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Statistical Methods I Test 1 [REVIEW]

25 Pages

Course Code
STA 2023

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Statistics Spring 2014
January 8, 2014
A) Types of statistical problems
a. Descriptive- Describing a set of measurements
using graphs and numbers
b. Inferential- use a simple to say something about a population
B) Elements of inferential problems
a. Population- set of objects of interest
b. Variable- measurements that are recorded for each object
c. Sample- subset of the population
d. Inference- decision, estimate or prediction
e. Measure of reliability- based on probability
C) Types of data
a. Qualitative- values of x are categories
i. Examples- gender, race
b. Quantitative- values of x are numerical measurements
Describing data
Graphical techniques sec 2.2
1. Histogram- bar graph
a. Relative frequency is the proportion of the data (data point over the total population).
2. Stem and leaf display
10 16 25 40
12 17 28 60
15 25 38 80
Stem leaf
3 8
4 0
6 0
8 0
*This is skewed to the right.
What is the stem unit? (Answer: Tenths)
0.348 0.536
0.349 0.548
0.467 0.621
0.498 0.638
Statistics Spring 2014
0.521 0.708
* you can round or you can make a two part leaf.
What is the stem unit? (Answer: hundredths)
0.328 0.378
0.329 0.379
0.356 0.382
0.358 0.385
0.369 0.39
Stem Leaf
32 8 9
35 6 8
36 9
37 8 9
38 5
39 0
Numerical measure of central tendency sec 2.4
1. mean: average
a. m=population mean
b. (X bar) =sample mean
c. n- sample size
2. median
a. the very middle age of all ages (middle of data)
i. first arrange data in numerical order
ii. then find the middle.
iii. n is odd: median=middle number
iv. n is even: median- average of the 2 middle numbers
Example of mean
To estimate the average cost of a new home in Orlando, 25 homes are selected and their average cost is found to
be $240,000
- Sample is 25 home
-Population is all homes
-Variable(x)is the cost of each house
-inference is to estimate the average cost of the homes in Orlando.
-m = unknown
Stem Leaf
7 0
Statistics Spring 2014
-(X bar) is 240,000
Shapes of distributions
-skewed right
-skewed left
How do the mean and median compare for each of these shapes?
-Mean and median are the same with the distribution is symmetric
mound shaped uniform
Skewed right: Skewed left:
X= age of student at UCF (skewed to the right)
X= grade on an easy test (skewed to the left)
X=number of days a UCF student hospitalized in the last year. (skewed to the right)
Sec. 2.5 measure of variability
1. Range= maximum-minimum
a. Data set: 1,3,3,3,10
i. Range= 9
ii. (X bar)= 4
iii. median= 3
b. Data set: 1,1,3,5,10
i. Range=9
ii. (x bar)=4
iii. median=3
2. Variance
a. S2 = sample variance
b. σ2 = population variance
3. Standard Deviation: positive square root of the variance
a. s= sample standard deviation
b. = population standard deviation. σ

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Statistics    Spring 2014 January 8, 2014 Ch.1 A) Types of statistical problems a. Descriptive­ Describing a set of measurements  using graphs and numbers b. Inferential­ use a simple to say something about a population B) Elements of inferential problems  a. Population­ set of objects of interest b. Variable­ measurements that are recorded for each object  c. Sample­ subset of the population d. Inference­ decision, estimate or prediction e. Measure of reliability­ based on probability C) Types of data a. Qualitative­ values of x are categories i. Examples­ gender, race b. Quantitative­ values of x are numerical measurements Ch.2  Describing data Graphical techniques sec 2.2  1. Histogram­ bar graph a. Relative frequency is the proportion of the data (data point over the total population).  