+
Chapter 8: Estimating with Confidence Section 8.3 Estimating a Population Mean To calculatea 95%confidenceintervalfor In Section 8.1,we estimatedthemean” “mystery The The valuecritical ChooseSRS ofan size constructinga confidenceintervalusing the samplemean The One conditionsare met, a level knownstandarddeviation estimate - Sample One ± z* - (critical value) (critical• (standarddeviationof statistic) Sample is found the from standardNormaldistribution. z n Interval for a Population Mean from a populationhavingunknownmean z x Interval for a Population Mean a Population for Interval σ C . As longas theandIndependent Normal z confidence interval for interval confidence * x n z * µ ( 240 240 230 , wetheuse familiarformula: n . . 79 79 . 99 µ , 250 (see page468)(see by 1 9 . . 96 8 . 59 µ ) 20 is 16 = 240.79. µ and
Estimating a Population Mean Population a Estimating + + Example – One Sample Z-Interval for a Population Mean A bottling machine is operating with a standard deviation of 0.12 ounce. Suppose that in an SRS of 36 bottles the machine inserted an average of 16.1 ounces into each bottle.
A) Estimate the mean number of ounces in all the bottles this machine fills Since x ,then x 16.1 + Example – One Sample Z-Interval for a Population Mean A bottling machine is operating with a standard deviation of 0.12 ounce. Suppose that in an SRS of 36 bottles the machine inserted an average of 16.1 ounces into each bottle.
B) Give an interval within which we are 95% certain that the mean lies. For samples of size 36, the sample means are approximately normally distributed with a standard deviation of 0.12 0.02 x n 36 We want to use * x z x 16.11.96(0.02) (16.0608,16.1392) • Set the expressionfor the marginof errorto less be than orequalto • Findvalue the critical • Get a reasonablevaluefor the populationstandard deviation To determinethe samplesize proportion. estimatinga meanin muchway thewe same didwhenestimatinga We determinea sample sizefor a desiredmarginof errorwhen The The marginof error mean Choosing the SampleSize andsolve for level earlieror pilotstudy. for a populationmeanwitha specified margin oferror Choosing Sample Size for a Desired Margin of Error When EstimatingWhen Error of Margin Desired a for Size Sample Choosing C µ . is n : ME z* of theinterval confidencefor the population from a standardfor Normalcurveconfidence n z 푛 * that willyielda level ≥ n 푧 푀퐸 ∗ 휎 2 C confidence interval confidence ME : σ from an ME µ
Estimating a Population Mean Population a Estimating + Researchers would like to estimate the mean cholesterol level level cholesterol mean the estimate wouldto like Researchers suggests that the standard deviation of cholesterol level 5mg/dl. is about level cholesterol of deviation standard the that suggests µ value of true the of (mg/dl) deciliter per within 1milligram to be estimate their like would They experiments. in laboratory used is often that monkey of variety Example:How ManyMonkeys? We willuse The critical valueconfidencefor 95% is 푛 at a 95% confidence level. level. confidence a 95% at Amonkey of involving this variety study previous ≥ n 1 . 96 96 σ 1 . 04 = 5bestguess asour forthe standarddeviation. ( 5 ) 2 monkeysto ensure the more than 1 mg/dl more 1mg/dl than at margin of erroris no We round up to 97 95% confidence. 95% z* = 1.96. = µ of a particular particular a of
Estimating a Population Mean Population a Estimating + + Checkpoint 1) At a certain plant, batteries are produced with a life expectancy that has a variance of 5.76 months squared. Suppose the mean life expectancy in an SRS of 64 batteries is 12.35 months. Find a 90% confidence interval estimate of life expectancy for all batteries produced at this plant. The standard deviation of the population is σ 5.76 2.4
2.4 So, the standard deviation of the sample mean is σ 0.3 x 64 * x z x 12.35 1.65(0.3) 12.35 0.4935 (11.8565,12.8435) + Checkpoint 2)To assess the accuracy of a laboratory scale, a standard weight known to weigh 10 grams is weighed repeatedly. The scale readings are Normally distributed with unknown mean (this mean is 10 grams if the scale has no bias). In previous studies, the standard deviation of the scale readings has been about 0.0002 gram. How many measurements must be averaged to get a margin of error of 0.0001 with 98% confidence? Show your work. 2.33 (0.0002) 2 n≥ 0.0001
