Problem 1. Agri-Beef Inc. operates cattle feed lots in several midwestern states

Question

Problem 1.    

Agri-Beef Inc. operates cattle feed lots in several midwestern states.
The company wishes to estimate the average daily weight gain
of cattle on their lots.  To do this a simple random sample
of cattle is taken and the daily weight gain is recorded.
The data below, in pounds, represent the results of the sample.     

  8.7

  6.1

  6.1

  9.8

  6.5

  8.3

  6.3

  5.8

  8.0

  8.7

  5.1

  7.0

  6.0

  7.9

  6.1

        a.    Construct a 95% confidence interval estimate for the true daily weight gain.
        b.    Construct a 99% confidence interval estimate for the true daily weight gain.
        c.    Discuss the differences between the two estimates found in parts a and b
                 and indicate the advantages and disadvantages of each.

Problem 2

A survey of 400 women for the American Orthopedic Foot and Ancle Society revealed that 38% wear flats to work.

    a.  Use this sample information to develop a 99% confidence interval for the population proportion of women who wear flats to work.

    b.  Suppose the society wished to estimate the proportion of women who wear athletic shoes to work and the proportion of women who wear flats to work within a margin of error of 0.01 (one percentage point) with 95% confidence.  Determine the sample size required.

Explanation / Answer

(1) (a) Given a=0.05, t(0.025, df=n-1=14) =2.14 (from student t table)

So 95% confidece interval is

xbar+/- t*s/vn

--> 7.093333+/- 2.14*1.368245/sqrt(15)

--> (6.337315, 7.849351)

(b) Given a=0.01, t(0.005, df=14)= 2.98

So 99% confidence interval is

xbar+/- t*s/vn

--> 7.093333+/- 2.98*1.368245/sqrt(15)

--> (6.040561, 8.146105)

(c) The width of the interval in part a is smaller than in part b

advantages: the width of interval in part a is smaller

disadvantages: there is more chances to reject the null hypothesis is part a.

(2) Given a=0.01, Z(0.005)=2.58 (from standard normal table)

So 99% confidece interval is

p +/-Z*sqrt(p*(1-p)/n)

--> 0.38+/- 2.58*sqrt(0.38*(1-0.38)/400)

--> (0.3173851, 0.4426149)

Given a=0.05, Z(0.025)=1.96

So n=(Z/E)^2*p*(1-p)

=(1.96/0.01)^2*0.38*(1-0.38)

=9050.81

Take n=9051

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