2. Stem and leaf display 10 16 25 40 12 17 28 60 15 25 38 80 Stem leaf 1 0 2 5 6 7 2 5 5 8 3 8 4 0 5 6 0 8 0 *This is skewed to the right. Example  What is the stem unit? (Answer: Tenths) 0.348 0.536 0.349 0.548 0.467 0.621 0.498 0.638 Statistics    Spring 2014 0.521 0.708 Stem Leaf 3 4 4 4 6 9 5 2 3 4 6 6 3 7 0 * you can round or you can make a two part leaf.  Example What is the stem unit? (Answer: hundredths) 0.328 0.378 0.329 0.379 0.356 0.382 0.358 0.385 0.369 0.39 Stem Leaf 32 8 9 35 6 8 36 9 37 8 9 38 5 39 0 Numerical measure of central tendency sec 2.4 1. mean: average  a. m=population mean b. (X bar) =sample mean  c. n­ sample size 2. median  a. the very middle age of all ages (middle of data) i. first arrange data in numerical order ii. then find the middle.  iii. n is odd: median=middle number iv. n is even: median­ average of the 2 middle numbers Example of mean  To estimate the average cost of a new home in Orlando, 25 homes are selected and their average cost is found to  be $240,000 ­ Sample is 25 home ­Population is all homes  ­Variable(x)is the cost of each house ­inference is to estimate the average cost of the homes in Orlando.  ­m = unknown Statistics    Spring 2014 ­(X bar) is 240,000 ­n=25  Shapes of distributions ­symmetric  ­skewed right ­skewed left How do the mean and median compare for each of these shapes? ­Mean and median are the same with the distribution is symmetric  mean=median mound shaped  uniform  Skewed right:                                                                      Skewed left:  Example X= age of student at UCF (skewed to the right) X= grade on an easy test (skewed to the left) X=number of days a UCF student hospitalized in the last year. (skewed to the right) Sec. 2.5 measure of variability 1. Range= maximum­minimum a. Data set: 1,3,3,3,10  i. Range= 9 ii. (X bar)= 4 iii. median= 3 b. Data set: 1,1,3,5,10 i. Range=9 ii. (x bar)=4 iii. median=3 2. Variance 2  a. S = sample variance b. σ = population variance 3. Standard Deviation: positive square root of the variance a. s= sample standard deviation b. σ= population standard deviation.  Statistics    Spring 2014 Example Deviation= Deviation squared = X x-(xbar) (x-(xbar))2 1 1-3=-3 9 1 -3 9 3 -1 1 5 1 1 10 6 36 (xbar)=4 sum (x-(xbar)=0 Sum (x-(xbar))^2= 56 (x−xbar) 2 S = 56/4=14   s =∑ n   S=standard deviation=3.74 Ti 83:  stat/edit/enter put data into list  Stat/calc/enter/ One variable statistics (which list?) The standard deviation of the sample isxs Work sheet ch2: 1.  a. 20 b. 1060 c. 840 *plot is skewed to the left.  2.  a.  January 13, 2014 Population= greek letters Properties of standard deviation 1. S ≥0thereisvariation 2. S=0this meansthedatadoesn tdeviate whichmeansits allthesame¿ 3. Thegreaterthecariation¿themean,thelargerthevalueof S. Interpretation of the standard Deviation (2.6)  Empirical Rule­ using only mound shaped data (normal distribution) Chebyshev’s rule­ used when the shape of the distribution is unknown 1. Empirical Rule ­If the distribution is mound shaped:  Statistics    Spring 2014 68% of the distn. Lies within 1 s.d of the mean about 95% of the distn. Lies within 2 s.d of the mean 2. Chebyshev’s Rule: Statistics    Spring 2014 ­At least  ¾ of a distribution will lie within two standard deviations of the mean ­At least 8/9 of a distribution will lie within three standard deviations of the mean. ­At least 0 of a distribution will lie within one standard deviation of the mena ­In general, at least 1­(1/k^2) of a distribution will lie within k standard deviations. Note  measures of relative standing sec. 2.7 percentileththe pth percentile is a # that has p% of the distribution to the left of it.  1. 50  percentile= mean th 2. If x=SAT scores with u=1500 and  σ =300, find the 84  percentile. Assure a mound shape. =1800 3. What percentile is a score of 2100? =97.5% percentile Z score­ the # of the standard deviations a score is from its mean. Can me + or ­.  