n 21.7156 n 22
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Estimating a Population Mean Population a Estimating +
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Estimating a Population Mean Population a Estimating + has the anDraw SRSofsize When When perform we inference about a population mean write write the subtracting 1 from the sample size distribution, the appropriate degrees of freedom are found by The distributionis close toNormal. haveapproximatelya distributionwithmean t t distribution distribution Distributions; Distributions; of Freedom Degrees t distribution distribution with The n with t from alargepopulationthathas aNormal Distributions; Degrees of Freedom of Degrees Distributions; t µ n degreesof freedom – and standarddeviation 1 distributionas longas the sampling t n - s x x 1 degrees 1 degrees of freedom as n n , making df = df= n σ . The statistic – 1. The statistic will n - µ 1. We will using using a t n - 1 . t
Estimating a Population Mean Population a Estimating + distributionswithdifferent degreesof freedom. We can use When When comparing the density curves of the standard Normal distribution and The t Distributions; Distributions; of Freedom Degrees invT on the calculatorto determinecritical values t distributions, several facts are apparent: Normalcurvemore ever closely. densitycurve approachesthe standard the standardNormal. inand thein tails less the centerdoes than distribution. greaterthan thatof the standardNormal curve. areshapeto in similarthe standardNormal As the degreesof freedom increase,the The The spreadofthe The densitycurvesthe of t distributionshavemore probability t distributionsis a bit t t* distributions for t t
Estimating a Population Mean Population a Estimating + Suppose Suppose you want to construct a 95% confidence interval for the 12. What critical mean Using the calculator to Find Critical µ of a Normal population based on an SRS of size The desired critical value is t* should should you use? as 11. degrees of freedom 0.975 and the use the area as In the calculator, t t* * = 2.201. Values n = we
Estimating a Population Mean Population a Estimating + Suppose you wantto construct aconfidenceSuppose 95% 3) 2) 1) should you use? should on anSRSofsizebased thefor meaninterval Critical to Find Calculator Graphing Using invT Choose 4: Go to 2 The desired critical value is (0.975, 11) t* nd Values VARS invT µ (0.5 + C/2, of a Normal populationof a Normal n = 12. What critical t n * = 2.201. - 1) t*
Estimating a Population Mean Population a Estimating + + Checkpoint Find the critical value t * that you would use for a confidence interval for a population mean µ in each of the following situations.
1. A 98% confidence interval based on n = 22 observations. t* = 2.518
2. A 90% confidence interval from an SRS of 10 observations. t* = 1.833
3. A 95% confidence interval from a sample of size 7. t* = 2.447 To construct a confidence interval for Definition: freedom freedom in place of the
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Estimating a Population Mean Population a Estimating + (2) (1) Use thiswhen:only interval • where • Choose• SRS of an size Random Independent Normal is nothan 1/10 more the of populationsize. the populationtimes asis atleast largeasthe 10 sample. population,wecheckshould the the populationdistributionis Normal reasonablyaccuratewhenwesample withoutreplacement from afinite individualobservationsareindependent. To keepthe calculations ( confidenceofinterest interval orarandomizedfor experiment. The The n conditionsbeforeestimate weapopulation mean. populationproportion. As before,havewe toverify three important reasoningandcomputational detail to the one One ≥ 30). t* one is the critical valueforthe : The populationa Normaldistributionhas orthe islarge samplesize : The data come arandom from sampleof size - - Sample sample : The methodfor calculating a confidenceintervalassumesthat Conditions for Inference about a Population Mean Populationa about Inference for Conditions The OneThe t interval for a population intervalfor meana population n t - Sample from a populationhavingunknownmean Interval Interval for aPopulation Mean µ is t Interval for a Population Mean a Population for Interval x t n 10%condition – 1 t * distribution. or s the samplesize islarge( x n : verifythatthe samplesize : - sample is similar inis similarboth n from the population z intervalfor a µ n . A level ≥ 30), C
