A random sample of n=900 observations is selected from a population with μ=100andσ=10

a. What are the largest and smallest values ofx¯ that you would expect to see?

b. How far, at the most, would you expect xto deviate from μ?

c. Did you have to know μto answer part b? Explain.

Short Answer

Expert verified

a.Thesmallestvalueofxis99,andthelargestvalueofxis101.b.Itwouldnotbeexpectedthatxdeviatesfromμc.No,becausepreviousansweronlydependentonthestandarddeviationofthesamplingdistributionofthesamplemean,butnotthemeanitself.

Step by step solution

01

Given information

Arandomsampleofn=900observationsisselectedfromapopulationwithμ=100andσ=10

02

Finding the largest and smallest values of x

a.

Here, the sample mean is μx=μand the sample standard deviation is σx=σn

Therefore,

mx=100,

σx=10900=1030=0.33

Since, almost all of the time, the sample mean will be within three standard deviations of the mean.

So,

μ±3σ=100±30.33=100-0.99,100+0.9999,101

Therefore, the smallest value of xis 99, and the largest value of xis 101.

03

Checking whether deviation of X from μ expect or not.

b.

No, because if more than three standard deviation then

3σx=313=1

Therefore, It would not be expected that that xdeviates fromμ.

04

Checking whether μ is necessary to answer part (b) or not.

c.

No, because previous answer only dependent on the standard deviation of the sampling distribution of the sample mean, but not the mean itself.

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Most popular questions from this chapter

Voltage sags and swells. Refer to the Electrical Engineering (Vol. 95, 2013) study of the power quality (sags and swells) of a transformer, Exercise 2.76 (p. 110). For transformers built for heavy industry, the distribution of the number of sags per week has a mean of 353 with a standard deviation of 30. Of interest is , that the sample means the number of sags per week for a random sample of 45 transformers.

a. FindEχ¯ and interpret its value.

b. FindVarχ¯.

c. Describe the shape of the sampling distribution ofχ¯.

d. How likely is it to observe a sample mean a number of sags per week that exceeds 400?

Question: The standard deviation (or, as it is usually called, the standard error) of the sampling distribution for the sample mean, x¯ , is equal to the standard deviation of the population from which the sample was selected, divided by the square root of the sample size. That is

σX¯=σn

  1. As the sample size is increased, what happens to the standard error of? Why is this property considered important?
  2. Suppose a sample statistic has a standard error that is not a function of the sample size. In other words, the standard error remains constant as n changes. What would this imply about the statistic as an estimator of a population parameter?
  3. Suppose another unbiased estimator (call it A) of the population mean is a sample statistic with a standard error equal to

σA=σn3

Which of the sample statistics,x¯or A, is preferable as an estimator of the population mean? Why?

  1. Suppose that the population standard deviation σis equal to 10 and that the sample size is 64. Calculate the standard errors of x¯and A. Assuming that the sampling distribution of A is approximately normal, interpret the standard errors. Why is the assumption of (approximate) normality unnecessary for the sampling distribution ofx¯?

Question: Refer to Exercise 5.3.

a. Find the sampling distribution of s2.

b. Find the population variance σ2.

c. Show that s2is an unbiased estimator of σ2.

d. Find the sampling distribution of the sample standard deviation s.

e. Show that s is a biased estimator of σ.

Refer to Exercise 5.3 and find E=(x)=μ. Then use the sampling distribution ofxfound in Exercise 5.3 to find the expected value ofx. Note thatE=(x)=μ.

Refer to Exercise 5.3.

  1. Show thatxis an unbiased estimator of.
  2. Findσx2.
  3. Find the probability that x will fall within2σxofμ.
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