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Chapter 6: Q7E (page 339)

Explain the difference between an interval estimator and a point estimator for 渭

Short Answer

Expert verified

A random variable is an estimator, as well as a value, is an estimation that represents the estimator's calculated number.

Step by step solution

01

Interval estimation

In comparison to point estimate, which is a specific number, interval estimation uses sample information to determine a range of potential (as well as likely) numbers for an unbiased estimator.

02

Difference between interval estimator and point estimator

A point estimatoris a single data point computed from the sampling that predicts a targeted population variable.

A population variable point estimate is a technique as well as an equation that informs us how to utilize sampling information to construct a specific figure that may be used as a goal variable estimation.

An interval estimator(also known as a confidence level) is a method that informs us on how to generate an interval that predicts the targeted variable using data samples.

The two types of population parameter estimates depending on sampling information are point as well as interval estimations. The point estimate is a simple process. The interval estimate, on the other hand, is a considerably more reliable as well as useful technique than the point estimates.

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Most popular questions from this chapter

Lobster trap placement. Refer to the Bulletin of MarineScience(April 2010) study of lobster trap placement,Exercise 6.29 (p. 348). Recall that you used a 95% confidenceinterval to estimate the mean trap spacing (in meters)for the population of red spiny lobster fishermen fishing inBaja California Sur, Mexico. How many teams of fishermenwould need to be sampled in order to reduce the width ofthe confidence interval to 5 meters? Use the sample standarddeviation from Exercise 6.29 in your calculation.

Question: Let t0 be a specific value of t. Use Table III in Appendix D to find t0 values such that the following statements are true.

a.P(t鈮�t0)=.025wheredf=11b.P(t鈮�t0)=.01wheredf=9c.P(t鈮�t0)=.005wheredf=6d.P(t鈮�t0)=.05wheredf=18

Question: A random sample of n measurements was selected from a population with unknown mean渭and known standard deviation蟽2. Calculate a 95% confidence interval for伪for each of the following situations:

a. n = 75, X = 28,蟽2= 12

b. n = 200, X= 102, 蟽2= 22

c. n = 100, X= 15,蟽2=.3

d. n = 100, X= 4.05, 蟽2= .83

e. Is the assumption that the underlying population of measurements is normally distributed necessary to ensure the validity of the confidence intervals in parts a鈥揹? Explain.

Evaporation from swimming pools. A new formula for estimating the water evaporation from occupied swimming pools was proposed and analyzed in the journal Heating Piping/Air Conditioning Engineering (April 2013). The key components of the new formula are number of pool occupants, area of pool鈥檚 water surface, and the density difference between room air temperature and the air at the pool鈥檚 surface. Data were collected from a wide range of pools for which the evaporation level was known. The new formula was applied to each pool in the sample, yielding an estimated evaporation level. The absolute value of the deviation between the actual and estimated evaporation level was then recorded as a percentage. The researchers reported the following summary statistics for absolute deviation percentage: x炉=18, s=20. Assume that the sample containedn=15 swimming pools

a. Estimate the true mean absolute deviation percentage for the new formula with a 90% confidence interval.

b. The American Society of Heating, Refrigerating, and Air-Conditioning Engineers (ASHRAE) handbook also provides a formula for estimating pool evaporation. Suppose the ASHRAE mean absolute deviation percentage is 渭=40%. (This value was reported in the article.) On average, is the new formula 鈥渂etter鈥� than the ASHRAE formula? Explain

Use Table III, Appendix D to determine thet0 values foreach of the following probability statements and their respectivedegrees of freedom (df ).

a.Pt猢�t0=.25withdf=15

b.Pt猢�t0=.1withdf=8

c.P-t0猢�t猢�t0=.01withdf=19

d.P-t0猢�t猢�t0=.05withdf=24

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