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InferenceaboutMean (σUnknown) Whenσisknown,thesamplingdistribu>onforasample meanisnormalifcondi>onsaresa>sfied. Formanyyears,itwasthoughtthatwhenσwas unknown,thiswass>llthecase.However,becauseof theincreasedvariabilityintroducedbynotknowingσ, thesamplingdistribu>onforasamplemeanwith unknownσisnotnormal. ThiswasdiscoveredbyW.S.Gosset,anIrishBrewery qualitycontrolinspectorin1908.Healsodiscovered thatwhenσisunknown,wecans>lldoinferenceusinga samplingdistribu>onmodelhediscovered,thetdistribu>on. Sec>on9.1,Page184 1 t-Distribu>on • Thet-Distribu>onisunimodalandsymmetric • IthasmoreareainthattailsandisflaSerinthe middlethanthenormaldistribu>on • Itisafamilyofdistribu>ons,withadifferentcurve foreachμ,σ,anddf(degreesoffreedom)df=n-1, wherenisthesamplesize. Cri$calValuesfor95%ConfidenceInterval Knownσ:1.960fromNormalDistribu>on Unknownσwith3df:3.18fromt-Distribu>on Unknownσwith10df:2.23fromt-Distribu>on Unknownσwith1000df:1.962fromt-Distribu>on Sec>on9.1,Page184 2 t-Distribu>on WhentoUseIt Is the population parameter of interest a mean? Yes Is the value of the population standard deviation, σ, an unknown value? Yes Use the t-Distribution Sec>on9.1,Page184 3 ConfidenceInterval (Unknownσ) Confidence Interval = Sample Mean ± Margin of Error Margin of Error = Critical Value x Standard Error. Confidence Interval = s x ± t(df ,α ) × n where s is the standard deviation of the sample and n is the sample size, and df is the degrees of freedom, n-1. € € Conditions: The population must be normal or the sample is large (n ≥ 30) Sec>on9.1,Page186 4 ConfidenceInterval(Unknownσ) Illustra>veProblem-TI83Add-inPrograms C.I.=SampleMean±Cri>calValue*StandardError PRGM–CRITVAL–ENTER 2:T-DIST CONFLEVEL=.95 df=19 Answer:2.0930 PRGM–STDERROR-ENTER 4:1-MEAN n=20 Sx=1.76 Answer:.3935 C.I.=6.87±2.0930*.3935=6.87±.8236 =(6.87-.8236,6.87+.8236)=(6.05,7.69) Sec>on9.1,Page186 5 ConfidenceInterval(Unknownσ) Illustra>veProblem-TI83BlackBoxProgram STAT–TESTS–8:Tinterval Inpt:Stats 6.87 x: Sx:1.76 n:20 C-Level:.95 Calculate Answer:(6.05,7.69) Sec>on9.1,Page186 6 Problems a. b. c. d. e. Findthe98%confidenceinterval. Findthecri>calvalue Findthemarginoferror. Findthestandarderror. Whatassump>onmustwemakeaboutthethe popula>ontohaveat-samplingdistribu>on. f. Whataretheproperwordstodescribethe confidenceinterval? g. Ifyouwantedtohaveamarginoferrorofone minuteandthe98%confidenceintervalforthis data,howlargemustthesamplebe? Problems,Page205 7 Problems a. b. c. d. e. Findthe93%confidenceinterval. Findthecri>calvalue Findthemarginoferror. Findthestandarderror. Whatassump>onmustwemakeaboutthepopula>on tohaveat-samplingdistribu>on. f. Whataretheproperwordstodescribetheconfidence interval? g. Ifyouwantedtohaveamarginoferrorof25lbs.anda confidencelevelof93%forthisdata,howmany studentswouldyouhavetomonitor? Sec>on9.1,Page205 8 HypothesesTest(Unknownσ) Illustra>veProblem-TI-83Add-In The EPA wanted to show that the mean carbon monoxide level is higher than 4.9 parts per million. Does a random sample of 22 readings (sample mean = 5.1, s=1.17) present sufficient evidence to support the EPA’s claim? Use α = .05. Assume the readings have approximately a normal distribution. Ho: µ = 4.9 (no higher than) Ha: µ > 4.9 (higher than) t-Distribution df = 21 P-value = .2158 µx = 4.9 x = 5.1 P − Value = P(x > 5.1 given µx = 4.9) € € € PRGM – TDIST – ENTER LOWER BOUND = 5.1 UPPER BOUND = 2ND EE99 MEAN = 4.9 SE(x ) = 1.17 / 22 € df = 21 Answer: .2158 (Evidence is not sufficient to support the EPA’s claim.) € Sec>on9.1,Page187 9 HypothesesTest(Unknownσ) Illustra>veProblem-TI-83BlackBox The EPA wanted to show that the mean carbon monoxide level is higher than 4.9 parts per million. Does a random sample of 22 readings (sample mean = 5.1, s=1.17) present sufficient evidence to support the EPA’s claim? Use α = .05. Assume the readings have approximately a normal distribution. Ho: µ = 4.9 (no higher than) Ha: µ > 4.9 (higher than) STAT − TESTS − TTest Inpt :Stats µ0 :4.9 x : 5.1 Sx :1.17 n = 22 µ :> µ0 Calculate Answer : p − value = .2158 There is not sufficient evidence to support the EPA claim. € Sec>on9.1,Page187 