Estimating a Population Mean Population a Estimating + Construct and interpret a 90% confidence interval for the mean 328.1 342.6 338.8 340.1 374.6 336.1 257.4 317.5 327.4 264.7 307.7 310.0 343.3 269.5 297.0 269.6 283.3 304.8 280.4 233.5 A manufacturer of high of 20 screens from a day’ssingle production: process. Here are the tension readings from a random sample millivolts measured by an electrical device with output readings in mesh, and too wrinkles. little allow will The tension is surface of the viewing screen. Too much tension tear will the the tension on the mesh of fine wires that lies behind the Example:Video TensionScreen tension tension µ of the all screens produced on this day. (mV). (mV). Some variation inherent is in the production - resolution video terminals must control
Estimating a Population Mean Population a Estimating + STATE: population of all screens produced that day.that produced screens of all population the from screens of 20 sample random Random one PLAN: level. confidence 90% a at daythis produced tension Example:Video TensionScreen - sample sample If the conditions are met, we can use a use can we met, are If the conditions We want to estimate the true mean mean true the We to estimate want : We are told that the data come from a from Wecome the data that : told are µ of all the video terminals terminals video the of all t interval to estimate toestimate interval µ .
Estimating a Population Mean Population a Estimating + Normal: Since the sample size is small (n < 30), we must check whether it’s reasonable to believe that the population distribution is Normal. Examine the distribution of the sample data.
These graphs give no reason to doubt the Normality of the population
Independent: Because we are sampling without replacement, we must check the 10% condition: we must assume that at least 10(20) = 200 video terminals were produced this day. true mean tension in the entire batch of video terminals produced that day.that produced terminals videoof batch entirethe in tensionmean true CONCLUDE: deviation of the 20 screens in the sample are: DO: We want to estimate the true mean tension terminals produced this dayat a 90% confidence level. Example:Video TensionScreen Using Using our calculator, findwe that the mean and standard We are 90% confident that the interval from 292.32 to 320.32 mV captures the captures 320.32 mV to 292.32 from intervalthe that confidentWeare 90% Therefore, theconfidenceinterval 90%for From the calculator, to find thevalue. critical Since x x 306 n = 20, wetheuse t . * 32 s mV and x n 306 ( 306 292 we findwe . . 32 32 . 32 t , distribution withdistribution 320.32) 1 14 t* . 729 = 1.729.= s x 36 36 20 . 21 µ . 21 of all the video mV µ df is: = 19
Estimating a Population Mean Population a Estimating + terminals produced that day. that produced terminals true interval CONCLUDE: Example:Video TensionScreen mean tension in the entire batch of video of batch entire in the tension mean from 292.32 to 320.32 mV captures the mVcaptures 320.32 to 292.32 from We are 90% confident that the the that Weconfident 90% are
Estimating a Population Mean Population a Estimating + + Example – Sample Size for a Mean Ball bearings are manufactured by a process that results in a standard deviation in diameter of 0.025 inch. What sample size should be chosen if we wish to be 99% confident of knowing the diameter to within +0.01 inch?
0.025 x n n 0.025 2.576 0.01 2 n z 2.576(0.025) n n 0.01 error n 41.5, son 42 + Checkpoint
Biologists studying the healing of skin wounds measured the rate at which new cells closed a cut made in the skin of an anesthetized newt. Here are data from a random sample of 18 newts, measured in micrometers (millionths of a meter) per hour.
29 27 34 40 22 28 14 35 26 35 12 30 23 18 11 22 23 33
We want to estimate the mean healing rate µ with a 95% confidence interval.
