% Z critical values 1.28 1.645 1.96 2.33 2.58 3.29 Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION 10-8 EXAMPLE Our researcher did fly to the South Pacific and collected 150 crabs. For each crab she recorded whether the left or right pincer was dominant and observed that 20 crabs were leftpincered. Calculate a 90% C.I. to estimate the true proportion of left-pincered crabs on the island. 1) Is the sample size large enough to use our method? Need to know if nπ and n(1 − π ) ≥ 10 but we don’t know π. Instead look to see that the number of successes and the number of failures in the sample both are greater than 10. Here she saw 20 successes and 130 failures So, yes p = 20 2) = 0.133, 150 .133(1 − .133 ) p(1 − p ) = = .0277 150 n 3) 90% Confidence Î z* = 1.645 4) The 90% C.I. then is Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION 10-9 p(1 − p ) = .133 ± 1.645(0.0277 ) n = .133 ± .0456 = (0.0874, 0.1786 ) p ± 1.645 The true proportion of Fiddler crabs on this island lies, with 90% confidence, between 8.4% and 17.9%. We interpret it as follows: Since this interval includes 10% there is no evidence that the true proportion on this island is any different than other populations. a) What if we had calculated a 95% C.I.? Would it be wider or shorter than the 90% C.I.? Wider since 1.96 > 1.645, i.e. 1.96(0.0277) = 0.0543 > 1.645(0.0277) = 0.0456 b) What if she had seen 13.3% based on a sample of 300 crabs? Would the 90% C.I. be wider or shorter than the one based on 150 crabs? Shorter because 1.645 .13(1 − .13 ) .13(1 − .13 ) = .0323 < 1.645 = .0456 300 150 (this interval does not include 10% Î based on a sample of 300 this island is different!!!) Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION 10 10- Defn: Confidence intervals can be written in the form point estimate ± MARGIN OF ERROR where the margin of error (ME) is the product of the critical value and the standard deviation of the point estimate. Suppose the scientist is still planning the fiddler crab experiment and wants to calculate a 95% confidence interval with a margin of error of no more than 2.5%. How big a sample size should she take? p(1 − p ) = 0.025 Margin of Error (ME) = 1.96 n Don’t have an estimate for p yet since she hasn’t sampled yet. She does have a guess though (10%) so we’ll use that. .1(1 − .1) = 0.025 for n. Now we need to solve 1.96 n 2 1 . 96 ⎛ ⎞ Get ⎜ ⎟ 0.1(1 − 0.1) = n = 553.19 ≈ 554 crabs ⎝ .025 ⎠ are needed. Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION 11 10- General equation to estimate the needed sampled size for a specified margin of error (ME) when estimating a population proportion is: ⎛ z ⎞ ~ ~ ⎟ n = π (1 − π )⎜⎜ ⎟ ME ⎠ ⎝ * 2 where • π~ is a guess as to the likely true proportion in the population (if completely unsure use 0.5) • z* is the z-score critical value for the confidence level desired • ME is the desired margin of error (in decimals) Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION 12 10- 2) INTERVAL ESTIMATION OF THE POPULATION MEAN µ Recall that we have a CLT for the sample mean so that if a sample is large enough the sample mean x is Normally distributed with a mean µ x = µ And a standard deviation σ x = σ . n ⇒ We should be able to construct a confidence interval estimate of the population mean µ in a way very similar to what we did for the population proportion: i.e. use point estimator ± z × std. dev. of estimator σ K for a 95% C.I. for µ. i.e. use x ± 1.96 n Problem? Almost never know σ either. Since s is an unbiased estimator of σ, can we use s K instead? x ± 1.96 n Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION 13 10- Sort of. If n is really large (> 120, say) then yes. Other wise we need to use a slightly different C.I. method. Use of z-scores in the above Confidence Intervals is x − µx x−µ = has a based on the fact that z = σx σ n standard normal distribution when x is normally distributed. x − µx x − µ . In fact to = SEM s n distinguish it from z-scores, it is called a T-SCORE. T-scores have what is known as a Student’s T-distribution with (n-1) degrees of freedom. . That isn’t true for t = Defn: STUDENT’S T-DISTRIBUTION 1) is unimodal and symmetric, with equal-sized tails. 