10 Problems In a study of computer use, 95 randomly selected Internet Canadian users were asked how much time they spend using the Internet in a typical week. The mean of the sample was 12.7 hours and the standard deviation is 4.1 hours. Is this convincing evidence that Canadians use the internet more than the USA average of 12.3 hours a week? a. b. c. d. State the correct hypothesis Find the p-value. State your conclusion. What is the name of the probability model used for the sampling distribution e. What is the mean of the sampling distribution? f. What is the value of the standard error? g. If your conclusion is in error, what type of error is it? Problems,Page209 11 Problems The recommended number of hours of sleep per night is 8 hours, but everybody “knows” that the average college student sleeps less than 7 hours. The number of hours slept last night by 10 randomly selected college students is listed here. 5.2 6.8 5.5 7.8 5.8 7.1 8.1 6.9 5.7 7.2 Is there convincing evidence that the students sleep less than 7 hours? Assume the population is normal. a. b. c. d. State the correct hypothesis. Find the p-value. State your conclusion. What is the name of the probability model used for the sampling distribution e. What is the mean of the sampling distribution? f. What is the value of the standard error? g. If your conclusion is in error, what type of error is it? Problems,Page205 12 Problems a. b. c. d. State the appropriate hypotheses. Find the p-value. State your conclusion. What is the name of the probability model used for the sampling distribution e. What is the mean of the sampling distribution? f. What is the value of the standard error? g. If your conclusion is in error, what type of error is it? Problems,Page206 13 InferenceaboutPropor>ons A binomial experiment is a experiment that has only two outcomes. Binomial experiments relate to categorical variables. Consider the variable, Supports Obama. The variable has two categories, yes and no. We will consider the yes category as a “success” and the no category a “failure”. Suppose we have a population of 1000 voters, and 550 support Obama. We define proportion of success for the population, p = 550/1000 = .55. The proportion of failures is q = 1-p = .45. Suppose we take a sample to estimate p, the true proportion of voters that support Obama. We take a random sample of 100 voters and 53 support Obama. Our sample proportion is p’ = 53/100=.53. Our sample q’ = 1-p’ = .47. Sec>on9.2,Page192 14 Problems Problems,Page206 15 SamplingDistribu>onfora Propor>on In practice, the following conditions will insure the sampling distribution for a proportion is normal. 1. The sample size is greater than 20 2. The product np and nq are both greater than 5. Where we do not know p, we substitute p’, np’ = # success in sample must be > 5 and nq’ =n(1p’) = # failures in sample must be > 5. 3. The sample consists of less than 10% of the population. Sec>on9.2,Page192 16 ConfidenceIntervalfora Propor>on,p Given a sample proportion p’, that has a normal sampling distribution, the confidence interval for the true population proportion, p, is: sample proportion ± margin of error sample proportion ± critical value × standard error of p’ p'±z(α ) × p'q' n where q’ = proportion of failures = 1-p’ and n is the sample size. € Sec>on9.2,Page193 17 ConfidenceIntervalforPropor>on Illustra>veExampleTI-83Add-inPrograms In a discussion about the cars that fellow students drive, several statements were made about types, ages, makes colors, and so on. Dana decided he wanted to estimate the proportion of convertibles students drive, so he randomly identified 200 cars in the student parking lot and found 17 to be convertibles. Find the 90% confidence Interval for p, the true proportion of convertibles. Check conditions for normal sampling condition: n = 200 > 20 # successes = 17 > 5 # failures = 200 – 17 = 183 > 5 Conditions are satisfied. PRGM: CRITVAL 1 – CONF LEVEL = .90 Answer: 1.6449 PRGM: STDERROR – 1:1 PROP – p’ = 17/200; n = 200 Answer: .0197 Confidence Interval = 17/200 ± 1.6449*.0197 = 0.0850 ± 0.0324 = (.0850 - .0324, .0850 + .0324)= (.0526, .1174) Sec>on9.2,Page193 18 ConfidenceIntervalforPropor>on Illustra>veExampleTI-83BlackBoxProgram In a discussion about the cars that fellow students drive, several statements were made about types, ages, makes colors, and so on. Dana decided he wanted to estimate the proportion of convertibles students drive, so he randomly identified 200 cars in the student parking lot and found 17 to be convertibles. Find the 90% confidence Interval for p, the true proportion of convertibles. STAT – TESTS – A:1 – PROPZInt x: 17 (The number of success in p’: must be integer) n: 200 C-Level: .90 Calculate Answer: (.0526, .1174) We are 90% confident that the true proportion of convertibles is in the interval. Sec>on9.2,Page193 19 Problems A bank randomly selected 250 checking account customers and found that 110 of them also had savings accounts at the same bank. a. Construct a 90% confidence interval for the true proportion of checking account customers who also have savings accounts. b. What is the name of the sampling distribution used for the confidence interval? c. Show that the necessary conditions for a sampling distribution are satisfied. d. What is the standard error of the sampling distribution?. Problems,Page206 20 Problems In a sample of 60 randomly selected students, only 22 favored the amount being budgeted for next year’s intramural and interscholastic sports. a. Construct the 99% confidence for the proportion of students who support the proposed budget amount. b. What is the name of the sampling distribution? c. Show that the necessary conditions for a sampling distribution are satisfied. d. What is the standard error of the sampling distribution?. Problems,Page206 21 ConfidenceIntervalforPropor>on RequiredSampleSize ME = Critical Value × Standard Error pq = z(α ) × n ME = z(α ) × p(1− p) n Solving for n : ⎛ z(α ) ⎞ 2 n =⎜ ⎟ × p(1− p) ⎝ ME ⎠ € If we have a good estimate of p, we use it. If we have no good estimate of p, we estimate p = .5. This will produce the largest sample for the given conditions. If p’ turns out to be different that p, our ME will be less than we initially required. Sec>on9.2,Page195 22 RequiredSampleSize Illustra>veProblemTI-83 Consider a manufacturer that purchases bolts from a supplier who claims the bolts are approximately 5% defective. How large a sample do we need to estimate the true proportion to be within ± .02 with 90% confidence? PRGM – SAMPLSIZ – 1:PROPORTION CONF LEVEL = .90 ME = .02 p EST = .05 (Problem estimate) Answer: 322 Sec>on9.2,Page195 23 Problems Problems,Page208 24 HypothesesTest-OnePropor>on Illustra>veProblem–TI-83Add-In Ho: p = .61 Ha: p > .61 p-value Sampling Distribution for p’ p’ P − Value = P( p' > .6714 given p = .61) € p=.61 p’ = 235/350 = .6714 PRGM – NORMDIST – 1 LOWER BOUND = .6714 – UPPERBOUND = 2ND EE99 MEAN = .61 ; σ p = .61(1− .61) /350) Answer: p-value = .0093 (Reject Ho, We proved that more than 61% sleep more that 7 or more hours) ' € Sec>on9.2,Page195 25 HypothesesTest-OnePropor>on Illustra>veProblem–TI-83BlackBox Ho: p = .61 Ha: p > .61 STAT – TESTS – 5:1-PropZTest po: .61 x: 235 (# of successes in p’; must be integer) n: 350 prop > po Calculate Answer: p-value = .0092 (Ho is rejected) Sec>on9.2,Page196 26 Problems A politician claims that she will receive at least 60% of the vote in an upcoming election. The results of a properly designed random sample of 100 voters showed that 57 of them will vote for her. Is the sample evidence sufficient to prove that her claim is false? a. Check the conditions for a normal sampling distribution. b. State the hypotheses. c. Find the p-value. d. State your conclusion. e. If you make an error in your conclusion, what type is it? f. Find the mean of the sampling distribution. g. Find the standard error of the sampling distribution. Problems,Page207 27 Problems a. Check the conditions for a normal sampling distribution. b. State the hypotheses. c. Find the p-value. d. State your conclusion e. If you make an error in your conclusion, what type is it? f. Find the mean of the sampling distribution. g. Find the standard error of the sampling distribution. Problems,Page207 28