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Final

# ADMS 2320 Formula Sheet for Final Exam Page 2.doc

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School
York University
Department
Course
Professor
Marwan Hassan
Semester
Winter

Description
no x y xy x y 2 2 Chapter 14 Analysis Of Variance (ANOVA): Three Interval Data Contingency table: Use to infer two nominal variables are related and differences E( p − p ) = p − p exist btw two or more populations of nominal variables. Cell Expected values > 5 Chapter 10 Estimator: appr. Pop. Parameters from Sample StatisXic. e.g. n1= 4, 1 n 2 4, n3= 4 2 μ H 0 The two variables (responses and employee group) are independent (no  Point Est.  single value – P(x)=0, sample n , Not reflect to Parameter 1alue25, x2= 3.25, x 3 5.75 relationship between 2 variables) Response Lo Mi Top  Interval Est.  range of values H 1 The two variables are dependent w d x − ( x) / n (2 variables are related) For 67 32 11 110  Desirable: UnbiasednesE ( X ) = μnsistencyV (X ) = σ n , / niance ) 2 ∑ ∑ 2 2 Against 63 18 9 90 and Relative Efficiency (smaller variance is means more efficient) s1= = 2.25, s2= 2.92 s3= 3.92 Observed Expected 130 50 Summation 200  Larger confidence level produces a wider confidence interval. σ n −1 Cell i Frequency Frequency Difference Component  Larger values of σ produce wider confidence intervals. X ± Z 9 +13 + 23 f e (f – e) (f – e) /e  Larger sample size narrow confidence interval. α / 2 n x = = 3.75 i i i i 2 i i 1 67 110(130)/200 = 71.50 -4.5 (-4.52 /71.5= 0.2832 Chapter 11 Hypothesis Test: Assume null H is true, find evident to infer alt. H 4 + 4 + 4 2 32 110(50)/200 = 27.50 4.5 4.5 /27.5= 0.7364 0 1 H : μ = μ = μ H : At least one mean is different 3 11 110(20)/200 = 11.00 0 0  If p-value is smaller than α  Reject the null hypothesis. 0 1 2 3 1 4 63 90(130)/200 = 58.50 4.5 4.5 /58.5= 0.3462  The smaller the p-value, the more evidence to support the alternative hySST =sis. n (x − x ) = 4(2.25 - 3.75) + 4(3.25 - 3.75) + 4(5.75 - 3.75) 18 90(50)/200 == 26.00 -4.5 (-4.5) /22.5= 0.9000  Type I Error  Rejec0 H (null hypothesis) 0hen H is true. P(Type I error) = α ∑ j j 6 9 90(20)/200 = 9.00 0 0  Type II Error  Do not rej0ct H wh0n H is false. P(Type II error) = β 2 Total 2.2658  Decreasing significance level α, increases the vaα  β  | α  β  SSE = ∑ (n −j1)s j =(4 1)(2.25) + (4 1)(2.92) + (4 1)(2.92) = 24.25 Rejection regionχ 2 > χ 2 = χ 2 = 5.99147  Increasing sample size n, the value of β will decn  β  ANOVA table α ,( r−1)(c−1) .05 ,2  Decreasing sample size decreases test-stat & increase p-value. n  - z  p  t  d. f. Sum of Squares Mean Square k 2  The power of a test is defined as 1– β  P(reject null hypothesis when it’s false) SST 26 2 ( f i e )i  Probability of Type II Error: Treatments k–1 = SST = 26.00 MST = = = 13 Value of Test Statisχ =: ∑ = 2.2658 E.g. α = 5%, n = 100, σ = 10, H : μ =1200, H : μ ≠ 200, given μ = 203 2 k −1 3 −1 i=1 e i x − μ x − 200 SSE 24.25 Conclusion: Do not Rejec0 H , there is not enough evidence to infer that responses R.R.: < −Z σ /2 x < 198< −1.96 Error n–k = SSE = 24.24 MSE = = = 2.6944 among the three groups of employees. σ / n 10 / 100 -4.96 -1.04 9 n − k 12 − 3 x − μ x − 200 Total n–1 = SS (Total) = or > Z α /2 x > 201> 1.96 11 50.25 σ / n 10 / 100 198 .04 − 203 Rej201 .96 − 203:> F = F = 4.26 = P( < Z < )α ,k −1,n − k .05 ,2,9 β = P(198 .04 < x < 201 .96, given μ = 203) MST 13 10 / 100 The Value of Test StatiF =c: = = 4.82 β = P(−4.96 < Z < −1.04) = .5 − .3508 = .1492 MSE 2.69 Chapter 12: Inference about a Population when σ is unknown  use t-test Conclusion: Reject0H , there is enough evidence to conclude that there are differences in the number of job offers between the three MBA majors.  t-test only valid when the histogram is NOT extremely nonnormal.  SST (Sum of Squares for Treatments) – between-treatments variation  Sampling distributionPof is appr. normal when np > 5 & n(1–p) ≥ 5  Larger SST  larger variation btw  support H | SST = 0  all X 1 X are equal Chapter 13 Comparing Two Populations: Dif. btw 2 mean/prop. | ratio of tw SSE (Sum of Squares for Error) – within
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