Soln_Assn1_ADM2304X_May11

Soln_Assn1_ADM2304X_May11 - Dr. Suren Phansalker ADM2304 X...

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Dr. Suren Phansalker ADM2304 X Assignment#1 Solution 115 Marks May 28, 2011 Qu.#1 (15 Marks) Here: 657 ˆ 0.8760 750 X p n == = , Assume LS = α = 0.05, CC = 1- α = 0.95 S1: 4 Marks 00 : ( 0.9) Hp p 0 : ( 0.9) a ≠= : Two Tailed Test {Not accurate Î Less than or More than} (You must test ‘p’ and not ‘ ’) ˆ p S2: 7 Marks N.B.: Here, n p = 50 (0.125) = 6.25 > 5.0, and 0 n q 0 = 50 (0.95) = 43.75 > 5.0. Thus “Normal” distribution is, strictly speaking reasonable. {3 Marks} 0.9(0.1) ˆ ( ) (Pr ) 0.0110 750 pq p SD oportion n σ = = {2 Marks} 0 ˆ 0.876 0.900 2.1909 ˆ ( ) 0.0110 Calc pp z p = {2 Marks} S5: 4 Marks p-Val = P[Z > {|z Calc | = 2.1909] x 2 = P[Z > 2.1909] x 2 = (1.0 – 0.9857) x 2 = 0.0286 Since {p-Val = 0.0286} < { α = 0.05} Î Reject H0 You could also use the equivalent critical value approach shown below. S3: LS = α = 0.05 and with the Two Tailed Test, / 2 0.025 1.96 Crit zzz α = S4: Since {|z Calc | = 2.1909} > {z Crit = 1.96} Î Reject H . {Just as before} 0 Qu.#2 (38 Marks) a. -- 6 Marks: 4 for one of the plots, 2 for comments. Descriptive Statistics: BMI Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 BMI 2500 0 28.006 0.120 5.993 8.230 23.960 28.000 32.040 Variable Maximum BMI 46.650 1
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Histogram (with Normal Curve) of BMI 45 40 35 30 25 20 15 10 200 150 100 50 0 BMI Frequency Mean 28.01 StDev 5.993 N 2500 Histogram (with Normal Curve) of BMI Or, 50 40 30 20 10 0 99.99 99 95 80 50 20 5 1 0.01 Percent Mean 28.01 StDev 5.993 N 2500 AD 0.242 P-Value 0.769 Probability Plot of BMI Normal - 95% CI Here: N = 2500 μ = 28.006 units of BMI σ = 5.993 units of BMI Here the Histogram as well as the Probability Plot indicates that the population is quite “Normal”. The same can be ascertained by drawing the Boxplot and finding that the population standard deviation, σ 6 R . b. -- 4 Marks for correct value Descriptive Statistics: H_Obese Variable N N* Mean SE Mean StDev Minimum Q1 Median H_Obese 2500 0 0.12720 0.00667 0.33326 0.00000 0.00000 0.00000 Variable Q3 Maximum H_Obese 0.00000 1.00000 2
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Since ‘1’ indicates BMI of 35 units or more, ‘p’ = Population Proportion of people having BMI of 35 or more units is found by simply obtaining the mean of these 2500 ‘0’s and ‘1’s. There are 318 adults out of 2500 with BMI 35 or more units. 318 0.1272 2500 p == c. --8 Marks: 6 for CI, 2 for MiniTab approach. 50, 28.752, 5.877, 1 0.90, 0.10 nX s C C αα = = = = Here: /2 0.10/ 2 0.05 ( 1) (49) (49) 1.6766 tn t t α −= = = 7.6173 ( 28.3318 1.6766 28.3318 1.8061 50 s Xt n n ± CI: CI: (26.5257, 30.1379) This CI is verified by using MiniTab as shown below.
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Soln_Assn1_ADM2304X_May11 - Dr. Suren Phansalker ADM2304 X...

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