1. Define the parameter of interest. Population mean healing rate. + Checkpoint 2. What inference method will you use? Check that the conditions for using this procedure are met.
One-sample t interval for µ.
Random: The description says that the newts were randomly chosen.
Normal: We do not know if the data are Normal and there are fewer than 30 observations, so we graph the data to check its shape. Create a histogram on the calculator to check.
Independent: We have data on 18 newts. There are more than 180 newts, so this condition is met. 3. Construct a 95% confidence interval for µ. Show your method. 8.32 25.67 2.110 18 (21.53,29.81)
4. Interpret your interval in context. We are 95% confident that the interval from 21.53 to 29.81 micrometers per hour captures the true mean healing time for newts. usingtheis violated. procedures involvedin theremain procedurefairlyaccuratewhena condition for An inferenceprocedureis called Definition: distributions when the population is not Normal. Larger samples improve the accuracy of critical values from the the population except when outliers or strong skewness are present. Fortunately, the The stated confidence level of a one they are affected by lack of Normality. the No population of real data is exactly Normal. The usefulness of exactly correct when the population distribution is exactly Normal. Using t procedures procedures in practice therefore depends on how strongly t Procedures Wisely Procedures t procedures procedures are quite robust against non robust if the probabilitycalculations - sample sample t interval for - µ Normality of is t
Estimating a Population Mean Population a Estimating + • • • Largesamples Samplesizeat least 15 Samplesizeless15 than distributionswhenthe sampleis large,roughly presenceof outliersor strong skewness. skewedor ifoutliersare present,notuse do Normal(roughlysinglesymmetric,no peak, If outliers).the data are clearly Except in the case of small samples, the condition that the data when when performing inference about a population mean. Normal. Here are practical guidelines for the Normal condition important than the condition that the population distribution is come from a random sample or randomized experiment is more Using Using OneUsing t : The Procedures Wisely Procedures t - procedurescan beevenfor used clearlyskewed Sample : The : Use : t t proceduresbecanexcept usedin the t Procedures: The Normal Condition Normal The Procedures: proceduresif the data appearcloseto t . n ≥ 30.
Estimating a Population Mean Population a Estimating + + Checkpoint 1) You construct three 88% confidence intervals as follows:
A) A t-interval with 6 degrees of freedom.
B) A t-interval with 2 degrees of freedom
C) A z-interval
Assuming the mean and standard deviation are the same for all three intervals, (A, B, and C) arrange the intervals in order, from narrowest to widest. C is more narrow than A, A is more narrow than B.
For t-intervals, critical values for any specific confidence level decrease as the degrees of freedom increases.
A z-interval is equivalent to a t-interval with “infinite” degrees of freedom. + Checkpoint 2) You are sampling from a population with a known standard deviation of 20 and want to construct a 95% confidence interval with a margin of error of no more than 4. What is the smallest sample that will produce such an interval?
We want a sample size n such that 20 (1.96) 4 n Solving as an inequality produces n = 96.04, so n should be 97. + Section 8.3 Estimating a Population Mean
Summary In this section, we learned that…
Confidence intervals for the mean µ of a Normal population are based on the sample mean of an SRS.
If we somehow know σ, we use the z critical value and the standard Normal distribution to help calculate confidence intervals.
The sample size needed to obtain a confidence interval with approximate margin of error ME for a population mean involves solving z * ME n In practice, we usually don’t know σ. Replace the standard deviation of the sampling distribution with the standard error and use the t distribution with n – 1 degrees of freedom (df). + Section 8.3 Estimating a Population Mean
Summary There is a t distribution for every positive degrees of freedom. All are symmetric distributions similar in shape to the standard Normal distribution. The t distribution approaches the standard Normal distribution as the number of degrees of freedom increases.
A level C confidence interval for the mean µ is given by the one-sample t interval s x t * x n
This inference procedure is approximately correct when these conditions are met: Random, Normal, Independent. The t procedures are relatively robust when the population is non-Normal, especially for larger sample sizes. The t procedures are not robust against outliers, however.