/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q9E Question: Consider the following... [FREE SOLUTION] | 91影视

91影视

Question: Consider the following probability distribution:

a. Calculate for this distribution.

b. Find the sampling distribution of the sample meanxfor a random sample of n = 3 measurements from this distribution, and show thatxis an unbiased estimator of .

c. Find the sampling distribution of the sample median x for a random sample of n = 3 measurements from this distribution, and show that the median is a biased estimator of .

d. If you wanted to use a sample of three measurements from this population to estimate , which estimator would you use? Why?

Short Answer

Expert verified

a) =5

b) xis not an unbiased estimator of role="math" localid="1658118180792" .

c) m is not an unbiased estimator of role="math" localid="1658118199646" .

d) None

Step by step solution

01

Calculation of the mean μ in part (a)

The calculation of the mean in case of the three values of x is shown below:

=xp(x)=213+413+913=23+43+93=153=5

Therefore the value ofis 5.

02

Determining whether x is an unbiased estimator of μ

The calculation of the probabilities of the means considering the three values of x is shown below:

Samples

Means

Probability

2,2,2

2

131313=127

2,4,2

2.67

131313=127

2,9,2

4.33

131313=127

2,2,4

2.67

data-custom-editor="chemistry" 131313=127

2,2,9

4.33

131313=127

2,4,4

3.33

131313=127

2,9,9

7

131313=127

4,4,4

4

131313=127

4,2,4

3.33

131313=127

4,9,4

5.67

131313=127

4,4,2

3.33

131313=127

4,4,9

5.67

131313=127

4,2,2

2.67

131313=127

4,9,9

7.33

131313=127

9,9,9

9

131313=127

9,2,9

6.67

131313=127

9,4,9

7.33

131313=127

9,9,2

6.67

131313=127

9,9,4

7.33

131313=127

9,2,2

4.33

131313=127

9,4,4

5.67

131313=127

2,4,9

5

131313=127

2,9,4

5

131313=127

4,2,9

5

131313=127

4,9,2

5

131313=127

9,2,4

5

131313=127

9,4,2

5

131313=127

Now the calculation ofis shown below:

Ex=xpx=2127+2.67127+4.33127+2.67127+4.33127+3.33127+7127+4127+3.33127+5.67127+3.33127+5.67127+2.67127+7.33127+9127+6.67127+7.33127+6.67127+7.33127+4.33127+5.67127+5127+5127+5127+5127+5127+5127=1272+2.67+4.33+2.67+4.33+3.33+7+4+3.33+5.67+3.33+5.67+2.67+7.33+9+6.67+7.33+6.67+7.33+4.33+5.67+5+5+5+5+5+5=10427=3.85

As E(x)is 3.85 and is 5,x is not an unbiased estimator of .

03

Determining whether m is an unbiased estimator of μ

The list of medians along with the associated probabilities is shown below:

Samples

Medians

Probability

2,2,2

2

2127=227

2,4,2

2

2127=227

2,9,2

2

2127=227

2,2,4

2

2127=227

2,2,9

2

2127=227

2,4,4

4

4127=427

2,9,9

9

4127=427

4,4,4

4

4127=427

4,2,4

4

4127=427

4,9,4

4

4127=427

4,4,2

4

4127=427

4,4,9

4

4127=427

4,2,2

2

2127=227

4,9,9

9

9127=927

9,9,9

9

9127=927

9,2,9

9

9127=927

9,4,9

9

9127=927

9,9,2

9

9127=927

9,9,4

9

9127=927

9,2,2

2

2127=227

9,4,4

4

4127=427

2,4,9

4

4127=427

2,9,4

4

4127=427

4,2,9

4

4127=427

4,9,2

4

4127=427

9,2,4

4

4127=427

9,4,2

4

4127=427

The summation of the medians are shown below:

E(m)=mp(m)=227+227+227+227+227+227+227+427+427+427+427+427+427+427+427+427+427+427+427+427+927+927+927+927+927+927=12927=4.78

As E(m)is 4.78 and is 5, m is not an unbiased estimator of .

04

Determining the better estimator of μ

It has been found from the above calculations that E(m)andxare not unbiased estimators of . Therefore, it can be envisaged that in this context, there are not proper estimators of .

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose a random sample of n measurements is selected from a population with u=100mean and variance role="math" localid="1657967387987" 2=100. For each of the following values of n, give the mean and standard deviation of the sampling distribution of the sample mean.

  1. role="math" localid="1657967260825" n=4
  2. n=25
  3. n=100
  4. n=50
  5. n=500
  6. n=1000

Question:Quality control. Refer to Exercise 5.68. The mean diameter of the bearings produced by the machine is supposed to be .5 inch. The company decides to use the sample mean from Exercise 5.68 to decide whether the process is in control (i.e., whether it is producing bearings with a mean diameter of .5 inch). The machine will be considered out of control if the mean of the sample of n = 25 diameters is less than .4994 inch or larger than .5006 inch. If the true mean diameter of the bearings produced by the machine is .501 inch, what is the approximate probability that the test will imply that the process is out of control?

Analysis of supplier lead time. Lead timeis the time betweena retailer placing an order and having the productavailable to satisfy customer demand. It includes time for placing the order, receiving the shipment from the supplier, inspecting the units received, and placing them in inventory. Interested in average lead time,, for a particular supplier of men鈥檚 apparel, the purchasing department of a national department store chain randomly sampled 50 of the supplier鈥檚 lead times and found= 44 days.

  1. Describe the shape of the sampling distribution ofx.
  2. If and are really 40 and 12, respectively, what is the probability that a second random sample of size 50 would yieldx greater than or equal to 44?
  3. Using the values for and in part b, what is the probability that a sample of size 50 would yield a sample mean within the interval 2n?

Downloading 鈥渁pps鈥 to your cell phone. Refer toExercise 4.173 (p. 282) and the August 2011 survey by thePew Internet & American Life Project. The study foundthat 40% of adult cell phone owners have downloadedan application (鈥渁pp鈥) to their cell phone. Assume thispercentage applies to the population of all adult cell phoneowners.

  1. In a random sample of 50 adult cell phone owners, howlikely is it to find that more than 60% have downloadedan 鈥渁pp鈥 to their cell phone?
  2. Refer to part a. Suppose you observe a sample proportionof .62. What inference can you make about the trueproportion of adult cell phone owners who have downloadedan 鈥渁pp鈥?
  3. Suppose the sample of 50 cell phone owners is obtainedat a convention for the International Association forthe Wireless Telecommunications Industry. How willyour answer to part b change, if at all?

Suppose a random sample of n = 500 measurements is selected from a binomial population with probability of success p. For each of the following values of p, give the mean and standard deviation of the sampling distribution of the sample proportion,p^.

  1. p= .1
  2. p= .5
  3. p= .7
See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.