z= x−μ σ What SAT score has a z score of 1.7? X=2010  x−1500 z= 300 January 15, 2014   Let x=GPA with  μ = 2.7 and  σ =.4 Assume a mound shape distribution ­What % of the GPAs are more than 2.3? use empirical rule because it’s a mound shape. about 84% has GPAs more than 2.3 th ­What GPA is the 16  perthntile? Or  What GPA has 16% to the left? 2.3 is also the 16  percentile ­What is the zscore for a GPA of 3.0? Or How many standard deviations away from the mean? .75 or ¾ or .3/.4 z=(3.0­2.7)/.4 = ¾ =.75 Don’t know the shape, what % of the GPAs are about 3.5? Statistics    Spring 2014 K=2          25(answer) CH3 Probability Use probability to make statistical inferences Assume coin is fair Toss the coin 10 times  Observe 9 heads Probability of a least 9 heas is .011 Conclusion? Possibly not fair 3.1 Experiment­ what you are asked to oberce  Roll a fair die one time Sample points­basic outcomes of the experiment 1,2,3,4,5,6 Sample space­ collection of all sample points Event (A,B,C..) –collection of some of the sample points A=even #, B=# ¿ 4 1 b 2 a b 3 b 4 a 5 6 a Sample space Venn Diagram­ picture of S and how events are related A 4,6 2 5,3 B Probability: measures likelihood an event will occur (A)≤1 Finding probability of event A 1. list sample space 2. assign probabilities to sample points 3. Determine points in A and sum a. P(A)= 3*1/6 =3/6=.5 b. P(B)= 3*1/6= 3/6 =.5  c. P(A ∪B¿=5/6(everythingoutsidethecircles) Statistics    Spring 2014 d. P(A ∩B¿=1/5(everythinginbetweenthetwocircles) Examples Order important? YES Sampleing with or without replacement? WITH Worksheet #1 toss a coin 3 times. A= at least one head B=at most one tail HHH THH HTH HHT HTT THT TTH TTT A=7/8 probability B= 4/8 probability  Worksheet #2  Six students consist of 4 males and 2 females. Randomly select 2 students for a scholarship. Probability of one  make and one female? (Without replacement) January 22, 2014 SI sessions: Monday 12:30pm­1:20pm BHC 128 Tuesday 12:00pm­12:50pm BA1 220 Wednesday 12:30 pm ­1:20pm HEC125 Thursday 1:30­2:20pm CB1 320 Exam Feb. 3 ch1­3 c  c P(A ∪  B )=  (.2+.3 ∪ .4+.3)= .9     ▯ don’t count the .3 twice P(A c ∩  B )= ( they both have .3 so…)= .3 P (Ac  ∩  B)= (.2, .3 ∩  .2, .1) =.2 P(A c ∪  B)= (.2+.3  ∪  .2+ .1)= .6  ▯ don’t count the .2 twice Sec 3.4 The additive rule and mutually exclusive events 1. Additive rule: a. P(A ∪  B)=   P(A) + P(B) ­  P(A∩  B) 2. Mutually Exclusive events A and B  a. P(A ∩B¿=0      Worksheet 5: A .25 .15 .45 B Statistics    Spring 2014 SS A ∩B  =.15 NS A c ∩B c SN A ∩B NN A c ∩B c a) Probability person must stop for at least one light? P(A ∪B )=.4+.3­.15 =.55 b) What is the probability they don’t’ have to stop for either light? B P(A ∪¿¿ c = .45 st c) What is the probability they have to stop at only the 1  light? P(A ∩B c )= .25 Sec. 3.5 Conditional Probability Example:  Roll a fair die once and let  A=(even #) and B= (# <4). Probability of A= .5 Probability of B=.5 1      B P(A)= 3/6= .5 2 A  B 3      B P(B)= 3/6 =.5 4 A   P(A ∩ B)=1/6 5 P(A ∪ B)=5/6 6 A     ­If you know that A has occurred, what is the probability of B occurring? P(B I A)= probability of B given the condition (A) Reduced sample space: 2 A  B P(B I A)= 1/3 4 A   6 A   ­ If you know that B has occurred, what is the probability of B occurring? P(A I B)= probability of A given the condition (B) Reduced sample space: 1      B 2 A  B P(A I B)= 1/3 3      B *P(A I B)= P(A ∩ B)/P(B)  =  1/6   = 1/3  3/6 Statistics    Spring 2014 Example:  if a heart is selected from a deck of 52, what is the probability it is a facec card? A=face card,  B=heart I ∩ 3/52  P(A B)= P(A B)/P(B)  = 13/52   = 3/13  Worsheet #6 A survey of 140 students was classifies into the following table: Fresh. Soph. Jr. Sr. Total Male 10 15 20 5 50 Female 30 10 38 12 90 Total 40 25 58 17 140 Let A=male          B=Freshman        C=Jr. , Sr.  • B and C are mutuality exclusive (can’t be in both groups).  