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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 as 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 of 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 part (c) and (d). Why are they different?

Short Answer

Expert verified

Part a. The answers can vary depending on the first number selected.

Part b. Mean \(=4.5\)

Standard Deviation \(=2.8733\)

Part c. Mean \(=4.5\)

Standard Deviation \(=1.4367\)

Part d. The answers can vary depending on the first number selected.

Part e. Sampling Error

Step by step solution

01

Part a. Step 1. Explanation

\(n=4\)

Now, generate four digits between \(0\) and \(9\) using random number table.

Here the initial \(4\) digits in row \(15\) in the random number table are selected as follows: \(0,7,8,5\)

Mean can be calculated as the summation of all values divided by the quantity of values

\(\bar{x}=\frac{0+7+8+5}{4}=\frac{20}{4}=5\)

After this, the above exercise can be repeated 50 times using different rows and columns or a random number generator can also be used to obtain the required results.

02

Part b. Step 1. Calculation

\(n=4\)

POPULATION PARAMETERS

In the given question the population consists of digits \(0,1,2,3,4,5,6,7,8,9\)

The population mean can be calculated as the summation of all values divided by the quantity of values.

\(\mu=\frac{0+1+2+3+4+5+6+7+8+9}{10}=\frac{45}{10}=4.5\)

The variance could be calculated as the sum of divisions already squared with respect to the mean divided by number of observations

\(\sigma =\sqrt{\frac{(0-4.5)^{2}+...+(9-4.5)^{2}}{10}}=\sqrt{\frac{82.5}{10}}\approx 2.8733\)

03

Part c. Step 1. Calculation

The mean of the distribution is sampling of the sample mean is same as the population mean

\(\mu_{x}=\mu=4.5\)

\(\sigma _{x}=\frac{\sigma}{\sqrt{4}}=\frac{2.8733}{\sqrt{4}}\approx1.4367\)

04

Part d. Step 1. Calculation

\(\bar{x}=\frac{5+4.75+3.75+...+4.25+3.75+4.5}{50}=4.54\)

The standard deviation is square root of the sum of divisions already squared with respect to the mean divided by \(n-1\)

\(s=\sqrt{\frac{(5-4.54)^{2}+...+(4.5-4.54)^{2}}{50-1}}\approx1.4596\)

05

Part e. Step 1. Explanation

The main difference between the results is parts (c) and (d) is because of sampling error. To be more precise, different samples would have different means and different standard deviations even though the values would be very close to the expected values.

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Most popular questions from this chapter

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.

Part (a): Is the sample mean an unbiased estimator of the population mean? Explain your answer.

Part (b): Is the sample median an unbiased estimator of the population mean? Explain your answer.

Early-Onset Dementia. Dementia is the loss of intellectual and social abilities severe enough to interfere with judgment, behavior, and daily functioning. Alzheimer's disease is the most common type of dementia. In the article "Living with Early Onset Dementia: Exploring the Experience and Developing Evidence-Based Guidelines for Practice" (Al=hcimer's Care Quarterly, Vol. 5, Issue 2, pp. 111-122), P. Harris and J. Keady explored the experience and struggles of people diagnosed with dementia and their families. If the mean age at diagnosis of all people with early-onset dementia is 55 years, find the probability that a random sample of 21 such people will have a mean age at diagnosis less than 52.5 years. Assume that the population standard deviation is 6.8 years. State any assumptions that you are making in solving this problem.

7.35 Refer to Exercise 7.5 on page 295 .

a. Use your answers from Exercise 7.5(b) to determine the mean, μi. of the variable x¯ for each of the possible sample sizes.

b. For each of the possible sample sizes, determine the mean, μ5, of the variable x¯, using only your answer from Exercise 7.5(a).

Population data: 1,2,3,4,5

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.

7.43 NBA Champs. Repeat parts (b) and (c) of Exercise 7.41 for samples of size 3. For part (b), use your answer to Exercise 7.13(b).

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