/*! 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} Q10 In Exercises 7鈥10, use the sam... [FREE SOLUTION] | 91影视

91影视

In Exercises 7鈥10, use the same population of {4, 5, 9} that was used in Examples 2 and 5. As in Examples 2 and 5, assume that samples of size n = 2 are randomly selected with replacement.

Sampling Distribution of the Sample Proportion

a. For the population, find the proportion of odd numbers.

b. Table 6-2 describes the sampling distribution of the sample mean. Construct a similar table representing the sampling distribution of the sample proportion of odd numbers. Then combine values of the sample proportion that are the same, as in Table 6-3. (Hint: See Example 2 on page 258 for Tables 6-2 and 6-3, which describe the sampling distribution of the sample mean.)

c. Find the mean of the sampling distribution of the sample proportion of odd numbers.

d. Based on the preceding results, is the sample proportion an unbiased estimator of the population proportion? Why or why not?

Short Answer

Expert verified

a. Population Proportion: 0.67

b. The following table represents the sampling distribution of the sample proportions.

Sample

Sample proportion

Probability

(4,4)

0

19

(4,5)

0.5

19

(4,9)

0.5

19

(5,4)

0.5

19

(5,5)

1

19

(5,9)

1

19

(9,4)

0.5

19

(9,5)

1

19

(9,9)

1

19

By combining all the same values of proportions, the following table is obtained.

Sample proportion

Probability

0.0

19

0.5

49

1.0

49

c.The mean of the sampling distribution of the sample proportions is equal to 0.67.

d. Since the mean value of the sampling distribution of the sample proportion is equal to the population proportion, the sample proportion of odd numbers can be considered an unbiased estimator of the population proportion of odd numbers.

Step by step solution

01

Given information

A population of ages of three children is considered. Samples of size equal to 2 are extracted from this population with replacement.

02

Population proportion

a.

The observations are {4,5,9}.

The total number of values (n) is equal to3.

The number of odd values (x) is equal to 2.

The population proportion of odd numbers is equal to

p=xn=23=0.67

Thus, the population proportion of odd numbers is equal to 0.67.

03

Sampling distribution of sample proportions

b.

All possible samples of size 2 selected with replacement are tabulated below.

(4,4)

(4,5)

(4,9)

(5,4)

(5,5)

(5,9)

(9,4)

(9,5)

(9,9)

The number of odd values in each of the nine samples is tabulated below.

Sample

Number of odd values

(4,4)

0

(4,5)

1

(4,9)

1

(5,4)

1

(5,5)

2

(5,9)

2

(9,4)

1

(9,5)

2

(9,9)

2

The following formula is used to compute the sample proportions:

p^=NumberofoddnumbersSampleSize

Since there are nine samples, the probability of the nine sample proportions is written as 19.

The following table shows all possible samples of size equal to 2, the corresponding sample proportions, and the probability values.

Sample

Sample proportion

Probability

(4,4)

p^1=02=0

19

(4,5)

p^2=12=0.5

19

(4,9)

p^3=12=0.5

19

(5,4)

p^4=12=0.5

19

(5,5)

p^5=22=1

19

(5,9)

p^6=22=1

19

(9,4)

p^7=12=0.5

19

(9,5)

p^8=22=1

19

(9,9)

p^9=22=1

19

By combining the values of proportions that are the same, the following probability values are obtained.

Sample Median

Probability

0

19

0.5

49

1

49

04

Mean of the sample proportions

c.

The mean of the sample proportions is computed below:

Meanofp^=p^1+p^2+.....+p^99=0+0.5+......+19=0.67

Thus, the mean of the sampling distribution of the sample proportion is equal to 0.67.

05

Unbiased estimator

d.

An unbiased estimator is a sample statistic whose sampling distribution has a mean value equal to the population parameter.

The mean value of the sampling distribution of the sample proportion (0.67) is not equal to the population proportion (0.67).

Thus, the sample proportion of odd numbers can be considered an unbiased estimator of the population proportion of odd numbers.

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

In Exercises 9鈥12, find the area of the shaded region. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1.

Critical Values. In Exercises 41鈥44, find the indicated critical value. Round results to two decimal places.

z0.15

In Exercises 21鈥24, use these parameters (based on Data Set 1 鈥淏ody Data鈥 in Appendix B):鈥⑩侻en鈥檚鈥俬eights鈥俛re鈥俷ormally鈥俤istributed鈥倃ith鈥俶ean鈥68.6鈥俰n.鈥俛nd鈥俿tandard鈥俤eviation鈥2.8鈥俰n.鈥⑩俉omen鈥檚 heights are normally distributed with mean 63.7 in. and standard deviation 2.9 in. Executive Jet Doorway the Gulfstream 100 is an executive jet that seats six, and it has a doorway height of 51.6 in.

a. What percentage of adult men can fit through the door without bending?

b. Does the door design with a height of 51.6 in. appear to be adequate? Why didn鈥檛 the engineers design a larger door?

c. What doorway height would allow 40% of men to fit without bending?

Continuous Uniform Distribution. In Exercises 5鈥8, refer to the continuous uniform distribution depicted in Figure 6-2 and described in Example 1. Assume that a passenger is randomly selected, and find the probability that the waiting time is within the given range.

Between 2 minutes and 3 minutes

Standard Normal DistributionIn Exercises 17鈥36, assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. In each case, draw a graph, then find the probability of the given bone density test scores. If using technology instead of Table A-2, round answers

to four decimal places.

Between -2.55 and -2.00.

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.