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Final

# Formula Sheet Final Exam Page 2.doc

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York University

Administrative Studies

ADMS 2320

all

Summer

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Chapter 5: Simple Random Sampling (equal chance) | Stratified Random Chapter 15 Analysis Of Variance (ANOVA): Three Interval Data
Sampling (male40% + female60%) | Cluster Sampling (divide into blocChapter 12: Inference about a Population when σ is unknown use t-test p 1 p ) =1p − p 1 2
Sample Error: diff. result for diff samples. Sample Size Sample t-test only valid when the histogram is NOT extremely nonnormal. n1= 4, n2= 4, n3= 4
Non-sampling Error: Due to data acquisition, non-response, and selection bias.
Sample Size Non-sample Error NOT Sampling distributiPn ois appr. normal when np > 5 & n(1–p) ≥ 5 x 1 2.25, x2= 3.25, x3= 5.75
2 2
Chapter 13 Comparing Two Populations: Dif. btw 2 mean/prop. | ratio o2 tw∑ S x − ( ∑ x) / n 2 2
Chapter 9 Sample Distribution: X is nXrmalis normal | If X is Non- Large d.f.= large sample sizes more infor. by producing more pow1rful tests = 2.25, s 2 2.92 s3= 3.92
normal X is appr. normal distribution for n230 | 2arger sample closer toe II error probabilities) and narrower confidence interval estimation. n −1
E(X ) = μ x = μ =V(X∑ =σ =σ xn σ xσ d.f 1σ2=σ ) ≥ 1.f2 (σ ≠σ ) so equal variances is more efficiency. 9 +13 + 23
normal. ˆ ˆ ˆ x = = 3.75
Sample Mean: Proportion:P 1 P 2 is appr. normal wh1n 1 5 4 + 4 + 4
Finite Population: σ 2 = x P (x) − μ 2 ˆ ˆ ˆ 2 p1(1− p ) 1 p2(1− p ) 2
Rule of thumb: Pop. Size > 20 Sample size omit correction factor n 11 − P ) 1 P ≥ 2 n 21 −V ( p 1 p ) 1 σ2 p1− 2 = H :0μ = 1 = μ 2 3 H 1 At least two means differ
( ) ( N − n ) /( N − 1) n1 n2 2
SST = ∑ n jx −jx ) = 4(2.25 - 3.75) + 4(3.25 - 3.75) + 4(5.75 - 3.75)
Z-test StatistZ =: X − μ P(μ − Z σ < X < μ + Z σ ) =1−α
E.g. σ = 2, μ=12 σ / n α /2 α /2 SSE = (n − 1)s 2 =(4 1)(2.25) + (4 1)(2.92) + (4 1)(2.92)
Prob. of 1 cells = 10 or less: x − μ 10 −12 n ∑ j j
Prob. Mean of 4 cells = 10 or less: < = P(Z ≤ −1) = .5 − 3413 = .1587 ANOVA table
Prob. of Total 4 cells = 40 or less: σ x − μ 2 10 −2 Source d. f. Sum of Squares Mean Square
Prob. All 4 cells = 10 or less:≤10) =4P < = P(Z ≤ −2) = .5 − 4771 = .0228 SST 26
[P(x ≤10) ] = (.187) = .000ˆ 2 / 4 Treatments k–1 = SST = 26.00 MST = = = 13
ProportionE (P ) = μ p (P) = σ ˆ = p(1 ˆ p) / n(1 − p) / n 2 k −1 3 −1
ˆ P P n
Z-Stat for ProportZ =: E.g. 80% agree(p), ask 350(n), P<75%=? n–k = SSE 24.25
ˆ p (1 − p ) / n Error 9 SSE = 24.24 MSE = = = 2.6944
P − p .75 − .80 n − k 12 − 3
P (P < .75 ) = P < 2 = P (2 < −2.34 ) = .5 − .4904 = .0096 n–1 = SS (Total) =
p(1 − p) / n (.80 )(1 − .82 ) / 350σ 1 σ 2 Total 11 50.25
V (X −1X ) =1σ x1−x2 = +
Two Mean: E(X − 1 ) = μ1 x −x = μ −1μ 2 n 1 n 2 Rejection region: F α ,k −1,n − k F .05 ,2,9= 4.26
1 2
Z-Stat for two Means: (X −1X ) − 2μ − σ )1 = 2 (σ / n ) + (σ / n ) The Value of Test Statistic:ST 13
Z = 2 x12x2 1 1 2 2 F = = = 4.82
(σ 1 n )1+ (σ / n2) 2 MSE 2.69
Conclusion: Reje0t H , there is enough evidence to conclude that there are
Chapter 10 Estimator: appr. Pop. Parameters from Sample Xtatistic. e.g. differences in the number of job offers between the three MBA majors.
μ SST (Sum of Squares for Treatments) – between-treatments variation
Point Est. single value – P(x)=0, sample n , Not reflect to Parameter value Larger SST larger variatiXn b support 1 | SST = 0 Xll
Interval Est. range of values are equal
Desirable: Unbiasedness ( ), Consistency ( , n , variance ) SSE (Sum

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