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DMD_StatisticalSampling_Maria

By Dean Russell,2014-06-13 09:27
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    Lingnan MBA Center

    Lingnan (University) College, Sun Yat-sen University

    Assignment Form

Course: DMD

Instructor: Fu Qing

    P190 Ex 4.6

    Homework:

    P193 Ex 4.19

Name: 陈明婧 Maria Chen

Student ID No.: 11210332

Class: 11PA

STUDENT DECLARATION

I declare that this assignment is my own work, which all sources of reference are acknowledged in full

    and it has not been submitted for any other course.

     Signature:…………………… Date:…………

    Lingnan MBA Center

    Lingnan (University) College, Sun Yat-sen University

P191 Ex 4.6

    Answer:

    From the exercise, we know:

    n=15, =11.32, s=4.4 x

    (a) Construct a 90% confidence interval for the mean amount invested by all

    institutional investors in the mutual fund.

    x follow t-Distribution with k = n-1 = 14 degrees of freedom, then c = 1.761,

    nL = c*s /= 1.761*4.4 /15= 2

    Therefore, the answer for (a) question is

    [-L, +L]= [11.32 - 2, 11.32 + 2] = [9.32, 13.32] xx

    (b) For confidence level = 95%, c=1.96, L=0.5 million,

    222222Therefore, n= cs/L = 1.96 * 4.4 /0.5 = 297.45

    Round this number up to 298

    Lingnan MBA Center

    Lingnan (University) College, Sun Yat-sen University

P193 Ex 4.19

    Answer:

    From the exercise, we know:

    = 95%, L = 5%

    (a) What is the size of the sample that would need to be surveyed?

    2222For= 95%, c = 1.96, n = c / 4 L = 1.96 / 4*0.05 = 384.16

    Round this number up to 385

    (b) The distribution of X is binomial distribution.

    (c) What is the expected value of X?

    E (X) = n*p = 500*(1-10%) = 500*90% = 450

    (d) What is the standard deviation of X?

    = 6.7 ;;;;npp(1)500*(110%)*10%x

    (e) What is the approximate probability that X is larger than 400? For this question, we Z follow Standard Normal Distribution, then P (X>400) = 1 P (Z<(400-450)/6.7) = 1-0 = 1=100%

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