P(A ∪ B)= (50+40 ­10)/ 140 = 80/140  ▯don’t count 10 twice! P(A ∩ B)= 10/140 10/140 P(A I B)= P(A ∩ B)/P(B)  = 40/140 = 10/40 10/140 P(B I A)= =¿  10/50 50/140 P(B I C)=0 January 27, 2014 P(AIB)= P(A∩B) P(B) P(A∩B) P(BIA)= P(A) Sec. 3.6 Multiplicative Rule and Independent Events 1. Multiplicative Rule a. P(A∩B)=P(A)P(BIA) b. ¿P(B)P(AIB) 2. Independent Events  a. A and B are independent events if and only if i. P(AIB)=P(A) ii. P(BIA)=P(B) P(A∩B)=P(A)P(B) iii. Worksheet: In the traffic light problem, are A and B independent events? P(A)=0.4                P(B)=0.3             P(A Statistics    Spring 2014 A .4 .3 B .2 .1 P(A ∩B¿=0.1    P(BIA)=0.2 Are A and B independent? No they are both dependent. Worksheet #7 Suppose a basketball player makes 80% of his free throws. If the throws are considered independent, what is the  probability he makes both shots?  HH      (Probability of each hit is 0.8)  You want to multiply them together to find the probability of them happening together.  P(A ∩B¿=Pst)P(B)   nd A is H on 1  and B is H on 2 .  (0.8) = .64 Exactly one? MH HM 2(.8) (.2) =3.2 At least one shot? .96 Worksheet #8 Now the throws are dependent. If he misses the 1  shot, he makes 70% of the second shots. If he makes the 1  t nd shot, he makes 90% of the 2  shots. Probability he makes both? A=make 1  shot P(A)=.8 B=Makes 2  shot  c P(A)=.8   P(BIA )=.7 P(BIA)=.9 P(A∩B)=P(A)P(BIA)=.8∗.9=¿ .72 SSS NSS  Example: Suppose 25% if Americans smoke. SNS  Randomly select 3 people. Find the probability… SSN  1. All 3 smoke: SNN  NSN  NNS NNN Statistics    Spring 2014 a. (.25)(.25)(.25)=0.0156 2. At least 1 is a smoker a. 1­P(no smokers in the sample) b. =1­(.75)(.75)(.75)=0.5781 3. Exactly 1 Smokes  a. 3((.25)(.75)(.75))=0.4219  Sec. 3.8 Counting Rules Methods for counting sample points without listing them.  • Multiplicative counting rule (With replacement) • Permutations (Without replacement­ order matters) • Combinations (Without replacement­ order doesn’t matter) (Omit partitions rule) 1.  Multiplicative Rule   n  elements  k n2 lements  n1 elements  a. Given k groups of elements  Then (n1) (n2) (n3)…(nk) is the number of ways of selecting one  member from each group.  Example: Style= 3  Color =10 M,A =2  1. How many different cars can you order if they come in manual or automatic, 10 different  colors, and 3 styles? (2)(10)(3) =60 2. Flip a coin 10 times.  How many sample points? Statistics    Spring 2014 a. 1024  January 29, 2014    . Permutation  An ordered arrangement of “r” distinct objects taken from one group containing “n” objects.  Example:  There are 4 teams, which will be ranked from best to worst? How many ways can you rank all 4 of them? P 4 4*3*2*1=24 How many ways can you rank the top 2? 4 P 2 4*3=12 n n! Formula:   Pr= (n−r ! (top number) M▯ ath ▯ rob▯selection 2 ▯ (bottom number) 3. Combinations:     Selecting r objects from n objects and order in not important.  ­always less combinations then there are permutations because order doesn’t matter n n! Formula:   C r r!(n−r)! (top number) M▯ ath ▯ rob▯selection 3 ▯ (bottom number) Worksheet examples: 1. How many different lotto picks from the numbers 1 to 53?  a. You can only pick 6 numbers  i. Combination from 53 to 6 ii. You can’t pick a number more then once (without replacement) iii. Order doesn’t matters iv. = 22957480 2. The international airline Assn. assigns three letter codes to represent airport locations. For example,  Orlando is MCO. How many codes are possible? a. You can repeat letters (with replacement)  i. Multiplicative (with replacement) ii. 3 combinations with 26 chances of it being a different letter iii. 263 = 17,576 3. Select a jury of 6 people from a group of 16 people. How many different jury selections are possible? a. Without replacement b. Order doesn’t matter  c. Combinations 16 to 6 i. =8008 4. Six poker chips are numbered 1 to 6. Select 2 chips and from a 2 digit number a. Order matters  b. Without replacement c. Permutation 6 to 2 Statistics 
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