Question

eBook Barron's reported that the average number of weeks an individual is unemployed is 21 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 21 weeks and that the population standard deviation is 6 weeks. Suppose you would like to select a sample of 45 unemployed individuals for a follow-up study. Use z- table. a. Shew the sampling distribution oほ, the sample mean average for a sample of 4S unemployed indviduals. Ea21 (to 1 decimal) or-L .89 : (to 2 decimals) Choose the correct graph A.

Answer 1

$\small \mu = 21, \sigma = 6, n =45$

a) E(x) = _$\mu$ = 21

$\small \sigma(\bar x)= \sigma/\sqrt{n}=6/\sqrt{45} = \textbf{0.89}$

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b) Sample mean within 1 week of population mean = $\small |\bar x -\mu | = \pm 1$

$\small \\ \\ = P(|\bar x -\mu | < 1) \\ \\ = P\left ( -1 <\bar x-\mu<1 \right ) \\ \\ = P\left ( \frac{-1}{0.89}<\frac{\bar x-\mu}{\sigma(\bar x)} <\frac{1}{0.89}\right ) \\ \\ =P(-1.1180<z<1.1180) \\ \\ = P(z<1.1180)-P(z<-1.1180) \\ \\ \textup{ Using excel function:} \\ \\ = NORM.S.DIST( 1.1180 ,1) - NORM.S.DIST( -1.1180 ,1) \\ \\= \textbf{0.7364}$

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C) Sample mean within 1/2 week of population mean = $\small |\bar x -\mu | = \pm 0.5$

$\small \\ \\ = P(|\bar x -\mu | < 0.5) \\ \\ = P\left ( -0.5<\bar x-\mu<0.5 \right ) \\ \\ = P\left ( \frac{-0.5}{0.89}<\frac{\bar x-\mu}{\sigma(\bar x)} <\frac{0.5}{0.89}\right ) \\ \\ =P(-0.559<z<0.559) \\ \\ = P(z<0.559)-P(z<-0.559) \\ \\ \textup{ Using excel function:} \\ \\ = NORM.S.DIST( 0.559 ,1) - NORM.S.DIST( -0.559 ,1) \\ \\= \textbf{0.4238 }$