Refer to Exercise 11.4. Suppose that the sample given

Chapter 11, Problem 41E

(choose chapter or problem)

Refer to Exercise 11.4. Suppose that the sample given there came from a large but finite population of inventory items.We wish to estimate the population mean of the audited values, using the fact that book values are known for every item on inventory. If the population contains N items and

                                \(E\left(Y_{1}\right)=\mu_{1} \beta_{0}+\beta_{1} x_{1}\)

then the population mean is given by

\(\mu y=\frac{1}{N} \sum_{i=1}^{N} \mu_{1}=\beta_{0}+\beta_{1}\left(\frac{1}{N}\right) \sum_{i=1}^{N} x_{i}=\beta_{0}+{ }_{1} \mu_{x}\)

a Using the least-squares estimators of \(\beta_{0}\) and \(\beta_{1}\), show that μY can be estimated by

                       \(\widehat{\mu_{y}}=y+\bar{\beta}_{1}\left(\mu_{x}-\bar{x}\right)\)

(Notice that y is adjusted up or down, depending on whether x is larger or smaller than μx .)

b Using the data of Exercise 11.4 and the fact that \(\mu_{x}=74.0\), estimate μY , the mean of the audited values, and place a 2-standard-deviation bound on the error of estimation. (Regard the xi -values as constants when computing the variance of μˆY .)

Equation transcription:

Text transcription:

E(Y{1})=\mu{1} \beta{0}+\beta{1} x{1}

muy=\frac{1}{N}sum{i=1}^{N}mu{1}=\beta{0}+\beta{1}(\frac{1}{N})sum{i=1}^{N} x{i}=\beta{0}+{1} \mu{x}

beta{0}

beta{1}

widehat{\mu{y}}=y+\bar{\beta}{1}(\mu{x}-\bar{x})

mu{x}=74.0