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Lecture 2

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Ryerson University
Quantitative Methods
QMS 202
Clare Chua

1/17/2011 Objective Learn how to coonnssttrrucctt and Chapter 9 intteerrprreett confidence interval Confidence Interval Estimation esttmamatess Confidence Interval Estimation for the Mean ( known) For the mean Confidence Interval Estimation for the Mean ( unknown) For the proportion Confidence Interval Estimation for the Proportion Determining Sample Size Frequently asked questions Confidence Interval Estimation 1. is known and is unknown is given and is not given Population Mean 2 Population Proportion - 2. What is Z scores/values? X Normal and/or nJ30? np J 5 and n(1-p) J 5? Go to this link to understand "z-score" : If no stop. If no stop. BF&list=QL&index=1 8 known? 8 unknown? Limits: p(1 p) mfu_in_order&list=UL p H z n Cut and paste the link on the URL box. p=x/n x H z 3. What are you estimating? Parameters 2 or 5 Limits: n Limitsx H t 4. How do you interpret CI? df=n-1 n Go to this link WE EEEK 11 WEEEEKK 2 vw Review 8 % Confidence Interval when is xivHnz:, / 2 =2 n We needed z value for a % confidence Interval. 90% confidence leveConfidence coefficient of 0.90??? 95% confidence leveConfidence coefficient oZ= ?????? 97% confidence leveConfidence coefficient oZ= ?????? 99% confidence leveConfidence coefficient oZ= ?????? % Confdence IIntterhen iis nott giiven : s x H t , ,n 1 We needed t value for a % confidence Interval.n H3 H2 H1 +1 +2 +3 Suppose we needed to find t, when sample size (denoted as n) is 25. 1 3 5 7 9 11 13 Xscale(=7, =2) H3 H2 H1 0 +1 +2 +3 Zscale(=0, =1) 90% confidence levConfidence coefficient ot = ?????? Convert any normal random variable X to standardized normal 95% confidence levConfidence coefficient ot = ?????? random varX 2 Z. 97% confidence levConfidence coefficient ot = ?????? Z 8 99% confidence level If X is normally distributed N(, ), the standardized variable Z has Confidence coefficient ot = ?????? standard deviation 1 (or =1), denoted N(0,1)r =0) and 1 1/17/2011 TEMPLATE : : Confidence Interval for population mean TEMPLATE :: Confidence Interval for population mean when is known when is unknown Define : Estimate Define : Estimate 95% CI : s 95% CI : x H t. ,n 1 x H Z . 2 n Note: 2 n Confidence intervalto find the Note: 1. Using the table to find t value and 1.8 Theearehre wayyo ndheth 33 use he fformuld heCII.fiind 20 .0 H 1.9566 Confidence interval 22555 .0 H 22.006440 50 1. Use the formula, find z value 25 2. Using the formula, find t value ufrom CASIO calculatorn using SPSS function: 2. Use the formula, find z value 255 .0 H 13 .6 IDF.T (./2, degree of freedom) 20 .0 H 0.5 using SPSS function: compute function. using SPSS IDF.Normal (prob, mean, stdev) 3. Using the INTR function of CASIO 3. Use the INTR function of CASIO 241 .3 @ @ 268 .6 Calculator to find Confidence Calculator to find Confidence Interval directly 19 .5 @ @ 20 .5 Interval directly Confidence Interval Estimation Interpreting the CI Population Mean 2 Population Proportion - Read page 369, .with 95% X Normal and/or nJ30? np J 5 and n(1-p) J 5? If no stop. If no stop. confidence, you conclude that the mean amount of all the sales 8 known? 8 unknown? iinvoiices ibei ttweenLII(llower Limits: interval) and UI (upper interval). p H z(1 p) p=x/n n Limits:H z n Limitx H t df=n-1 n Referring to population WEEEEK 11 WEEEEKK 2 Confidence Interval Estimation for the proportion The unknown population proportion is represented by 5 Confidence Interval Estimation The point estimate for 5 is the sample for the p propportion prropoortion, deenotedd as p, whhere p==X/nn Where n = sample size (number of observed items/subjects in a sample) We extends the concept of CI to categorical data And X=number of items in the sample having the characteristic of interest Here you are concerned with estimating the proportion of items in a population having a certain characteristic of interest (e.g. people who have an iphone) 2 1/17/2011 TEMPLATE : Confidence Interval for population proportion, - Note Define : est - = population proportion of __________________ State the n and p values p(1 p) p H z Check the conditions: np > 5 and n(1-p) > 5 n If conditions are satisfied, proceed to find CI
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