2) has a mean of 0 3) has a standard deviation that depends on the sample size, n, used to calculate x and s. Hence every sample size has a different t-distribution which is denoted as t(n-1). n-1 are called the DEGREES OF FREEDOM (df) for the distribution. Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION 14 10- T-distributions look like a standard normal distribution but are more variable, i.e. slightly wider and flatter. How wider and flatter depends on the sample size involved. As n increases, the shape becomes more and more like a Normal distribution. x−µ is more variable than z = x − µ but s n σ n how much more depends on the sample size. For large sample sizes they have almost the identical frequency distributions! i.e. t = T(3 df) T(15 df) N(0,1) -1.96 -2.13 -3.18 +1.96 +2.13 +3.18 Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION 15 10- So, C.I.s for estimating the population mean µ use the t-score for (n-1) degrees of freedom in place of the z-score: CONFIDENCE INTERVALS FOR THE POPULATION MEAN µ have the general formula ⎛ s ⎞ x ± ( t critical value)⎜ ⎟ ⎝ n⎠ where the t critical value is based on n-1 df. This interval is appropriate under the following: 1) sampling is random, and 2) either a) the sample size is large so we can use the CLT or b) the original population has a frequency distribution that is bell-curve shaped How do we find the t critical values that we need? Once we do that, how do we use this equation? Since there are different values for t-score for each sample size we need an entire table to list them for Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION 16 10- some typical confidence levels. (Recall that before I could list the 6 commonly used z values. Now I need 6 numbers for each and every sample size!). The table is at the end of this topic. EXAMPLE Suppose we wish to determine if the average cortex weight for rats is larger than usual when the rats are raised in an enriched environment. Assign 10 randomly selected newborn rats to an enriched environment and raise them to adult stage. The cortexes had an average weight of x = 565 mg and a standard deviation of s = 170 mg. Calculate a 90% C. I. for the true mean cortex weight of rat raised in enriched environments. 1) since n is small we should check to see if the frequency distribution of the population sampled is a bell curve. • Can’t since we didn’t raise the entire population! • Can we use the 10 sample values, i.e. do a boxplot and make sure there are no extreme outliers or obvious skew (hard with 10 numbers)? Since we weren’t given the raw data we can’t do this either. Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION 17 10- • It is not unreasonable to assume that weight is normally distributed or at least approximately so. 2) n=10 so n -1 = 9 90% confidence level + 9 df gives t-score = 1.83 (note that it’s larger than z=1.645) ⎛ s ⎞ ⎛ 170 ⎞ 565 1 . 83 = ± ⎟ = 565 ± 98.4 ⎟ ⎜ ⎝ n⎠ ⎝ 10 ⎠ 3) x ± t ⎜ “We are 90% confident that the average cortex weight of rats which are raised in enriched environments has a value within the interval 467 mg and 663 mg.” Suppose the researcher had been interested in comparing enriched environments to deprived ones. So, a sample of 10 rats were placed in a deprived environment but otherwise treated identically. At the end of the experiment the calculated 90% C.I. for the deprived rats was 389 mg to 460 mg. Is this sufficient evidence to argue that an enriched environment increases cortex weights? Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION 18 10- Sort of. The comparison is valid BUT what is your confidence level now? Not possible to know exactly when you do this type of comparison using 2 confidence intervals. (it’s actually around 82% not 90%) We’ll learn later a better method for comparing 2 populations which will also give us a way of determining our confidence in the method. EXAMPLE As part of a larger experiment on the use of exercise to reduce stress and susceptibility to illness, researchers gathered baseline data on the levels of Human Beta-Endorphin (HBE) in sedentary people. Higher levels of HBE are associated with higher stress levels. They randomly selected 25 people getting yearly physicals at a nearby hospital and measured HBE levels (pg/ml). The sample mean level was 42.5 and a standard deviation of 16.8 pg/ml. Construct a 95% CI of the population mean HBE level. Population: HBE levels in all sedentary people? Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION 19 10- Sample: HBE levels in 11 people at a nearby hospital Can we use the equation we learned? Don’t know the distribution of the original population but we will assume that it is not too skewed. Then n=25 should be larger enough for the CLT to hold. Calculations: For n − 1 = 24 and 95% confidence: t = 2.06 ⎛ s ⎞ ⎛ 16.8 ⎞ Then x ± t ⎜ ⎟ = 42.5 ± 6.8 ⎟ = 42.5 ± 2.06⎜ ⎝ n⎠ ⎝ 25 ⎠ Conclusion: “We are 95% confident that the average HBE level in sedentary people is 42.8±6.8 pg/ml.” QUESTION: Suppose I had decided that I could report an 80% confidence interval rather than a 95% interval. A) which interval would be wider? B) which interval is more likely to include the true mean that I am trying to estimate? Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION 20 10- Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION 21 10- Example of Mis-Statements and Some Computations Using a N(0,1) Random Variable Distributions N(0,1) -3 -2 -1 Quantiles 100.0% maximum 99.5% 97.5% 90.0% 75.0% quartile 50.0% median 25.0% quartile 10.0% 2.5% 0.5% 0.0% minimum 0 3.635 2.563 1.949 1.298 0.677 0.012 -0.678 -1.297 -1.976 -2.506 -3.447 1 2 3 Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION 22 10- Take a sample of size 15 from this population and calculate a 95% CI using the method we just learned: -0.3272061, -2.3610111, -0.7944772, -1.6243058, 1.34490057, -0.6010952, -0.8823261, 0.50218859, 0.29919709, -0.5588271, 0.36830855, 0.46442333, -0.3233998, 0.5258819, 0.28472816 -2 -1 Moments Mean Std Dev Std Err Mean upper 95% Mean lower 95% Mean n 0 1 -0.245535 0.9435046 0.2436118 0.2769608 -0.76803 15 Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION 23 10- So, the 95% confidence interval is (-0.768, +0.277). Correctly Stated Interpretation: 1) We are 95% confident that the population mean, µ, has a value that is between -0.768 and +0.277. 2) There’s a 5% chance that our interval will not cover the population mean, µ. It’s the probability that the method results in a CI that covers the true parameter value (µ) that is being addressed here, not the particular sample NOR the population values. For example, suppose we were able to take 200 samples of 15 values from the population. For each one we calculate x , s , and a 95% CI. Now, since we know a N(0,1) random variable has a mean of µ = 0 , we can determine whether the CI covers the value of 0 or not. Using JMP, I simulated exactly the repeated sampling and got 9 out of the 200 (4.5%) did not cover µ = 0 . The next table shows some of the results. Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION 24 10- Sample Mean Sample Standard Deviation 95% CI Lower Bound 95% CI Upper Bound CI Cover µ=0? -1.1049107 0.84677211 -2.1310767 1.24204895 1.79642982 2.10221624 -1.1471177 1.08150832 -1.1533823 -0.3213116 -1.5549539 0.36021018 1.25636375 -1.9770196 0.54086785 -1.5792089 2.1220965 -0.8529584 0.41356674 0.55320446 2.16646888 1.35168506 -0.7047083 -0.0917601 0.32999897 0.51051004 0.21348212 -1.3225274 -0.4963301 0.25010729 -1.2455287 -0.5720726 2.26894349 0.31171105 -0.2931029 0.25012456 -0.7918701 -0.0735042 0.85730542 0.5527533 0.42753442 0.33801653 0.78304642 0.3464639 0.75434897 0.16443923 0.2669313 0.36706328 0.35510663 1.24317913 2.0860707 0.99486637 1.03681304 1.36476692 1.02076565 1.17371463 1.13639355 0.91286459 2.56664539 1.26726012 1.23548621 1.2013229 0.97791792 1.59662907 1.29138324 0.41676529 1.01401071 1.28280408 1.29482828 0.97027273 2.53712976 0.98406263 1.47509377 2.02997874 0.46674714 1.75337152 1.30972571 1.94757715 1.16635039 0.57662726 -2.0218808 0.12179875 -3.8105442 0.49895779 0.17851219 1.74952917 -1.7196284 0.29423588 -1.9150103 -2.9876656 -6.0291305 -1.773566 -0.9673791 -4.9041535 -1.6484567 -4.0965765 -0.3152253 -2.8108583 -5.0913401 -2.1647982 -0.4833855 -1.2248963 -2.8021336 -3.5161889 -2.4397426 -0.3833626 -1.9613545 -4.0738685 -3.2734606 -1.8309207 -6.6871308 -2.682677 -0.894818 -4.0421603 -1.294176 -3.5104833 -3.6009524 -4.2506417 -1.6442674 -0.6839892 -0.1879405 1.57174548 -0.4516092 1.98514011 