THE CENTRAL LIMIT THEOREMSampling Distribution of x- normally distributed populationNormal PopulationsNon-normal PopulationsThe Central Limit Theorem (for the sample mean x)The Importance of the Central Limit TheoremHow Large Should n Be?SummaryThe Central Limit Theorem (for the sample proportion p)Slide 10Slide 11Population Parameters and Sample StatisticsExampleGraphicallyExample (cont.)Example 2Example 2(cont.)Example 3Example 3 (cont.)Example 4Example 5Population X: amount of salmon in a can E(x)=6.05 oz, SD(x) = .18 ozPowerPoint PresentationExample 6Example 7Example 7 (cont.)THE CENTRAL LIMIT THEOREMThe World is Normal TheoremSampling Distribution of x- normally distributed populationn=10/10Population distribution:N( , )Sampling distribution of x:N( ,  /10)Normal PopulationsImportant Fact: If the population is normally distributed, then the sampling distribution of x is normally distributed for any sample size n.Previous slideNon-normal PopulationsWhat can we say about the shape of the sampling distribution of x when the population from which the sample is selected is not normal?The Central Limit Theorem(for the sample mean x)If a random sample of n observations is selected from a population (any population), then when n is sufficiently large, the sampling distribution of x will be approximately normal.(The larger the sample size, the better will be the normal approximation to the sampling distribution of x.)The Importance of the Central Limit TheoremWhen we select simple random samples of size n, the sample means we find will vary from sample to sample. We can model the distribution of these sample means with a probability model that is,Nnms� �� �� �How Large Should n Be? For the purpose of applying the central limit theorem, we will consider a sample size to be large when n > 30.SummaryPopulation: mean ; stand dev. ; shape of population dist. is unknown; value of  is unknown; select random sample of size n;Sampling distribution of x:mean ; stand. dev. /n;always true!By the Central Limit Theorem:the shape of the sampling distribution is approx normal, that isx ~ N(, /n)The Central Limit Theorem(for the sample proportion p)If a random sample of n observations is selected from a population (any population), and x “successes” are observed, then when n is sufficiently large, the sampling distribution of the sample proportion p will be approximately a normal distribution.The Importance of the Central Limit TheoremWhen we select simple random samples of size n, the sample proportions p that we obtain will vary from sample to sample. We can model the distribution of these sample proportions with a probability model that is(1 ),p pN pn-� �� �� �How Large Should n Be? For the purpose of applying the central limit theorem, we will consider a sample size to be large when np > 10 and nq > 10Population Parameters and Sample StatisticsˆpThe value of a population parameter is a fixed number, it is NOT random; its value is not known.The value of a sample statistic is calculated from sample dataThe value of a sample statistic will vary from sample to sample (sampling distributions)Population parameterValueSample statistic used to estimatepproportion of population with a certain characteristicUnknownµmean value of a population variableUnknownxExample( )48A random sample of =64 observations isdrawn from a population with mean =15and standard deviation =4.a. ( ) 15; ( ) .5b. The shape of the sampling distribution model for is approx. noSD XnnE X SD Xxmsm= = = = =( )rmal (by the CLT) with mean E(X) 15 and ( ) .5. The answerdepends on the sample size since ( ) .SD XnSD XSD X= ==GraphicallyShape of population dist. not knownExample (cont.)15.5 15 .5.5 .5( )c. 15.5;1This means that =15.5 is one standarddeviation above the mean ( ) 15xSD XxzxE Xm--== = = ==Example 2The probability distribution of 6-month incomes of account executives has mean $20,000 and standard deviation $5,000. a) A single executive’s income is $20,000. Can it be said that this executive’s income exceeds 50% of all account executive incomes?ANSWER No. P(X<$20,000)=? No information given about shape of distribution of X; we do not know the median of 6-mo incomes.Example 2(cont.)b) n=64 account executives are randomly selected. What is the probability that the sample mean exceeds $20,500? ( )( )5,0006420,000 20,500 20,000625 625( ) $20,000, ( ) 625By CLT, ~ (20,000,625)( 20,500)( .8) 1 .7881 .2119SD xnXE x SD xX NP X PP z- -= = = => = > => = - =answer E(x) = $20,000, SD(x) = $5,000Example 3A sample of size n=16 is drawn from a normally distributed population with mean E(x)=20 and SD(x)=8.( )81620 24 202 216 20 24 202 2~ (20,8); ~ (20, )) ( 24) ( ) ( 2)1 .9772 .0228) (16 24)( 2 2) .9772 .0228 .9544XX N X Na P X P P zb P X P zP z- -- -� = � = � =- =� � = � � =- � � = - =Example 3 (cont.)c. Do we need the Central Limit Theorem to solve part a or part b?NO. We are given that the population is normal, so the sampling distribution of the mean will also be normal for any sample size n. The CLT is not needed.Example 4 Battery life X~N(20, 10). Guarantee: avg. battery life in a case of 24 exceeds 16 hrs. Find the probability that a randomly selected case meets the guarantee.102420 16 202.04 2.04( ) 20; ( ) 2.04. ~ (20, 2.04)( 16) ( ) ( 1.96).1 .0250 .9750XE x SD x X NP X P P z- -= = => = > = >- =- =Example 5Cans of salmon are supposed to have a net weight of 6 oz. The canner says that the net weight is a random variable with mean =6.05 oz. and stand. dev. =.18 oz.Suppose you take a random sample of 36 cans and calculate the sample mean weight to be 5.97 oz.Find the probability that the mean weight of the sample is less than or equal to 5.97 oz.Population X: amount of salmon in a canE(x)=6.05 oz, SD(x) = .18 ozX sampling dist: E(x)=6.05 SD(x)=.18/6=.03By the CLT, X sampling dist is approx. normalP(X  5.97) = P(z  [5.97-6.05]/.03)=P(z  -.08/.03)=P(z  -2.67)= .0038How could you use this answer?Suppose you work for a “consumer watchdog” groupIf you sampled the weights of 36 cans and obtained a sample mean x  5.97 oz., what would you think?Since P( x  5.97) = .0038, either–you observed a “rare” event (recall: 5.97 oz is 2.67 stand. dev. below the mean) and the mean fill E(x) is in fact 6.05 oz. (the value