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7.58 Class Project Simulation. This exercise can be done individually or, better yet, as a class project.
a. Use a random-number table or random-number generator to obtain a sample (with replacement) of four digits between 0 and 9 .
Do so a total of 50 times and compute the mean of each sample.
b. Theoretically, what are the mean and standard deviation of all possible sample means for samples of size 4?
c. Roughly what would you expect the mean and standard deviation of the 50 sample means you obtained in part (a) to be? Explain your answers.
d. Determine the mean and standard deviation of the 50 sample means you obtained in part (a).
e. Compare your answers in parts (c) and (d). Why are they different?

Short Answer

Expert verified

(a) Depending on the first number chosen, the solutions may differ.
(b) The mean and standard deviation of all possible sample means for samples of size 4are 4.5and 2.8733.

(c) The mean and standard deviation for sample mean 50are 4.5and 1.4367.

(d) Depending on the first number chosen, the solutions may differ.

(e) Parts (c) and (d) have the most significant differences due to sampling error.

Step by step solution

01

Part (a) Step 1: Given information

To use a random-number table or random-number generator to obtain a sample (with replacement) of four digits between 0and 9.

02

Part (a) Step 2: Explanation

Let, the given value:
n=4
Using the random number table, produce four digits between 0and 9. In the random number table, the first four digits in row 15 are chosen as follows:
0,7,8,5
The sum of all values divided by the number of values can be used to compute the mean.
x¯=0+7+8+54

=204

=5

Repeat the previous exercise 50times with different rows and columns, or use a random number generator to get the desired results.

As a result, depending on the first number chosen, the solutions may differ.

03

Part (b) Step 1: Given information

To find the mean and standard deviation of all possible sample means for samples of size 4.

04

Part (b) Step 2: Explanation

Let, the given value:

n=4

The population in the provided question is made up of digits 0,1,2,3,4,5,6,7,8,9. The population mean can be found by multiplying the total of all values by the number of values.
μ=0+1+2+3+4+5+6+7+8+910

=4510

=4.5

The variance can be computed by dividing the number of observations by the sum of divisions already squared with regard to the mean.
σ=(0-4.5)2+…+(9-4.5)210

=82.510

≈2.8733

As a result, the mean and standard deviation are 4.5 and 2.8733.

05

Part (c) Step 1: Given information

To calculate approximately the mean and standard deviation of the 50 sample means that obtained in part (a).

06

Part (c) Step 2: Explanation

The sample mean is the same as the population mean because the mean of the distribution is sampling.
The mean is determined as:
μx¯=μ

=4.5

And the standard deviation is determined as:

σx¯=σn

=2.87334

≈1.4367

As a result, the mean and standard deviation for sample size 50 are 4.5and 1.4367.

07

Part (d) Step 1: Given information

To determine the mean and standard deviation of the 50 sample means that obtained in part (a).

08

Part (d) Step 2: Explanation

Determine the standard deviation as follows:

x¯=5+4.75+3.75+….+4.25+3.75+4.550

=4.54

The square root of the sum of divisions already squared with regard to the mean divided by n-1 is the standard deviation.
s=(5-4.54)2+…+(4.5-4.54)250-1

≈1.4596

Depending on the first number chosen, the solutions may differ.

09

Part (e) Step 1: Given information

To compare your answers in parts (c) and (d). And determine that how they different.

10

Part (e) Step 2: Explanation

Since,the parts (c) and (d) have the most significant differences due to sampling error.
To be more specific, even though the numbers are fairly near to the expected values, various samples will have different means and standard deviations.

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Most popular questions from this chapter

Does the sample size have an effect on the mean of all possible sample mean? Explain your answer.

Population data: 1,2,3

Part (a): Find the mean, μ,of the variable.

Part (b): For each of the possible sample sizes, construct a table similar to Table 7.2on the page 293and draw a dotplot for the sampling for the sampling distribution of the sample mean similar to Fig 7.1on page 293.

Part (c): Construct a graph similar to Fig 7.3and interpret your results.

Part (d): For each of the possible sample sizes, find the probability that the sample mean will equal the population mean.

Part (e): For each of the possible sample sizes, find the probability that the sampling error made in estimating the population mean by the sample mean will be 0.5or less, that is, that the absolute value of the difference between the sample mean and the population mean is at most 0.5.

Explain why increasing the sample size tends to result in a smaller sampling error when a sample means is used to estimate a population mean.

7.54 Unbiased and Biased Estimators. A statistic is said to be an unbiased estimator of a parameter if the mean of all its possible values equals the parameter. otherwise, it is said to be a biased estimator. An unbiased estimator yields, on average, the correct value of the parameter, whereas a biased estimator does not.
a. Is the sample mean an unbiased estimator of the population mean? Explain your answer.
b. Is the sample median an unbiased estimator of the population median? (Hint: Refer to Example 7.2 on pages 292-293. Consider samples of size 2.)

Baby Weight. The paper "Are Babies Normal?" by T. Clemons and M. Pagano (The American Statistician, Vol. 53, No, 4. pp. 298-302) focused on birth weights of babies. According to the article, the mean birth weight is 3369 grams (7 pounds, 6.5 ounces) with a standard deviation of 581 grams.
a. Identify the population and variable.
b. For samples of size 200, find the mean and standard deviation of all possible sample mean weights.
c. Repeat part (b) for samples of size 400.

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