3.41434745 2.4549033 -0.574607 1.86878077 -0.3917544 2.34504248 2.91922279 2.49398633 3.48010655 0.95011428 2.73019243 0.93815856 4.55941826 1.10494137 5.91847361 3.2712071 4.81632325 3.92826642 1.39271708 3.33266868 3.09974055 1.40438269 2.38831878 1.42881374 2.2808003 2.33113532 4.19607344 1.53853185 5.43270497 4.66558242 0.70797014 4.01073245 2.01721219 4.10363339 3.35887822 1.78949576 No No No No No No No No No Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION 25 0.54496347 0.26360623 -1.7795686 1.76889744 1.36987175 0.15408404 0.61575948 1.75964955 -0.032808 0.29596459 0.71309248 0.14647825 -1.5130959 0.25016094 0.34062799 0.38727695 -0.6162605 0.7944431 0.00054833 -1.2182523 -0.3770921 -0.3548987 0.18431648 0.17146672 2.26729026 0.87556293 1.11354895 -1.2177139 0.24642509 0.2851893 1.26836219 -1.0806759 -1.0910837 -0.7211335 0.19407641 0.28889596 -0.1802052 -0.6870411 -0.2942918 -0.7288637 -1.3148595 1.29699205 -0.5270651 0.09414593 2.34976751 1.48932661 1.00724524 1.39430358 2.02759735 1.26877604 0.85684775 1.81143194 1.0357546 0.75204033 0.69665007 0.7872054 1.04700543 1.02961537 1.21204694 1.05993052 2.27295095 0.77982765 1.61345228 2.40089814 0.66754035 1.14986562 2.13065507 1.01970509 1.15701845 1.69744134 1.98171209 1.61473751 1.06505164 1.53116947 1.20047293 2.21306374 2.19968934 1.28751128 0.88624608 1.25641273 1.35738017 1.78051269 1.42903957 1.34804592 0.7970715 1.37794775 1.48434652 1.73039274 -4.4947866 -2.9306817 -3.9398947 -1.2215863 -2.9788921 -2.5671699 -1.2219962 -2.1254856 -2.2542807 -1.3170015 -0.7810733 -1.5419094 -3.7586992 -1.9581444 -2.2589541 -1.8860479 -5.4912554 -0.8781209 -3.4599626 -6.3676666 -1.8088238 -2.8211151 -4.3854841 -2.0155832 -0.2142675 -2.7650867 -3.1368008 -4.6809814 -2.0378835 -2.9988426 -1.3063962 -5.8272255 -5.8089481 -3.4825706 -1.7067324 -2.4058413 -3.0914961 -4.505861 -3.3592768 -3.6201346 -3.0244079 -1.6584119 -3.7106718 -3.6171774 5.58471354 3.45789412 0.38075764 4.75938119 5.71863555 2.875338 2.45351513 5.64478466 2.18866463 1.90893068 2.20725829 1.83486592 0.73250742 2.45846629 2.94021013 2.66060182 4.25873447 2.46700707 3.46105932 3.93116209 1.05463954 2.1113178 4.7541171 2.35851664 4.74884803 4.51621252 5.36389866 2.24555357 2.53073368 3.56922119 3.84312054 3.66587375 3.62678073 2.04030352 2.0948852 2.98363327 2.73108572 3.13177879 2.77069327 2.16240729 0.39468881 4.25239604 2.65654154 3.80546925 Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes 10- Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION 26 1.14444656 … … 1.65451125 -0.1351413 -1.8632532 -0.5569952 -2.2413172 -0.0982971 0.36542458 -0.3203501 -1.2338743 0.40171205 -0.8315755 0.97514472 0.16588811 -0.5426113 0.00646725 1.18488525 1.10724653 1.30641932 2.00831787 0.88297557 1.75641268 1.60158032 2.2890512 2.11831252 1.29986472 1.42429477 0.66079362 0.72570411 0.81747025 0.75995975 1.27561648 0.72405208 0.80901499 -1.3968796 -3.4763009 -3.2327707 -2.6529022 -2.0289355 -5.6303838 -3.9920433 -7.1508438 -4.6416256 -2.422508 -3.3751586 -2.6511356 -1.1547685 -2.5848748 -0.6548068 -2.5700371 -2.0955485 -1.7286973 3.68577266 1.27331435 2.37121085 5.96192468 1.75865294 1.90387729 2.878053 2.66820932 4.44503137 3.15335712 2.73445833 0.18338709 1.95819256 0.92172379 2.60509628 2.90181336 1.01032599 1.74163183 10- Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Must be very careful that you state the interpretation correctly. Mis-statement #1: we are confident that 95% of the data fall between -0.768 and +0.277. Here, only 4 out of 15 (26.67%) fall within that interval! -0.3272061, -2.3610111, -0.7944772, -1.6243058, 1.34490057, -0.6010952, -0.8823261, 0.50218859, 0.29919709, -0.5588271, 0.36830855, 0.46442333, -0.3233998, 0.5258819, 0.28472816 Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION 27 10- Mis-statement #2: we are confident that 95% of the population falls between -0.768 and +0.277. Let’s calculate the Pr(-0.768 < Z < +0.277), i.e. the probability that a randomly selected value falls in the interval (or equivalently the proportion of the population values falling within the interval). Pr(-0.768 < Z < +0.277) = Pr(Z < +0.277) - Pr(Z < -0.768) = 0.6091 - 0.2212 = 0.3879 Here, only 39% of the population values fall between -0.768 and +0.277.