e71_hw5_scan - Stats for Strategy Whitten HW 5 Solution A....

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Unformatted text preview: Stats for Strategy Whitten HW 5 Solution A. Text Exercises Exercise 7.54 The two samples are independent since the two groups of employees have no obvious relation- ship and there is no pairing mechanism . Variables: x 1 = employee’s satisfaction score with flat screen x 2 = employee’s satisfaction score with standard monitor Parameters: μ 1 = mean satisfaction score for all employees with flat screen μ 2 = mean satisfaction score for all employees with standard monitor (b) From MINITAB , a 95% confidence interval for ( μ 1- μ 2 ) is (0.05578 to 2.74422) points Interpretation : We are 95% confident that employees with flat screens average between 0.06 and 2.74 points higher satisfaction than employees with standard monitors. (c) (1) Parameters μ 1 and μ 2 are defined above. H A : μ 1 6 = μ 2 or ( μ 1- μ 2 ) 6 = 0 H : μ 1 = μ 2 or ( μ 1- μ 2 ) = 0 (2) From MINITAB , P-value = 0.043 (3) Reject H since P-value = 0 . 043 < . 05 = α . (4) There is sufficient evidence to show that mean employee satisfaction differs between types of computer monitor. (d) Yes. The hypothesis test rejects the null hypothesis ( μ 1- μ 2 ) = 0 while the confidence interval contains all positive numbers so that the value 0 is not plausible for ( μ 1- μ 2 ). Exercise 7.55 If you assigned the next 10 employees to flat screens and the following 10 to standard monitors, you might bias the experiment due to time differences : At the end of the test period the flat screen users will all have greater usage times, which may somehow influence employees’ satisfaction. So the random assignment described in the exercise makes for a better plan....
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This note was uploaded on 06/23/2008 for the course BUSINESS e71 taught by Professor Blake during the Spring '07 term at University of Iowa.

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e71_hw5_scan - Stats for Strategy Whitten HW 5 Solution A....

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