parameter calculated from sample data Confidence Level: number that provides information on how much “confidence” we have in the method used to construct a confidence interval estimate SO WHY DO WE NEED IT? To estimate an unknown population parameter.
assigned treatments) σ unknown Normal (or approx. normal) distribution • Given • Large sample size (n≥30) • Check graph of data main difference is sigma another main difference is that when n is under 30 you must automatically use t t-test
is known, the formula is: CI: ⨉ ± z* (ϭ/√n) Statistic Critical Value Standard Deviation of Statistic Margin of Error Confidence Interval: statistic ± z critical value (standard deviation of statistic)
is unknown, the formula is: Confidence Interval: statistic ± t critical value (standard deviation of statistic) same as z in terms of location of important terms
(s/√n) 1. ⨉ is sample mean from random sample 2. sample size n is large (n≥30) OR the population distribution is normal 3. population standard deviation is unknown
a 95% confidence interval for the population mean from a sample size of 35 where we know the sample mean is 100 and the population deviation is 12. We are going to use a Z-Interval test because sigma is known CALC: STAT>TESTS>7:ZInterval Since we know all information, we got to STATS in ZInterval table (left) and just insert information where necessary. We then press Calculate and get interval answer (right) When you only have the data and not the mean or n, just go to CALC: STAT>EDIT>L1 and type in values (left). The process will be the same, you just press DATA on the Zinterval table (right)
develop a 95% confidence interval for the population mean from a sample size of 40 women where we know the sample mean is 76.3 and the population deviation is 12.5. no context in problem by the way....
± 1.960 (12.5/√40) CI: 76.3 ± 3.87 CI: (72.3, 80.17) 95% confidence goes with 1.960 z critical value Calculations Assumptions -SRS -Normal because n≥30 -sigma known Conclusions We are 95% confident that the true population mean of women ___ is between 72.3 and 80.17.
for critical value z ϭ is 5.0 CI: ⨉ ± z* (ϭ/√n) 1.96 (5.0/√n) 1.96 (5.0/√n) = 1 5.0/√n = .510 5 = .510 (√n) 9.8 = √n (9.8)² = (√n)² 96.04 = n 97 = n Assume Margin of Error= 1 in order to solve for n Margin of Error .510=1/1.96 9.8=5/.510 ALWAYS round up!
the margin of error, to 1. This gives you: B = 1.96 (ϭ/√n) which is just: 1 = 1.96 (ϭ/√n) The result for solving variable n is: n= (1.96ϭ/B)² or just n= (1.96ϭ/1)² which solves n as n= (1.96(5)/1)² n=(9.8/1)² n=96.04 which rounds to 97 Determining Sample Size: 2nd/Easier Option Problem: 95% confident so 1.96 for critical value z ϭ is 5.0 Basically the formula is: (confidence level)(ϭ) B ( ) 2 n=
confidence level, z will be 1.645 and B=1 because we assume the margin of error is 1 so n= ((1.645)(9)/(1))² = (14.805)² n= 219.188 3. D: Assumptions: Have an SRS from population, σ known, Normal because large sample size (n≥30) 35>30 so check 4. B: Zinterval: (5.829, 6.2448) 5. C: CI: ⨉ ± t* (s/√n) CI: 67.5 ± 1.676 (9.3/√51) CI: 67.5 ± 2.18258 CI: (65.318, 69.682) df= 50, s=9.3, t=1.676 Extra Credit: To estimate an unknown population parameter calculated t critical value with table with df as 50 (51-1) at 90% level, should get 1.676
e n ce I nte r va l. " P re n h a l l. N.p. , n . d . We b. 2 3 M a y 2 0 1 1 . < htt p : //w w w.pre n h a l l.co m /e s m /a p p / ca lc _ v 2/ca lc u lato r/m e d ia li b /Te c h n o lo g y/ D o c u m e nt s / T I- 8 3/d e s c _ p a g e s /z _ co n f _ i nte r. ht m l > M a s s ey, Tiffa n y. "C o n f i d e n ce I nte r va l N ote s. " A P S tat i st ic s: B e l l 7. M H S M at h D e p a r t m e nt. M a u r y H i g h S c h o o l, N o r fo l k , VA. 2 0 1 0 -2 0 1 1 . L e ct u re s. Slide 13 Rest of Slides
francescoronel
utokyoissr2360
matleenalaakso
bernhardsvt
pokotyamu
jamashita
signer
asial_edu
codeforeveryone
yasslab
matthewcrist
pauljervisheath
chriscoyier
caitiem20
jeffersonlam
dotmariusz
tanoku
kevinliebholz
eileencodes
ufuk
marktimemedia
ddemaree
